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

BTGmoderatorDC wrote: ↑

Sun Jan 17, 2021 8:43 pm

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


OA D

Solution:

First , we can simplify the expression as 3(ADD) = SUMS. Therefore, instead of looking at it as a sum, let’s look at it as a product.

Since A is even and it can’t be 0, let’s say A is 2. However, when we multiply a number in the 200’s by 3, the product can’t be a 4-digit number (since the product will be less than 900). We see that A can’t be 2. So let’s say A is 4. The product of a number in the 400’s and 3 is a 4-digit number. In that case, S, the thousands digit of the product, must be 1. Since S is also the units digit of the product, we see the D must be 7. So let’s see if it works:

3(477) = 1431

However, this doesn’t work since we would have U = 4, but A is already 4. We see that A can’t be 4. So let’s say A = 6. The product of a number in the 600’s and 3 is a 4-digit number. In that case, S is either 1 or 2. If S = 1, then D has to be 7 also. If S = 2, then D has to be 4. Let’s see which one works:

3(677) = 2031 (This doesn’t work; we see that the units digit is 1, but the thousands digit is not.)

3(644) = 1932 (This doesn’t work, either; we see that the units digit is 2, but the thousands digit is not.)

Now we are left to try A = 8. If that is the case, then S must be 2 and D must be 4. Let’s see if it works:

3(844) = 2532

We see that this works indeed!. So U = 5.

Answer: D