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
Source: Veritas Prep
In the addition problem ADD + ADD + ADD = SUMS, each of A, D, S, U, and M represents a different digit, and A is even.
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Let us take \(D = 1; S = 3 \& M = 3;\) Since \(S \& M\) are different digits \(D =1\) is not possibleBTGmoderatorDC wrote: ↑Sun Jan 17, 2021 8:43 pmIn 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
Source: Veritas Prep
Let us take \(D = 2; S = 6 \& M = 6;\) Since \(S \& M\) are different digits \(D =2\) is not possible
Let us take \(D = 3; S = 9 \& M = 9;\) Since \(S \& M\) are different digits \(D =3\) is not possible
Let us take \(D = 4; S = 2 \& M = 3;\) Since \(S \& M\) are different digits \(D =4\) is possible
\(A44+A44+A44=2U32\)
There is a carry over of \(1\) for hundredth digit.
Since \(A\) is even; \(A = \{2,4,6,8\}; 3A = \{6,12,18,24\}; 3A+1 = \{7,13,19,25\}\)
Since \(3A+1 = 2U;\) only \(A = 8\) is possible
\(844+844+844 = 2532\)
\(U = 5\)
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Solution:BTGmoderatorDC wrote: ↑Sun Jan 17, 2021 8:43 pmIn 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
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
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