Prove that the difference between numbers with the same sum of digits is a multiple of 9.

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Madilyn Shah · Accepted Answer

Step 1Well, the difference is not exactly 9, but it is a multiple of 9. Consider two n-digit numbers A=a1a2…an and B=b1b2…bn such that∑i=1nai=∑i=1nbi⇒∑i=1n(ai−bi)=0You are free to assume the first few digits of the number (or any other for that matter) to be zero, in order to produce numbers that have less than n digits. The proof still goes through.A−B=10n−1(a1−b1)+10n−2(a2−b2)+…+(an−bn)Using ∑i=1n(ai−bi)=0 we can simplify the above expression to get:A−B=(10n−1−1)(a1−b1)+(10n−2−1)(a2−b2)+(101−1)(an−1−bn−1)Now, 10k−1 is always a multiple of 9. In addition, each of the terms in the expression of A−B above, is a multiple of 10k−1 for some k≥1,k∈Z. As a result, each term of the sum is a multiple of 9, and therefore A−B itself is also a multiple of 9.Why is 10k−1 a multiple of 9 - you may ask? The binomial expansion suffices!10k−1=(9+1)k−1=sum of multiples of  9=multiple of  9I leave the details of the above argument to you. Lastly, this works for all n∈N, and so we have proved it for all numbers.The difference between two integers with the same sum of digits, is divisible by 9.

Prove that the difference between numbers with the

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