Divisibility and Multiples | FIO

Question 20

Sreelatha says, "I have a number that is divisible by 9. If I reverse its digits, it will still be divisible by 9".

(i) Examine if her conjecture is true for any multiple of 9.

(ii) Are any other digit shuffles possible such that the number formed is still a multiple of 9?

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Solution
Understand the Question
  • Divisibility Rule for 9: A number is divisible by 99 if and only if the sum of its digits is divisible by 99.
  • For example, for the number 4545: 4+5=94 + 5 = 9 Since 99 is divisible by 99, 4545 is divisible by 99.
  • Similarly, for 108108: 1+0+8=91 + 0 + 8 = 9 Since 99 is divisible by 99, 108108 is divisible by 99.
  • Since reversing or rearranging the digits of a number does not change the set of digits being added, the sum of digits remains identical.

(i) Examine if her conjecture is true for any multiple of 9.

Step 1 · Examine Reversing the Digits of a Multiple of 9

Diagram 1

Let a multiple of 99 be NN with digits dk,dk1,,d1,d0d_k, d_{k-1}, \dots, d_1, d_0: N=dk10k+dk110k1++d110+d0N = d_k \cdot 10^k + d_{k-1} \cdot 10^{k-1} + \dots + d_1 \cdot 10 + d_0

The sum of its digits is: S=dk+dk1++d1+d0S = d_k + d_{k-1} + \dots + d_1 + d_0

Since NN is divisible by 99, SS must be divisible by 99.

Reversing the digits forms a new number NN': N=d010k+d110k1++dk110+dkN' = d_0 \cdot 10^k + d_1 \cdot 10^{k-1} + \dots + d_{k-1} \cdot 10 + d_k

The sum of the digits of NN' is: S=d0+d1++dk1+dkS' = d_0 + d_1 + \dots + d_{k-1} + d_k

Since addition is commutative, S=SS' = S. Because SS is divisible by 99, SS' is also divisible by 99, which implies NN' is divisible by 99.

Example: Consider the number 126126 (126=9×14126 = 9 \times 14): Sum of digits=1+2+6=9\text{Sum of digits} = 1 + 2 + 6 = 9

Reversing the digits gives 621621 (621=9×69621 = 9 \times 69): Sum of digits=6+2+1=9\text{Sum of digits} = 6 + 2 + 1 = 9

Since 99 is divisible by 99, 621621 is also divisible by 99.

Answer

(i) Yes, Sreelatha's conjecture is true for any multiple of 99.

(ii) Are any other digit shuffles possible such that the number formed is still a multiple of 9?

Step 1 · Analyze Arbitrary Digit Shuffles

A digit shuffle rearranges the positions of the digits without changing their values.

Let the sum of digits of the original multiple of 99 be SS. Any permutation or shuffle of its digits will result in a new number whose sum of digits is still SS.

Since SS is divisible by 99, any shuffled number is also divisible by 99.

Examples with 126126: Original sum=1+2+6=9\text{Original sum} = 1 + 2 + 6 = 9

  • Shuffling to form 261261: Sum=2+6+1=9    261=9×29\text{Sum} = 2 + 6 + 1 = 9 \implies 261 = 9 \times 29
  • Shuffling to form 612612: Sum=6+1+2=9    612=9×68\text{Sum} = 6 + 1 + 2 = 9 \implies 612 = 9 \times 68
Answer

(ii) Yes, any digit shuffle forms a number that is still a multiple of 99.

Common Mistakes
  • Assuming Divisibility Depends on Place Value: Thinking that changing place values affects divisibility by 99. For divisibility by 99, only the total sum of the digits matters.
  • Applying This Rule to Other Numbers: Generalizing this property to other divisors like 2,4,5,2, 4, 5, or 1010, where place value (specifically the last digit) determines divisibility.

More questions in FIO

Q1

The sum of four consecutive numbers is 34. What are these numbers?

Q2

Suppose pp is the greatest of five consecutive numbers. Describe the other four numbers in terms of pp.

Q3

For each statement below, determine whether it is always true, sometimes true, or never true. Explain your answer. Mention examples and non-examples as appropriate. Justify your claim using algebra.

(i) The sum of two even numbers is a multiple of 3.

(ii) If a number is not divisible by 18, then it is also not divisible by 9.

(iii) If two numbers are not divisible by 6, then their sum is not divisible by 6.

(iv) The sum of a multiple of 6 and a multiple of 9 is a multiple of 3.

(v) The sum of a multiple of 6 and a multiple of 3 is a multiple of 9.

Q4

Find a few numbers that leave a remainder of 2 when divided by 3 and a remainder of 2 when divided by 4. Write an algebraic expression to describe all such numbers.

Q5

“I hold some pebbles, not too many, When I group them in 3’s, one stays with me. Try pairing them up — it simply won’t do, A stubborn odd pebble remains in my view. Group them by 5, yet one’s still around, But grouping by seven, perfection is found. More than one hundred would be far too bold, Can you tell me the number of pebbles I hold?”

Q6

Tathagat has written several numbers that leave a remainder of 2 when divided by 6. He claims, “If you add any three such numbers, the sum will always be a multiple of 6.” Is Tathagat’s claim true?

Q7

When divided by 7, the number 661 leaves a remainder of 3, and 4779 leaves a remainder of 5. Without calculating, can you say what remainders the following expressions will leave when divided by 7? Show the solution both algebraically and visually.

(i) 4779+6614779 + 661

(ii) 47796614779 - 661

Q8

Find a number that leaves a remainder of 2 when divided by 3, a remainder of 3 when divided by 4, and a remainder of 4 when divided by 5. What is the smallest such number? Can you give a simple explanation of why it is the smallest?

Q9

Find, without dividing, whether the following numbers are divisible by 9.

(i) 123 (ii) 405 (iii) 8888 (iv) 93547 (v) 358095

Q10

Find the smallest multiple of 9 with no odd digits.

Q11

Find the multiple of 9 that is closest to the number 6000.

Q12

How many multiples of 9 are there between the numbers 4300 and 4400?

Q13

The digital root of an 8-digit number is 5. What will be the digital root of 10 more than that number?

Q14

Write any number. Generate a sequence of numbers by repeatedly adding 11. What would be the digital roots of this sequence of numbers? Share your observations.

Q15

What will be the digital root of the number 9a+36b+139a + 36b + 13?

Q16

Make conjectures by examining if there are any patterns or relations between (i) the parity of a number and its digital root. (ii) the digital root of a number and the remainder obtained when the number is divided by 3 or 9.

Q17

If 31z531z5 is a multiple of 99, where zz is a digit, what is the value of zz? Explain why there are two answers to this problem.

Q18

"I take a number that leaves a remainder of 8 when divided by 12. I take another number which is 4 short of a multiple of 12. Their sum will always be a multiple of 8", claims Snehal. Examine his claim and justify your conclusion.

Q19

When is the sum of two multiples of 3, a multiple of 6 and when is it not? Explain the different possible cases, and generalise the pattern.

Q20

Sreelatha says, "I have a number that is divisible by 9. If I reverse its digits, it will still be divisible by 9".

(i) Examine if her conjecture is true for any multiple of 9.

(ii) Are any other digit shuffles possible such that the number formed is still a multiple of 9?

Q21

If 48a23b48a23b is a multiple of 1818, list all possible pairs of values for aa and bb.

Q22

If 3p7q83p7q8 is divisible by 44, list all possible pairs of values for pp and qq.

Q23

Find three consecutive numbers such that the first number is a multiple of 2, the second number is a multiple of 3, and the third number is a multiple of 4.

Are there more such numbers? How often do they occur?

Q24

Write five multiples of 36 between 45,000 and 47,000. Share your approach with the class.

Q25

The middle number in the sequence of 5 consecutive even numbers is 5p5p. Express the other four numbers in sequence in terms of pp.

Q26

Write a 6-digit number that it is divisible by 15, such that when the digits are reversed, it is divisible by 6.

Q27

Deepak claims, "There are some multiples of 11 which, when doubled, are still multiples of 11. But other multiples of 11 don't remain multiples of 11 when doubled". Examine if his conjecture is true; explain your conclusion.

Q28

Determine whether the statements below are 'Always True', 'Sometimes True', or 'Never True'. Explain your reasoning.

(i) The product of a multiple of 6 and a multiple of 3 is a multiple of 9.

(ii) The sum of three consecutive even numbers will be divisible by 6.

(iii) If abcdefabcdef is a multiple of 6, then badcefbadcef will be a multiple of 6.

(iv) 8(7b3)4(11b+1)8(7b - 3) - 4(11b + 1) is a multiple of 12.

Q29

Choose any 3 numbers. When is their sum divisible by 3? Explore all possible cases and generalise.

Q30

Is the product of two consecutive integers always multiple of 2? Why? What about the product of these consecutive integers? Is it always a multiple of 6? Why or why not? What can you say about the product of 4 consecutive integers? What about the product of five consecutive integers?

Q31

Solve the cryptarithms —

(i) EF×E=GGG\text{EF} \times \text{E} = \text{GGG}

(ii) WOW×5=MEOW\text{WOW} \times 5 = \text{MEOW}

Q32

Which of the following Venn diagrams captures the relationship between the multiples of 4, 8, and 32?

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