Maths Olympiad Prep

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Problem 1204

AMC 12 late, AIME early
Combinatorics Difficulty 5.0 Prove it Romanian Mathematical Olympiad · Romania

How many four digit numbers abcdabcd simultaneously satisfy the equalities a+b=c+da + b = c + d and a2+b2=c2+d2a^2 + b^2 = c^2 + d^2?

This one wants a proof. Work it on paper, then read the official solution and mark yourself. Be honest about it: the record is only any use to you if it is.

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Official solution

From a+b=c+da + b = c + d we get (a+b)2=(c+d)2(a + b)^2 = (c + d)^2, hence ab=cdab = cd and furthermore a2−2ab+b2=c2−2cd+d2a^2 - 2ab + b^2 = c^2 - 2cd + d^2. As (a−b)2=(c−d)2(a - b)^2 = (c - d)^2, we have ∣a−b∣=∣c−d∣|a - b| = |c - d|, which implies a−b=c−da - b = c - d or a−b=d−ca - b = d - c. Recall that a+b=c+da + b = c + d, so either a=c,b=da = c, b = d or a=d,b=ca = d, b = c.
The numbers must have one of the forms aaaa‾\overline{aaaa}, abba‾\overline{abba} or abab‾\overline{abab}, with a≠0a \neq 0 and a≠ba \neq b. In all, there are 9+9⋅9+9⋅9=9+81+81=1719 + 9 \cdot 9 + 9 \cdot 9 = 9 + 81 + 81 = 171 numbers.

Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty, ordering) added by this project.