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A + B + C + D = 27

C = 3A

C is a single digit, and must be a multiple of three, so it has to be one of the numbers:

3, 6, 9

Let's start with our highest possibillity:

If A = 3, C = 9

B + D = 27 - (3 + 9)

B + D = 27 - 12 = 15

B = 8, D = 7 OR B = 7, D = 8

The number then can be 3897 or 3798

Now let's try a smaller value for A and C:

if A = 2, C = 6:

B + D = 27 - (2 + 6)

B + D = 19

But that can't be, because B + D are both single digits, and there are no two single digit numbers that add up to 19. The same will be true if A = 1, so the two combinations given above are the only ones.

Q: What four digit number when added has a sum of 27 when added but repeats no digets the number in the tens place is 3times greater than the number in the thousands place?

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