Sreelatha says, "I have a number that is divisible by 9. If I reverse its digits, it will still be divisible by 9".
Examine if her conjecture is true for any multiple of 9.
Yes, Sreelatha's conjecture is true for any multiple of 9. A number is divisible by 9 if and only if the sum of its digits is divisible by 9. When the digits are reversed, the sum of digits remains exactly the same, so the reversed number is also divisible by 9.
Are any other digit shuffles possible such that the number formed is still a multiple of 9?
Yes, any permutation (rearrangement) of the digits will result in a number that is still divisible by 9. Since the sum of digits remains unchanged regardless of how the digits are arranged, any number formed by shuffling the digits of a multiple of 9 will also be divisible by 9.
Explanation
(i) The chapter proves that divisibility by 9 depends only on the sum of digits. Since reversing digits does not change the sum, the reversed number maintains divisibility by 9. (ii) The chapter's divisibility rule for 9 depends only on the sum of digits, not on their order. Therefore, any rearrangement of digits preserves divisibility by 9.