Bunuel wrote:

In the addition problem ADD + ADD + ADD = SUMS, each of A, D, S, U, and M represents a different digit, and A is even. What is the value of the digit U?

A. 2

B. 3

C. 4

D. 5

E. 6

ADD + ADD + ADD = SUMS

Addition of three 3-digit numbers giving sum of 4-digit number can yield S as only 1 or 2

multiplication of 3 * D = Last digit 1 ...D=7

multiplication of 3 * D = Last digit 2 ...D=4

CASE 1: S = 1 and D = 7

now A is even can take 2,4,6,8,0

we know

multiplication of 3 * A + 2 = 1U

0 and 2 not possible (no carry over to form 4-digit) (0*3 = 0, 2*3 = 6)

Putting A= 8 gives U= 6 but S= 2 ..not possible

Putting A= 6 gives U= 0 but S= 2 ..not possible

Putting A= 4 gives U= 4 and S= 1 ..not possible (unique digits)

CASE 2: S = 2 and D = 4

we know

multiplication of 3 * A + 2 = 1U

now A is even can take 2,4,6,8,0

A cannot take 2,4 (already taken-unique digits)

Putting A= 6 gives U= 9 but S= 1 ..not possible

Putting A= 8 gives U= 5 and S= 2 ..possible

Hence A = 8, D = 4, S = 2 and U = 5

D

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