a, b, c, d, e, and f each represent a digit

B for me.

be = 24, only 6 * 4 will give 24

Hence, c + f must be less than 10. As per question, c + f must result in units digit of 0. Hence c & f must be 0.

Since a + d + 1 (from b + e) is 10, a + d = 9. Hence total of a + b + c + d + e + f = 19

hansoo wrote:In the below addition a, b, c, d, e and f each represent a digit. What is the value of a + b + c + d + e + f ?

(1) At least three of the digits are greater than 3.

(2) be = 24

(1) we can pick multiple sets of numbers to make this work, for example: 444+556 or 450+550. In the first case the digits sum to 28; in the second they sum to 19. Insufficient.

(2) If b and e are digits with a product of 24, they must be 3/8 or 4/6.

If they were 3/8, then they sum to 11; there would be no way to get a "0" in the summation as the second last digit in that case. Therefore, b and e must be 4/6.

If b and e are 4/6 and must produce a "0" underneath, then we can't carry anything over from the right column. Accordingly, c and f must be 0 and 0.

We know that when we add b and e, we'll carry a 1 over to the left column. Therefore, a+c must equal 9 to produce "10" in the summation.

So:

a+c=9
b+e=10
d+f = 0

a+b+c+d+e+f = 19... sufficient.

(2) is sufficient, (1) isn't: choose B.

Great explanations! The official answer is B.

Statement 2 is the more difficult statement to evaluate. As already mentioned, the key is recognize that b & e can only be 4 & 6 or 3 & 8, and that 3 & 8 don't sum to 10.

Here's a explanation to supplement the great explanations already given: