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?
The divisibility rule for 9 states that a number is divisible by 9 if the sum of its digits is divisible by 9.
Step 1 — The Divisibility Rule for 9 Let us first understand the divisibility rule for 9. A number is divisible by 9 if its digit sum is divisible by 9. For example, consider the number 45. The sum of its digits is: Since 9 is divisible by 9, 45 is also divisible by 9. Another example is 108. The sum of its digits is: Since 9 is divisible by 9, 108 is also divisible by 9.
Step 2 — Examining Sreelatha's Conjecture Sreelatha says reversing digits keeps a multiple of 9 divisible by 9. Let us take any number . Let be a multiple of 9. Let its digits be . The number can be written as: The sum of its digits is . Since is divisible by 9, must be divisible by 9. Now, let us reverse the digits of . This forms a new number, let's call it . The digits of are . The number can be written as: The sum of the digits of is . We can see that is exactly the same as . Since is divisible by 9, must also be divisible by 9. Therefore, (with reversed digits) must also be divisible by 9. Sreelatha's conjecture is true.
Let's use an example to make this clear. Consider the number . The sum of its digits is: Since 9 is divisible by 9, 126 is divisible by 9 (). Now, let us reverse its digits. The new number formed is . The sum of its digits is: Since 9 is divisible by 9, 621 is divisible by 9 (). This example confirms our finding.

Step 3 — Other Digit Shuffles A digit shuffle means rearranging a number's digits. For example, for 126, we can get 216, 162, 612. When we shuffle digits, the actual digits do not change. Only their positions change. This means the sum of the digits remains the same. Let the original number be . Its sum of digits is . If is a multiple of 9, then is divisible by 9. Any number formed by shuffling 's digits will have sum . Since is divisible by 9, the shuffled number is also divisible by 9. So, yes, any digit shuffle forms a multiple of 9.
Let's use an example. Consider the number again. Its sum of digits is: We know 126 is divisible by 9. Let us shuffle its digits to form . The sum of its digits is: Since 9 is divisible by 9, 261 is divisible by 9 (). Another shuffle: . The sum of its digits is: Since 9 is divisible by 9, 612 is divisible by 9 (). These examples show that any shuffle works.
Answer
(i) Sreelatha's conjecture is true. Reversing digits does not change their sum. A number is divisible by 9 if its digit sum is divisible by 9. (ii) Yes, any other digit shuffle is possible. Any rearrangement of digits keeps the sum of the digits the same. If the original sum was divisible by 9, the new sum will also be divisible by 9.
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