Number Play

Is This a Multiple Of? — Part 1

Is This a Multiple Of? — Part 1

Welcome to the fascinating world of "Number Play"! In this first part, we'll dive into how numbers behave, especially when we add or subtract them in sequences. We will uncover hidden patterns and learn to predict properties like whether a result is even or odd, or if it's divisible by 4, without doing heavy calculations.

This skill is like being a detective for numbers. Instead of just seeing 3 + 4 + 5 = 12, you'll start to see the underlying rules that govern all numbers. Let's begin our investigation!

Concept Introduction

Imagine you are arranging oranges in a pyramid for a display at a fruit stand. You start with 5 oranges in the bottom row, then a row of 4 on top of it, and a row of 3 on top of that. The total number of oranges is 3 + 4 + 5 = 12. This is a sum of consecutive numbers — numbers that follow each other in order.

What if you used 4, 5, 6, and 7 oranges? The total is 4 + 5 + 6 + 7 = 22. Notice anything about the results 12 and 22? They are both even! This chapter explores such intriguing patterns. We will ask questions like: Can any number be written as a sum of consecutive numbers? Do these sums follow specific rules? Let's use the power of algebra to find out.

{{FORMULA: expr=N = 2k | symbols=N:An even number, k:any integer}}

Definitions & Formulas

Before we start playing with numbers, let's define the key terms we'll be using. Understanding these concepts is crucial for mastering the patterns we are about to explore.

Derivation & Logic: The Unchanging Parity

The NCERT text highlights a very powerful idea: when you take a set of numbers and combine them with + or - signs, the parity (even or odd) of the result stays the same, no matter which combination of signs you use! Let's understand why this happens.

The Logic of Parity in Addition and Subtraction

1. First, let's recall the basic rules of parity for addition and subtraction (±).

   odd ± odd = even
   

   even ± even = even
   

   odd ± even = odd
   

   These rules are the foundation for everything that follows.

   {{VISUAL: diagram: Parity arithmetic using colored blocks. Two red blocks combining to form a blue block (odd+odd=even). Two blue blocks combining to form a blue block (even+even=even). A red and blue block combining to form a red block (odd+even=odd).}}

2. Now, consider two expressions: a + b and a - b. Let's find the difference between them.

   (a + b) - (a - b) = a + b - a + b = 2b
   

   The difference is 2b, which is always an even number because it has a factor of 2.

3. If the difference between two numbers is even, they must have the same parity. Think about it: an odd number minus an even number is odd. An even number minus an odd number is odd. Only even - even or odd - odd gives an even result.

4. This means a + b and a - b are either both even or both odd. They always share the same parity.

5. We can extend this idea. The parity of a ± b is always the same. Therefore, the parity of (a ± b) + c is the same as the parity of (a ± b) - c.

6. By extending this logic, we can conclude that all 8 expressions formed by a ± b ± c ± d must have the same parity. If you calculate just one of them and find it's even, you can be certain that the other seven are also even!

{{KEY: type=concept | title=The Law of Constant Parity | text=For any set of numbers, all expressions formed by placing '+' or '–' signs between them (e.g., a ± b ± c ± d) will have the same parity. If one result is even, all results will be even.}}

Solved Examples

Let's apply these concepts to solve some problems, starting from easy and moving to more challenging ones.

Example 1: Parity of an Expression (Easy)

Given: The consecutive numbers 3, 4, 5, and 6.

To Find: The value and parity of the expression 3 + 4 – 5 + 6.

Solution:

1. First, perform the additions.

   3 + 4 = 7
   

   7 + 6 = 13
   

2. Now, perform the subtraction.

   13 – 5 = 8
   

3. The result is 8. To check its parity, we see if it is divisible by 2.

   8 ÷ 2 = 4
   

   Since 8 is divisible by 2, it is an even number.

Final Answer: The value of the expression is 8, which is an even number.

Example 2: Proving Parity with Algebra (Medium)

Given: Four consecutive numbers, which can be represented as n, n+1, n+2, and n+3.

To Find: Prove that the result of any expression n ± (n+1) ± (n+2) ± (n+3) is always even.

Solution:

1. According to the Law of Constant Parity, we only need to check the parity of one of the possible expressions. The simplest one is the sum of all four numbers.

   Sum = n + (n+1) + (n+2) + (n+3)
   

2. Combine the like terms (n's and constants).

   Sum = 4n + (1+2+3) = 4n + 6
   

3. Factor out the common factor, which is 2.

   Sum = 2(2n + 3)
   

4. Since the entire expression 2(2n + 3) has a factor of 2, it is guaranteed to be an even number for any integer value of n.

5. Because one of the expressions is always even, and all 8 expressions must have the same parity, we can conclude that all possible results are always even.

Final Answer: The expression simplifies to 2(2n + 3), which is always a multiple of 2 and therefore always even. By the Law of Constant Parity, all expressions formed with these numbers and ± signs will also be even.

Example 3: Sum of Evens and Divisibility by 4 (Hard)

Given: Two even numbers that are not multiples of 4.

To Find: Show that their sum is always a multiple of 4.

Solution:

1. First, let's represent the two numbers algebraically. An even number that is not a multiple of 4 can be written in the form 4k + 2, where k is any integer.

2. Let the first number be 4p + 2 and the second number be 4q + 2, where p and q are integers.

   {{VISUAL: diagram: Visual representation of adding two even numbers that are not multiples of 4. Shows a group of p rows of 4 dots plus 2 extra dots, added to a group of q rows of 4 dots plus 2 extra dots. The result is (p+q) full rows of 4 dots, plus one extra row formed by the 2+2 extra dots, totaling (p+q+1) rows of 4 dots.}}

3. Now, let's add them together.

   Sum = (4p + 2) + (4q + 2)
   

4. Group the terms with a factor of 4 and the constant terms.

   Sum = 4p + 4q + 2 + 2
   

   Sum = 4p + 4q + 4
   

5. Factor out the common factor, which is 4.

   Sum = 4(p + q + 1)
   

6. Since the entire expression 4(p + q + 1) has a factor of 4, the sum is always a multiple of 4.

Final Answer: The sum of two even numbers not divisible by 4 is 4(p + q + 1), which is always a multiple of 4.

Example 4: Working Backwards with Divisibility (Tricky)

Given: The sum of two even numbers, a and b, is a multiple of 4. The number a is also a multiple of 4.

To Find: What can you conclude about the number b?

Solution:

1. Let's translate the given information into algebraic expressions.

   • a + b is a multiple of 4, so we can write a + b = 4m for some integer m.
   • a is a multiple of 4, so we can write a = 4k for some integer k.

2. Substitute the expression for a into the first equation.

   (4k) + b = 4m
   

3. To find out about b, we need to isolate it. Subtract 4k from both sides of the equation.

   b = 4m - 4k
   

4. Factor out the common factor of 4 on the right side.

   b = 4(m - k)
   

5. The result shows that b can be expressed as 4 times some integer (m - k). This is the definition of a multiple of 4.

   {{VISUAL: diagram: A large rectangle representing a multiple of 4 (say, 20 dots arranged in 5 rows of 4). It is shown partitioned into two smaller regions. One region represents 'a' (also a multiple of 4, say 12 dots in 3 rows of 4). The remaining region 'b' is shown to also consist of full rows of 4 (8 dots in 2 rows of 4), illustrating that 'b' must also be a multiple of 4.}}

Final Answer: The number b must also be a multiple of 4.

Tips & Tricks

Use these shortcuts to solve problems related to parity and divisibility faster.

Common Mistakes

Many students make small errors when dealing with these concepts. Here are a few common pitfalls to avoid.

Brain-Teaser Questions

Ready for a challenge? Test your understanding with these slightly trickier questions.

1. Can the number 64 be expressed as a sum of consecutive natural numbers? Why or why not?

   💡 Answer: No. Numbers that are powers of 2 (like 2, 4, 8, 16, 32, 64, ...) cannot be expressed as a sum of consecutive natural numbers. This is a special property of these numbers related to their odd factors.

2. If you take any 5 numbers a, b, c, d, e, will all 16 expressions formed by a ± b ± c ± d ± e have the same parity?

   💡 Answer: Yes. The logic we used for 4 numbers extends to any quantity of numbers. Since x+y and x-y always have the same parity, we can apply this rule repeatedly. The parity of a±b±c±d±e will be the same as the parity of the simple sum a+b+c+d+e.

3. The sum of three consecutive integers is 21. Is the sum of the next three consecutive integers a multiple of 3?

   💡 Answer: Yes. The first three integers are 6, 7, 8. The next three are 9, 10, 11. Their sum is 9+10+11 = 30, which is a multiple of 3. In general, if n+(n+1)+(n+2) = 3n+3 is a multiple of 3, the next three are (n+3)+(n+4)+(n+5) = 3n+12 = 3(n+4), which is also always a multiple of 3.

Mini Cheatsheet

Here's a quick summary of the key ideas from this page. Screenshot this for your last-minute revision!

Is This a Multiple Of? — Part 2

Page 2: Is This a Multiple Of? — Part 2

Welcome back! In the previous section, we explored how numbers can be combined in different ways. Now, we'll dive deeper into the fascinating world of divisibility and remainders. We will learn to act like mathematical detectives, investigating statements about numbers to determine if they are always, sometimes, or never true. This skill is not just for exams; it's the foundation of logical reasoning and problem-solving.

Imagine you are a baker organizing a large order of cupcakes. You have several boxes that can hold 6 cupcakes each, and several others that hold 9 cupcakes each. If you fill up a certain number of each type of box completely, will the total number of cupcakes you've packed always be perfectly divisible by 18? Or only sometimes? What about being divisible by 3? Using the power of algebra and logic, we can answer these questions without ever packing a single cupcake. This section will give you the tools to analyze such problems with confidence.

{{FORMULA: expr=a = bq + r | symbols=a:Dividend, b:Divisor, q:Quotient, r:Remainder}}

Definitions & Formulas

To analyze divisibility, we need a clear set of definitions. The relationship between a number, its divisor, and the remainder is formally known as the Division Algorithm.

Logic: Proving Properties of Divisibility

How can we be sure a rule about numbers is always true? We can't test every number in existence! This is where algebra comes in. It allows us to use variables to represent any number and prove a property for all cases. Let's prove a fundamental rule from our NCERT text.

Property: If a number a divides two other numbers M and N separately, then a must also divide their sum (M + N) and their difference (M - N).

Proof:

1. First, let's translate the given information into algebra. If a divides M, it means M is a multiple of a. We can write this as:

   M = a × k  (where k is some integer)
   

2. Similarly, if a divides N, it means N is also a multiple of a. We can write this as:

   N = a × j  (where j is some integer)
   

3. Now, let's find their sum, M + N, by substituting the expressions from steps 1 and 2.

   M + N = (a × k) + (a × j)
   

4. We can see that a is a common factor in both terms. Let's take a common using the distributive property.

   M + N = a × (k + j)
   

5. Look at the result! The sum M + N is expressed as a multiplied by (k + j). Since k and j are integers, their sum (k + j) is also an integer. This means M + N is a multiple of a. Therefore, a divides M + N.

6. The same logic applies to the difference, M - N.

   M - N = (a × k) - (a × j) = a × (k - j)
   

   Since (k - j) is an integer, M - N is also a multiple of a.

This algebraic proof confirms that the statement is always true.

Solved Examples

Let's apply this logical thinking to determine if statements are Always True, Sometimes True, or Never True.

Example 1: Divisibility by Factors (Easy)

Given: A number N is divisible by 12.

To Find: Is N always divisible by 4?

Solution:

1. "Divisible by 12" means N is a multiple of 12. Algebraically, this is:

   N = 12 × m  (for some integer m)
   

2. We want to check if N is divisible by 4. Let's see if we can rewrite the expression for N to show a factor of 4. We know that 12 = 4 × 3.

   N = (4 × 3) × m
   

3. Using the associative property of multiplication, we can regroup the terms.

   N = 4 × (3 × m)
   

4. The expression 4 × (3 × m) clearly shows that N is a multiple of 4, because (3 × m) is an integer.

Final Answer: The statement is Always True.

Example 2: Combined Divisibility with LCM (Medium)

Given: A number A is divisible by both 6 and 4.

To Find: Is A always divisible by 24?

Solution:

1. Let's test this with an example. The number 12 is divisible by 6 (since 12 = 6 × 2) and also by 4 (since 12 = 4 × 3).

   12 ÷ 6 = 2
   12 ÷ 4 = 3
   

2. Now, let's check if 12 is divisible by 24.

   12 ÷ 24 = 0.5
   

   The result is not an integer. So, 12 is not divisible by 24. We have found a counter-example.

3. Let's consider another number. The number 24 is divisible by 6 (24 = 6 × 4) and by 4 (24 = 4 × 6). It is also divisible by 24.

4. Since the property holds for A = 24 but fails for A = 12, it doesn't always happen.

   Rule: If a number is divisible by k and m, it must be divisible by their Least Common Multiple (LCM). The LCM of 6 and 4 is 12, not 24. So, A must be divisible by 12, but not necessarily by 24.

Final Answer: The statement is Sometimes True.

{{KEY: type=rule | title=The LCM Rule for Divisibility | text=If a number N is divisible by two integers 'a' and 'b', then N is ALWAYS divisible by the LCM(a, b). It is only divisible by the product (a × b) if 'a' and 'b' are co-prime (their HCF is 1).}}

Example 3: Working with Remainders (Hard)

Given: A number X leaves a remainder of 3 when divided by 5.

To Find: Which of these algebraic expressions correctly represents all such numbers: 3k + 5, 5k + 3, or 5k - 3? Justify your choice.

Solution:

1. Let's analyze the definition of a remainder using the Division Algorithm: Dividend = Divisor × Quotient + Remainder.

   a = bq + r
   

2. In our problem, the dividend is X, the divisor is 5, and the remainder is 3. The quotient can be any integer, which we can represent with the variable k.

   X = 5 × k + 3
   

   This matches the expression 5k + 3. Let's test it.

   • If k = 0, X = 5(0) + 3 = 3. (3 ÷ 5 gives quotient 0, remainder 3)
   • If k = 1, X = 5(1) + 3 = 8. (8 ÷ 5 gives quotient 1, remainder 3)
   • If k = 2, X = 5(2) + 3 = 13. (13 ÷ 5 gives quotient 2, remainder 3) This expression works perfectly.

3. Now let's check the other options. Consider 3k + 5.

   • If k = 1, 3(1) + 5 = 8. This number works.
   • If k = 2, 3(2) + 5 = 11. Let's check 11 ÷ 5. It gives a remainder of 1, not 3. So, 3k + 5 is incorrect.

4. Consider 5k - 3.

   • If k = 1, 5(1) - 3 = 2. When 2 is divided by 5, the remainder is 2, not 3. So, this is also incorrect.

   Note: The expression 5k - 2 also works! 5k - 2 = 5(k-1) + 5 - 2 = 5(k-1) + 3. This generates the same set of numbers as 5k+3, just with a different starting value of k. The text mentions this as a valid alternative. But 5k+3 is the most direct representation.

Final Answer: The expression 5k + 3 correctly represents all numbers that leave a remainder of 3 when divided by 5.

Example 4: Parity and Multiples (Tricky)

Given: The statement "When you add an even number to an odd number, the result is a multiple of 6."

To Find: Determine if this statement is Always True, Sometimes True, or Never True.

Solution:

1. First, let's represent an even number and an odd number algebraically.

   • An even number is a multiple of 2, so we can write it as 2m.
   • An odd number is one more than an even number, so we can write it as 2n + 1. Here, m and n can be any integers.

2. Now, let's add them together.

   Sum = (2m) + (2n + 1)
   

3. Let's simplify the expression.

   Sum = 2m + 2n + 1 = 2(m + n) + 1
   

4. The result 2(m + n) + 1 is in the form of 2 × (an integer) + 1. By definition, this is always an odd number.

5. Now, let's analyze the other side of the statement: "a multiple of 6". Multiples of 6 are 6, 12, 18, 24, ... All multiples of 6 can be written as 6j, which is the same as 2 × (3j). This means all multiples of 6 are even.

6. The statement claims that an odd number (the sum) can be equal to an even number (a multiple of 6). This is a contradiction. An odd number can never be an even number.

Final Answer: The statement is Never True.

Tips & Tricks

Here are a few shortcuts to help you analyze divisibility problems faster.

Common Mistakes to Avoid

Pay close attention to these common errors when working with divisibility.

Brain-Teaser Questions

Test your understanding with these slightly trickier questions!

1. If a number K is a multiple of 18, is it always true that K is a multiple of 12?

   💡 Answer: Sometimes True. A number divisible by 18 must have prime factors of 2 × 3². A number divisible by 12 needs prime factors 2² × 3. Since a multiple of 18 isn't guaranteed to have 2² as a factor, it's not always divisible by 12. For example, K=18 is not divisible by 12, but K=36 is.

2. A number is written in the form 8k - 3. What is the remainder when this number is divided by 8?

   💡 Answer: The remainder is 5. We can rewrite the expression: 8k - 3 = 8(k-1) + 8 - 3 = 8(k-1) + 5. This is in the standard form (Divisor × Quotient) + Remainder.

3. If a number is divisible by x and y, where x and y are two different prime numbers, is the number always divisible by x × y?

   💡 Answer: Always True. The LCM of two distinct prime numbers is always their product. For example, if a number is divisible by 3 and 5, it must be divisible by LCM(3,5) = 15.

Mini Cheatsheet

Here's a quick summary of the key ideas from this page. Screenshot this for your last-minute revision!

Checking Divisibility Quickly — Part 1

Checking Divisibility Quickly — Part 1

Welcome back! Have you ever looked at a large number, like 73,548, and wondered if it can be divided perfectly by 3 or 9 without leaving a remainder? Doing the full division can be time-consuming. What if there were "shortcuts" or rules to find out in just a few seconds?

These shortcuts, known as divisibility rules, are not magic. They are clever tricks based on the structure of our number system. Imagine a baker with 1,290 cookies. Can they be packed perfectly into boxes of 10 without any left over? A quick look at the last digit tells us the answer is yes! In this lesson, we will explore these rules for numbers like 2, 5, 10, 3, and 9. More importantly, we'll use simple algebra to understand why these shortcuts work, transforming them from mere tricks into powerful mathematical tools.

{{FORMULA: expr=N = ... + 1000d + 100c + 10b + a | symbols=N:any number, a:units digit, b:tens digit, c:hundreds digit, d:thousands digit}}

General Form of a Number

To understand the logic behind divisibility rules, we first need to represent any number in a general algebraic form. This is based on its place value. For instance, the number 4075 means 4 thousands, 0 hundreds, 7 tens, and 5 units.

We can write this using variables. A number ...dcba represents:

... + 1000 × d + 100 × c + 10 × b + a

Here, the letters a, b, c, d, etc., stand for the digits in the units, tens, hundreds, thousands place, and so on.

Logic & Derivation: Why the Rules Work

Let's use the general form of a number to prove why the divisibility rules are always true.

1. Divisibility by 10, 5, and 2

These three rules are related because 10, 5, and 2 are all factors of 10.

1. Represent the number: Let's take any number N. We can write it in its expanded form:

   N = (... + 1000d + 100c + 10b) + a
   

2. Factor out 10: Notice that every term except the last one (a) is a multiple of 10. We can rewrite the expression in the parenthesis as:

   10 × (... + 100d + 10c + b)
   

   So, the whole number N is:

   N = 10 × (some integer) + a
   

3. Analyze for Divisibility by 10: The first part of the expression, 10 × (some integer), is always divisible by 10. For the entire number N to be divisible by 10, the remaining part, a, must also be divisible by 10. Since a is a single digit (0-9), the only way it can be divisible by 10 is if a is 0.

4. Analyze for Divisibility by 5 and 2: The first part, 10 × (some integer), is also always divisible by 5 and 2 (since 10 = 2 × 5). Therefore, for the whole number N to be divisible by 5 or 2, the units digit a must be divisible by 5 or 2.

   • For divisibility by 5, a must be 0 or 5.
   • For divisibility by 2, a must be an even digit: 0, 2, 4, 6, or 8.

2. Divisibility by 9 and 3

The rule for 9 is fascinating and relies on a clever algebraic trick.

1. Represent the number: Again, we start with the expanded form. Let's use a 4-digit number dcba for simplicity.

   N = 1000d + 100c + 10b + a
   

2. Rewrite the place values: The key insight is that every power of 10 is just 1 more than a multiple of 9.

   • 10 = 9 + 1
   • 100 = 99 + 1
   • 1000 = 999 + 1 Now, substitute these into our expanded form.

3. Substitute and regroup:

   N = (999 + 1)d + (99 + 1)c + (9 + 1)b + a
   

   Distribute the digits d, c, and b:

   N = 999d + d + 99c + c + 9b + b + a
   

   Now, group all the multiples of 9 together:

   N = (999d + 99c + 9b) + (d + c + b + a)
   

4. Analyze for Divisibility by 9: Look at the two parts. The first part, (999d + 99c + 9b), is always a multiple of 9, because every term inside it is a multiple of 9. For the entire number N to be divisible by 9, the second part, (d + c + b + a), must also be divisible by 9. This second part is simply the sum of the digits!

5. Extend to Divisibility by 3: Since 9 is a multiple of 3, the same logic applies. The first part, (999d + 99c + 9b), is also a multiple of 3. Therefore, for the number N to be divisible by 3, the sum of its digits (d + c + b + a) must be divisible by 3.

{{KEY: type=concept | title=The "Sum of Digits" Rule | text=A number is divisible by 9 if the sum of its digits is divisible by 9. The remainder when a number is divided by 9 is the same as the remainder when the sum of its digits is divided by 9.}}

Solved Examples

Example 1: Basic Divisibility Check (Easy)

Given: The number 34,785.

To Find: Is this number divisible by 2, 5, or 10?

Solution:

1. The rule for divisibility by 2, 5, and 10 depends only on the units digit. The units digit of 34,785 is 5.

2. Check for 2: Is the units digit (5) an even number (0, 2, 4, 6, 8)? No. So, it is not divisible by 2.

3. Check for 5: Is the units digit (5) either 0 or 5? Yes. So, it is divisible by 5.

4. Check for 10: Is the units digit (5) equal to 0? No. So, it is not divisible by 10.

Final Answer: The number 34,785 is divisible by 5, but not by 2 or 10.

Example 2: Sum of Digits for Divisibility by 9 (Medium)

Given: The number 68,759,124.

To Find: Is this number divisible by 9?

Solution:

1. To check for divisibility by 9, we need to find the sum of all its digits.

2. Add the digits of the number:

   Sum = 6 + 8 + 7 + 5 + 9 + 1 + 2 + 4
   

3. Calculate the sum:

   Sum = 42
   

4. Now, we check if this sum (42) is divisible by 9. We know that 9 × 4 = 36 and 9 × 5 = 45. So, 42 is not a multiple of 9.

Final Answer: No, the number 68,759,124 is not divisible by 9.

Example 3: Finding a Missing Digit (Hard)

Given: The number 471z82 is divisible by 9, where z is a single digit.

To Find: The value of the digit z.

Solution:

1. For a number to be divisible by 9, the sum of its digits must be divisible by 9.

2. Calculate the sum of the known digits:

   Sum of known digits = 4 + 7 + 1 + 8 + 2
   

   Sum of known digits = 22
   

3. The total sum of all digits is 22 + z. This total sum must be a multiple of 9.

4. Let's find the multiples of 9 that are close to 22.

   • 9 × 1 = 9
   • 9 × 2 = 18
   • 9 × 3 = 27
   • 9 × 4 = 36

5. The sum 22 + z must be one of these multiples. Since z is a single digit (0 to 9), 22 + z will be between 22 + 0 = 22 and 22 + 9 = 31. The only multiple of 9 in this range is 27.

6. Set the total sum equal to 27 and solve for z.

   22 + z = 27
   

   z = 27 - 22
   

   z = 5
   

Final Answer: The value of the missing digit z is 5.

Example 4: Finding the Remainder (Tricky)

Given: The number 8,123,456,789.

To Find: The remainder when this number is divided by 9, without performing the actual division.

Solution:

1. A key property of the divisibility rule for 9 is that the remainder of a number divided by 9 is the same as the remainder of the sum of its digits divided by 9.

2. First, let's find the sum of the digits of the number.

   Sum = 8 + 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9
   

3. Calculate this sum. We can pair numbers to make it easier: (1+9) + (2+8) + (3+7) + (4+6) + 5 + 8 = 10 + 10 + 10 + 10 + 5 + 8.

   Sum = 53
   

4. Now, instead of dividing the original huge number by 9, we just need to find the remainder when 53 is divided by 9.

   53 ÷ 9
   

5. We know that 9 × 5 = 45. The next multiple is 9 × 6 = 54. So, 53 is 53 - 45 = 8 more than a multiple of 9.

   Remainder = 8
   

   Alternatively, we can apply the rule again to 53. The sum of digits of 53 is 5 + 3 = 8. Since 8 is a single digit, it is the final remainder.

Final Answer: The remainder when 8,123,456,789 is divided by 9 is 8.

Tips & Tricks

Common Mistakes

Brain-Teaser Questions

1. A number 72k is divisible by 2. Another number 45k is divisible by 5. What is the value of the digit k?

   💡 Answer: For 72k to be divisible by 2, k must be an even digit {0, 2, 4, 6, 8}. For 45k to be divisible by 5, k must be {0, 5}. The only digit that satisfies both conditions is 0. So, k = 0.

2. What is the smallest positive number that must be added to 1,234,567 to make it divisible by 9?

   💡 Answer: First, find the sum of digits of 1,234,567: 1+2+3+4+5+6+7 = 28. Now, find the remainder when 28 is divided by 9. 28 = (9 × 3) + 1. The remainder is 1. To become divisible by 9, we need to reach the next multiple of 9, which is 36. But we are looking for the smallest number to add. We need to get rid of the remainder of 1. To make the sum a multiple of 9 (like 27 or 36), the easiest way is to add a number to cancel the remainder. The next multiple of 9 after 28 is 36. The number to add is 36 - 28 = 8. So, we must add 8.

3. Find the smallest 4-digit number that is divisible by both 3 and 10.

   💡 Answer: For a number to be divisible by 10, its last digit must be 0. So the number must be of the form xyz0. The smallest 4-digit number starts with 1, so our number looks like 1yz0. For the number to be divisible by 3, the sum of its digits (1 + y + z + 0) must be a multiple of 3. To make the number as small as possible, we should make y and z as small as possible. Let's try y=0. The sum becomes 1 + 0 + z = 1 + z. The smallest multiple of 3 greater than or equal to 1 is 3. So, 1 + z = 3, which means z = 2. The number is 1020.

Mini Cheatsheet

Checking Divisibility Quickly — Part 2

{{FORMULA: expr=Divisibility by C = p₁ × p₂ | symbols=C:composite number, p₁,p₂:co-prime factors, Condition:A number is divisible by C if it is divisible by both p₁ and p₂}}

Checking Divisibility Quickly — Part 2

Welcome back! In the previous section, we revisited the divisibility rules for prime numbers like 2, 5, and also for 3 and 9 which have special patterns. But what about larger, non-prime numbers like 6, 12, 15, or even 72? Do we need to memorize a unique rule for every number?

Thankfully, no! We can cleverly combine the rules we already know. This section will teach you how to test for divisibility by any composite number by breaking it down into its smaller, friendlier parts. We'll also formalize the "sum of digits" trick into a powerful concept called the digital root.

Concept Introduction: The Warehouse Manager's Dilemma

Imagine you are a manager at a large warehouse that ships out boxes of chocolates. Today's order is for 1,350 boxes. The client wants them packed into larger cartons. One client wants cartons that hold exactly 15 boxes each, and another client wants cartons that hold 18 boxes each.

Without doing long division, can you quickly figure out if the 1,350 boxes can be packed perfectly into cartons of 15 without any leftovers? What about into cartons of 18?

This is a divisibility problem. If 1350 is divisible by 15, the packing is perfect. If it's divisible by 18, that packing is also perfect. By breaking 15 into its factors (3 and 5) and 18 into its factors (2 and 9), you can use the simple rules you already know to find the answer in seconds, saving time and avoiding packing errors. This is the power of composite divisibility rules.

Definitions & Formulas

Let's define the key terms for this section. Understanding these will make the logic much clearer.

The Logic Behind Composite Divisibility

Why does breaking a number into co-prime factors work? Let's understand the logic for checking divisibility by 6.

1. A number N is divisible by 6 if N can be written as 6 × k for some whole number k.

2. We know that 6 can be written as a product of its prime factors: 2 × 3. So, we can substitute this into our equation.

   N = (2 × 3) × k
   

3. Using the associative property of multiplication, we can group the terms differently.

   N = 2 × (3k)
   

   This shows that N is a multiple of 2, which means N is divisible by 2.

4. Similarly, we can regroup it as:

   N = 3 × (2k)
   

   This shows that N is a multiple of 3, which means N is divisible by 3.

5. Therefore, for a number N to be divisible by 6, it must be divisible by both 2 and 3. These factors, 2 and 3, are co-prime.

This logic extends to any composite number. You break it down into a set of factors that are all co-prime to each other.

{{KEY: type=concept | title=The Co-prime Factor Rule | text=A number is divisible by a composite number 'C' if and only if it is divisible by each of 'C's co-prime factors. For example, to test for 36, use its co-prime factors 4 and 9, not 6 and 6, or 3 and 12, because they are not co-prime.}}

Solved Examples

Let's apply these concepts to solve some problems, starting from easy and moving to tricky.

Example 1: Basic Divisibility Check (Easy)

Given: The number 9,732.

To Find: Is 9,732 divisible by 12?

Solution:

1. First, find a pair of co-prime factors for 12. The factors are 3 and 4 (since HCF(3, 4) = 1). We need to check for divisibility by both 3 and 4.

2. Check for divisibility by 3: Sum the digits of 9,732.

   9 + 7 + 3 + 2 = 21
   

   Since 21 is divisible by 3 (21 ÷ 3 = 7), the number 9,732 is divisible by 3.

3. Check for divisibility by 4: Look at the last two digits of the number, which are '32'.

   32 ÷ 4 = 8
   

   Since 32 is divisible by 4, the number 9,732 is divisible by 4.

4. Because 9,732 is divisible by both 3 and 4, it must be divisible by 12.

Final Answer: Yes, 9,732 is divisible by 12.

Example 2: Finding a Missing Digit (Medium)

Given: The number 6701*, where * is a missing digit.

To Find: The smallest whole number to replace * so that the number is divisible by 6.

Solution:

1. The co-prime factors for 6 are 2 and 3. The number 6701* must be divisible by both.

2. Check for divisibility by 2: For a number to be divisible by 2, its last digit (*) must be an even number. So, * can be 0, 2, 4, 6, or 8.

3. Check for divisibility by 3: The sum of the digits must be a multiple of 3.

   6 + 7 + 0 + 1 + * = 14 + *
   

   We need 14 + * to be a multiple of 3. Let's test the possible even values for *.

4. If * = 0, sum is 14 + 0 = 14 (not a multiple of 3). If * = 2, sum is 14 + 2 = 16 (not a multiple of 3). If * = 4, sum is 14 + 4 = 18 (this is a multiple of 3!).

5. We are looking for the smallest whole number. We found that * = 4 makes the number divisible by both 2 and 3.

Final Answer: The smallest whole number to replace * is 4.

{{VISUAL: diagram: A factor tree showing 72 breaking down into 8 and 9, and then 8 breaking into 2x2x2 and 9 into 3x3, to visually represent co-prime factors 8 and 9.}}

Example 3: Multi-step Composite Divisibility (Hard)

Given: A 4-digit number 35AB.

To Find: The values of A and B such that the number is divisible by both 5 and 9. What is the number?

Solution:

1. This problem requires us to satisfy two separate divisibility rules. Let's tackle them one by one.

2. Check for divisibility by 5: A number is divisible by 5 if its last digit is 0 or 5. So, B must be either 0 or 5.

3. Check for divisibility by 9: The sum of the digits must be divisible by 9.

   3 + 5 + A + B = 8 + A + B
   

   This sum, 8 + A + B, must be a multiple of 9.

4. Let's consider the two possible cases for B.

   Case 1: B = 0 The sum becomes 8 + A + 0 = 8 + A. For this to be a multiple of 9, A must be 1 (since 8 + 1 = 9). A cannot be 10 as it must be a single digit. So, if B = 0, then A = 1. The number is 3510.

   Case 2: B = 5 The sum becomes 8 + A + 5 = 13 + A. For this to be a multiple of 9, the closest multiple of 9 is 18.

   13 + A = 18  =>  A = 5
   

   So, if B = 5, then A = 5. The number is 3555.

5. The problem asks for the number, implying there might be a single solution or both are valid. Both 3510 and 3555 are divisible by 5 and 9. Let's re-read. It asks for "the values," plural. So both pairs are solutions. Let's state the resulting numbers.

Final Answer: The possible numbers are 3510 (A=1, B=0) and 3555 (A=5, B=5).

Example 4: The Digital Root Remainder (Tricky)

Given: The number 4,819,275.

To Find: What is the remainder when this number is divided by 9, without performing the actual division?

Solution:

1. As we learned from the NCERT text, the remainder when a number is divided by 9 is the same as the remainder when its sum of digits is divided by 9. This is the principle of digital roots.

2. First, sum all the digits of the number.

   4 + 8 + 1 + 9 + 2 + 7 + 5 = 36
   

3. Now, we check if this sum, 36, is divisible by 9.

   36 ÷ 9 = 4
   

   The sum of the digits is perfectly divisible by 9.

4. This means the original number, 4,819,275, is also perfectly divisible by 9. A number that is perfectly divisible leaves a remainder of 0.

5. Alternative using Digital Root: The digital root of 36 is 3 + 6 = 9. When the digital root is 9, the number is divisible by 9, and the remainder is 0.

Final Answer: The remainder is 0.

Tips & Tricks

Here are some shortcuts to make checking divisibility even faster.

Common Mistakes

Be careful! Many students get tripped up by these common errors.

Brain-Teaser Questions

Ready for a challenge? Test your understanding with these higher-order problems.

1. A six-digit number is formed by repeating a three-digit number, like 481481. This number is always divisible by which three distinct prime numbers?

   💡 Answer: Any number abcabc can be written as abc × 1001. The prime factors of 1001 are 7, 11, and 13. So, any such number is always divisible by 7, 11, and 13.

2. If the 5-digit number 72X3X is divisible by 12, what is the value of the digit X? (Note: X represents the same digit in both places).

   💡 Answer: For divisibility by 12, it must be divisible by 3 and 4. For 4: The last two digits, 3X, must be divisible by 4. This means X can be 2 (for 32) or 6 (for 36). For 3: The sum 7+2+X+3+X = 12 + 2X must be divisible by 3. If X=2, sum is 12 + 2(2) = 16 (not divisible by 3). If X=6, sum is 12 + 2(6) = 24 (divisible by 3). So, X must be 6.

3. What is the remainder when the huge number formed by writing the numbers 1 to 40 side-by-side (12345...3940) is divided by 9?

   💡 Answer: We only need the sum of all digits from 1 to 40. The sum of digits from 1 to 9 is 45. The sum of digits for numbers 10-19 is (1×10 + 45) = 55. For 20-29 it's (2×10 + 45) = 65. For 30-39 it's (3×10 + 45) = 75. The digits in 40 are 4+0=4. Total sum = 45 (for 1-9) + 55 (for 10-19) + 65 (for 20-29) + 75 (for 30-39) + 4 (for 40) = 244. The remainder of 244 ÷ 9 is the same as the remainder of (2+4+4) ÷ 9, which is 10 ÷ 9. The remainder is 1.

Mini Cheatsheet

Here's a quick summary of the key rules from this page. Screenshot this for your last-minute revision!

Digits in Disguise & Summary & Quick Revision

Page 5: Digits in Disguise & Chapter Summary

Welcome to the final page of our journey through "Number Play"! So far, we've uncovered the secret patterns behind numbers, mastering the rules of divisibility. Now, it's time to put all that knowledge to work in a fun and challenging way. We'll become number detectives, solving puzzles where digits are hidden behind letters.

These puzzles, called cryptarithms, are more than just a game. They sharpen your logical reasoning, improve your arithmetic skills, and reinforce your understanding of number properties. Think of it as a final boss battle where you must use every skill you've learned in this chapter to emerge victorious!

Concept Introduction: The Art of Cryptarithmetic

Imagine you are a detective trying to crack a secret code. The only clues you have are the structure of the message and the rules of the language. Cryptarithms are exactly like that, but the language is mathematics! They are puzzles where letters stand for unique digits, and you must solve an arithmetic problem like SEND + MORE = MONEY.

To solve it, you can't just guess randomly. You need to use logic. For instance, in the example above, the 'M' in MONEY must be 1, because it's a carry-over from the previous column (S + M). The maximum sum of two single digits plus a carry-over (9 + 9 + 1 = 19) can only produce a carry of 1. This is the kind of deductive reasoning we will use to unravel these "digits in disguise".

{{KEY: type=concept | title=The Core Rules of Cryptarithms | text=Every cryptarithm puzzle follows three golden rules: 1. Each letter represents a unique digit from 0 to 9. 2. If a letter appears more than once, it represents the same digit everywhere. 3. The first digit of any number in the puzzle cannot be zero.}}

Definitions & Puzzle Rules

Before we start solving, let's define the key terms and rules that govern these puzzles.

The Logic: How to Solve a Cryptarithm

Solving a cryptarithm is a systematic process of deduction. There's no single formula, but there is a reliable strategy.

1. Analyze the Structure: First, understand the arithmetic operation involved—is it addition, subtraction, or multiplication? Write the problem in a vertical column format if it isn't already.

2. Start with the Units Column: The rightmost column (the units place) is often the best starting point. It's the least likely to be affected by unknown carry-overs from a previous column.

3. Hunt for Carry-overs: Look for columns that must produce a carry. For example, in an addition problem, if the sum has more digits than the numbers being added, the leading digit of the sum is almost always a carry-over of 1.

4. Use Number Properties: Apply your knowledge of arithmetic.

   • Adding two even numbers results in an even number.
   • A number multiplied by 5 must end in 0 or 5.
   • A number multiplied by an even number is always even.

5. Test and Eliminate: When you have a few possibilities for a letter, test them one by one. If a choice leads to a contradiction (e.g., two different letters having the same digit value), eliminate it and try the next possibility.

6. Verify Uniqueness: Once you find a potential solution, double-check that every letter corresponds to a different digit.

Solved Examples

Let's apply this logic to a few puzzles, starting from simple and moving to more complex ones.

Example 1: Simple Addition (Easy)

Given: The cryptarithm AB + 37 = 6A.

To Find: The values of the digits A and B.

Solution:

1. Let's write the problem in column format:

     A B
   + 3 7
   -----
     6 A
   

2. Focus on the units column (B + 7 = A). This means the units digit of B + 7 is A. There might be a carry-over to the tens column.

3. Now, look at the tens column (A + 3 + carry = 6).

   • Case 1: No carry-over from the units column. Then A + 3 = 6. This gives A = 3.
   • Case 2: A carry-over of 1 from the units column. Then A + 3 + 1 = 6. This gives A = 2.

4. Let's test Case 1: If A = 3. Substitute A = 3 into the units column equation: B + 7 = A → B + 7 = 3. This is not possible for a single digit B. The sum must be 13 for the units digit to be 3. So, B + 7 = 13.

5. From B + 7 = 13, we find B.

B = 13 - 7 = 6

6. This implies B = 6 and there is a carry-over of 1 to the tens column. Let's check if this contradicts our tens column logic. Our tens column equation was A + 3 + carry = 6. Substituting A = 3 and carry = 1: 3 + 3 + 1 = 7. This result is 7, but the sum in the tens place is 6. So A = 3 is incorrect.

7. Let's test Case 2: If A = 2. This assumes there was a carry-over of 1 from the units column. So, the tens column is A + 3 + 1 = 6, which gives 2 + 3 + 1 = 6. This is correct!

8. Now find B using A = 2 and the fact that there was a carry of 1. The equation for the units column is B + 7, and its result must have a units digit of A=2 and produce a carry of 1. This means B + 7 must equal 12.

B + 7 = 12
B = 5

9. Our solution is A = 2 and B = 5. Let's check: 25 + 37 = 62. This matches the 6A format, where A=2.

Final Answer: A = 2, B = 5

Example 2: Multiplication Puzzle (Medium)

Given: The cryptarithm JK × 6 = KKK.

To Find: The values of the digits J and K.

Solution:

1. The equation is JK × 6 = KKK. The number KKK can be written as K × 111.

KKK = K × 100 + K × 10 + K = K × (100 + 10 + 1) = 111 × K

2. So, our equation becomes (10J + K) × 6 = 111 × K. We know that 111 = 3 × 37.

(10J + K) × 6 = 3 × 37 × K

3. Divide both sides by 3.

(10J + K) × 2 = 37 × K

4. Expand the left side.

20J + 2K = 37K

5. Rearrange the equation to isolate J.

20J = 37K - 2K
20J = 35K

6. Simplify the equation by dividing both sides by their greatest common divisor, which is 5.

4J = 7K

7. Since J and K are single digits, and 4 and 7 are co-prime, the only possible solution is J = 7 and K = 4. J and K cannot be 0 because KKK is a 3-digit number and JK is a 2-digit number.

8. Let's check the solution: J = 7, K = 4. The original equation is JK × 6 = KKK. 74 × 6 = 444. This is correct.

Final Answer: J = 7, K = 4

Example 3: Repeated Addition (Hard)

Given: The cryptarithm ON + ON + ON = PO.

To Find: The values of the digits O, N, and P.

Solution:

1. The problem can be rewritten as a multiplication:

3 × (ON) = PO

2. Let's write this in column format:

     O N
   ×   3
   -----
     P O
   

3. Focus on the units column: 3 × N must result in a number ending with the digit O. 3 × N = ...O

4. Focus on the tens column: 3 × O plus any carry-over from the units column must equal P. 3 × O + (carry) = P.

5. Let's list the possibilities for 3 × N. The result's last digit is O.

   • If N=1, 3×1=3, so O=3.
   • If N=2, 3×2=6, so O=6.
   • If N=3, 3×3=9, so O=9.
   • If N=4, 3×4=12, so O=2 (and carry=1).
   • If N=5, 3×5=15, so O=5. This is impossible, as O and N must be unique.
   • If N=6, 3×6=18, so O=8 (and carry=1).
   • If N=7, 3×7=21, so O=1 (and carry=2).
   • If N=8, 3×8=24, so O=4 (and carry=2).
   • If N=9, 3×9=27, so O=7 (and carry=2).
   • N cannot be 0, because then 3×0=0, so O=0, but O and N must be unique.

6. Now we test these pairs in the tens column equation: 3 × O + (carry) = P.

   • Try N=1, O=3 (carry=0): 3 × 3 + 0 = 9. So P=9. Let's check: 3 × 31 = 93. This matches PO where P=9 and O=3. All digits are unique. This is a possible solution.

   • Try N=2, O=6 (carry=0): 3 × 6 + 0 = 18. So P=18. P must be a single digit. Impossible.

   • Try N=4, O=2 (carry=1): 3 × 2 + 1 = 7. So P=7. Let's check: 3 × 24 = 72. This matches PO where P=7 and O=2. All digits are unique. This is another possible solution.

   • Try N=6, O=8 (carry=1): 3 × 8 + 1 = 25. P=25. Impossible.

   • Try N=7, O=1 (carry=2): 3 × 1 + 2 = 5. So P=5. Let's check: 3 × 17 = 51. This matches PO where P=5 and O=1. All digits are unique. This is a third possible solution.

   • Try N=8, O=4 (carry=2): 3 × 4 + 2 = 14. P=14. Impossible.

   • Try N=9, O=7 (carry=2): 3 × 7 + 2 = 23. P=23. Impossible.

7. The problem has multiple valid solutions based on the constraints. A standard cryptarithm usually has one unique solution. Let's re-read the NCERT source. ON + ON + ON = PO is presented without a unique solution constraint. Let's provide one of the valid ones. Let's pick the first one we found.

Final Answer: One possible solution is O = 3, N = 1, P = 9. (The puzzle is 31+31+31 = 93).

Example 4: A Tricky Multiplication (Tricky)

Given: The cryptarithm BYE × 6 = RAY.

To Find: The values of B, Y, E, R, A.

Solution:

1. Let's analyze the structure. A 3-digit number BYE multiplied by 6 results in another 3-digit number RAY.

2. This gives a strong clue about the first digit, B. If B were 2, the number would be at least 200. 200 × 6 = 1200, which is a 4-digit number. Therefore, B must be 1.

B = 1

3. The puzzle is now 1YE × 6 = RAY. R is the leading digit of the result, so R cannot be 0. Also, R must be the first digit of 1YE × 6. 100 × 6 = 600. 199 × 6 = 1194. The product is a 3-digit number, so 1YE must be less than 167 (167 × 6 = 1002).

4. Let's look at the units column: E × 6 must end in the digit Y. E × 6 = ...Y.

5. And the hundreds column: 6 × 1 plus a carry-over from the tens column (C₂) equals R. 6 + C₂ = R.

6. Let's list possibilities for E and the resulting Y. Y must be an even digit because any number multiplied by 6 is even. E and Y must be different. B=1.

   • E=0: 0×6=0. Y=0. Not unique.
   • E=2: 2×6=12. Y=2. Not unique.
   • E=3: 3×6=18. Y=8. (Carry C₁=1).
   • E=4: 4×6=24. Y=4. Not unique.
   • E=5: 5×6=30. Y=0. (Carry C₁=3).
   • E=6: 6×6=36. Y=6. Not unique.
   • E=7: 7×6=42. Y=2. (Carry C₁=4).
   • E=8: 8×6=48. Y=8. Not unique.
   • E=9: 9×6=54. Y=4. (Carry C₁=5).

7. We have three potential pairs for (E, Y) with their carry C₁:

   • Case A: E=3, Y=8, C₁=1
   • Case B: E=5, Y=0, C₁=3
   • Case C: E=7, Y=2, C₁=4
   • Case D: E=9, Y=4, C₁=5

8. Now let's check the tens column. The equation is 6 × Y + C₁ = ...A, which produces a carry C₂.

   • Case A (E=3, Y=8, C₁=1): 6 × 8 + 1 = 49. So A=9 and carry C₂=4. Now check the hundreds column: 6 × B + C₂ = R → 6 × 1 + 4 = 10. So R=10. Impossible, R is a digit.

   • Case B (E=5, Y=0, C₁=3): 6 × 0 + 3 = 3. So A=3 and carry C₂=0. Hundreds column: 6 × 1 + 0 = 6. So R=6. Let's check this solution: B=1, Y=0, E=5, R=6, A=3. 105 × 6 = 630. This matches RAY → 630. All digits are unique (1,0,5,6,3). This is a valid solution.

   • Since we have found a valid solution, we can stop, but let's quickly check the others for completeness.

   • Case C (E=7, Y=2, C₁=4): 6 × 2 + 4 = 16. So A=6 and carry C₂=1. Hundreds column: 6 × 1 + 1 = 7. So R=7. Check B=1, Y=2, E=7, R=7, A=6. R and E are not unique. Invalid.

   • Case D (E=9, Y=4, C₁=5): 6 × 4 + 5 = 29. So A=9 and carry C₂=2. Hundreds column: 6 × 1 + 2 = 8. So R=8. Check B=1, Y=4, E=9, R=8, A=9. A and E are not unique. Invalid.

Final Answer: B=1, Y=0, E=5, R=6, A=3

Tips & Tricks

Common Mistakes to Avoid

Brain-Teaser Questions

1. Solve the classic cryptarithm: SEND + MORE = MONEY

   💡 Answer: M=1, O=0, S=9, E=5, N=6, D=7, R=8, Y=2. The puzzle is 9567 + 1085 = 10652. The key is realizing M=1 from the carry, which then helps deduce O=0 and S=9.

2. Find the number ABC such that A! + B! + C! = ABC. (Hint: n! means factorial, e.g., 5! = 5×4×3×2×1 = 120)

   💡 Answer: 145. 1! + 4! + 5! = 1 + 24 + 120 = 145. The digit 7! is already 5040, so the digits can't be very large. You can quickly test combinations of small digits.

3. Solve the multiplication puzzle: EF × E = GGG

   💡 Answer: E=7, F=4, G=3. The puzzle is 74 × 7 = 518. Whoops, that's not GGG. Let's re-solve. GGG = G × 111 = G × 3 × 37. So EF × E must be a multiple of 37. Since 37 is prime, either E or EF must be a multiple of 37. E is a single digit, so it can't be. Therefore, EF must be a multiple of 37. The two-digit multiples of 37 are 37 and 74. Case 1: EF = 37. Then E=3, F=7. The puzzle is 37 × 3 = 111. Here G=1. This works! Case 2: EF = 74. Then E=7, F=4. The puzzle is 74 × 7 = 518. This doesn't result in GGG. So the only solution is E=3, F=7, G=1. The puzzle is 37 × 3 = 111.

Mini Cheatsheet: Chapter 5 Revision

Here is a summary of the most important rules from the entire "Number Play" chapter. Screenshot this for your last-minute revision!