AB and CD are 2-digit numbers. Multiplying AB with CD results in a 3-digit number DEF. Adding DEF to another 3-digit number GHI results in 975. Further A, B, C, D. E, F, G, H, I are distinct digits. If E= 0, F=8, then what is A+B+C equal to?

1. A. 6
2. B. 7
3. C. 8
4. D. 9

Answer: A

Explanation

We are given: 1) AB × CD = DEF, 2) DEF + GHI = 975, and A, B, C, D, E, F, G, H, I are distinct digits. Also, E=0 and F=8. Substituting E=0, F=8 into DEF + GHI = 975 gives D08 + GHI = 975.
From the units column: 8 + I = 15 (since I is a digit), so I=7 (carry 1).
From the tens column: 0 + H + 1 (carry) = 7, so H=6.
From the hundreds column: D + G = 9.
Used digits so far: E=0, F=8, I=7, H=6. Remaining digits for D, G are from {1,2,3,4,5,9}. Possible (D,G) pairs summing to 9 are (4,5) or (5,4).
Now consider AB × CD = DEF, which is AB × C_ = D08. If D=5, then AB × C_ = 508. A number ending in 5 (C5) multiplied by any number cannot result in a product ending in 8. So D cannot be 5. Thus, D=4 and G=5. So DEF = 408.
Now we have AB × C_ = 408. Used digits: 0, 4, 5, 6, 7, 8. Remaining for A, B, C are {1,2,3,9}. Since C_ is a 2-digit number ending in 4, C must be one of {1,3,9}. Also, B must be 2 (since _2 * _4 ends in 8). So B=2.
Now, A2 × C4 = 408. If C=1, A2 × 14 = 408 => A2 = 29.14 (not integer). If C=3, A2 × 34 = 408 => A2 = 12. This works! So A=1, B=2, C=3. All digits {1,2,3,4,0,8,5,6,7} are distinct. Therefore, A+B+C = 1+2+3 = 6. This is a complex number puzzle requiring careful deduction, typical of CSAT.