A and B are distinct nonzero digits such that 2(ABA)+3(BB)+1(A)=BBB. (ABA refers to the three-digit integer whose digits are A, B, A, etc.) If T = AB, the least positive integer that is a multiple of T+3, whose sum of the digits is T-3, let S be the closed integer to the least positive integer that is a multiple of T+3, whose sum of the digits is T-3.

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With the given constraints A=2, B=7, so that T=27. Then T+3=30. S=k(T+3) or 30k, and we are looking for the value of S such that the sum of its digits is T-3=24. S=6990. {233*30=6990, so S is a multiple of T+3=30, and the sum of the digits is 6+9+9+0=24 which is T-3=24].

(1) We need to determine A and B. We are given 2(ABA)+3(BB)+A=BBB with the understanding that ABA is a three-digit number.

Rewrite as 2(100A+10B+A)+3(10B+B)+A=100B+10B+B.

202A+20B+33B+A=111B
203A=58B
7A=2B ==> A=2 and B=7. (A and B are natural numbers less than 10.)

(2) Then T=AB=27 so that T+3=30 and T-3=24.

(3) We are looking for an S, S=k(T-3)=30k such that the sum of the digits is 24. A few guesses at S yields that the sum of digits creates a pattern.

3,6,9,3,6,9,3,6,9,3,6,9,12,6,9,12,6,9,12,...

We are looking for the first instance where the pattern of three numbers is interrupted by a fourth which is 24. (In the pattern shown, 3,6,9 is eventually broken by 12. ) This finally happens at 6990.

So, S=6990.

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