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Number theory Difficulty 6.6 National olympiad Prove it Belarus

a) Given the ten digits from 00 to 99, prove that three numbers AA, BB, and CC can be formed by combining these digits, provided that each digit is used exactly once and A+B=CA+B=C. Notice that 00 may not be the first digit of any of the numbers.

b) Find all possible values of the sum of the digits of CC.

Solution

a)
For example, 765+324=1089765 + 324 = 1089.

b)
Since any integer YY is congruent modulo 99 to the sum of its digits, we have
A+B+CS(A)+S(B)+S(C)=1+2+3+4+5+6+7+8+9+0=450(mod9). A + B + C \equiv S(A) + S(B) + S(C) = 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 0 = 45 \equiv 0 \pmod{9}.
By condition, we have A+B=CA + B = C, so 2C0(mod9)2C \equiv 0 \pmod{9}, whence it follows that C0(mod9)C \equiv 0 \pmod{9}, and, therefore, S(C)0(mod9)S(C) \equiv 0 \pmod{9}, i.e. the sum of the digits of CC is divisible by 99.

Note that S(C)=S(A+B)S(A)+S(B)S(C) = S(A+B) \leq S(A) + S(B). Indeed, if the sum of the digits of AA and BB does not exceed 99 in all number positions, then when we add AA and BB there is no 'carry' from any number position, so S(A+B)=S(A)+S(B)S(A+B) = S(A) + S(B). Otherwise, since the greatest possible carry is 11, if there is a carry from some number position, then the sum S(A+B)S(A+B) decreases by 99.

Thus, S(C)S(A)+S(B)S(C) \leq S(A) + S(B) and 1818 are smaller than 22.522.5 and are divisible by 99. The following examples show that S(C)S(C) can be 99 (the digits are 1+0+8+91+0+8+9); 765+324=1089765 + 324 = 1089 (the sum of the digits is equal to 1818).

Therefore, the possible values for the sum of the digits of CC are 99 or 1818.

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