Thursday, April 24, 2008

IDEAs

Here's another S.A.T. problem.

Problem:

In the correctly worked addition problem to the left, A, B, D, E, I, and S each represent a different digit. What is the smallest possible value of D?


Solution:

You don't need to know what every digit in the problem is. You only need to know which digits have to be assigned to which letters in order for the addition problem to work.

Starting in the one's column, E + A = A, or
E + A = A + a carry digit. The most two one-digit numbers can add to is 18, so the carry digit would have to be 10. If E + A = 10 + A, then E = 10, but we know that E < 10 (all of the variables are single digits). Thus, E + A = A, and E = 0. Filling that value into the puzzle we get: The value of A doesn't matter. We can forget about the two right-most columns of the addition and look at two left columns. B + S = 10I + D. The most that B and S can add up to is 17 (with B and S equaling 9 and 8), and the least they can sum to is 3 (if B and S equal 1 and 2). However, if B + S < 10, I = 0. We can't have another 0, so B + S must be greater than 10. Therefore, I = 1. Since we've used up 0 and 1, smallest digit we have left for D is 2. We can use 4 for B and 8 for S to make B + S = 12. If we use these values to fill in the puzzle, it all adds up. The smallest number D can be is 2.

4 comments:

Joe Shields said...

Keep up the good work!

Jacqueline Granche said...

Thanks, Joe!

Anonymous said...

why doesnt 499+599=1098 with D=0 work?

Jacqueline Granche said...

Anonymous, that would work for BEE + SEE, but the formula was BEE + SEA, and A can't equal E.