In the addition problem ADD + ADD + ADD = SUMS, each of A, D, S, U, an

Question

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

niks18 · Accepted Answer

+ADD 
 +ADD 
 +ADD 
----------------
 SUMS

so unit's digit of  3D  is  S  and  M  and given that every digit is distinct, this implies that  D>=4  to have a carry forward of  1  or  2

So starting with  D=4  we can calculate values of all variables and test the given conditions to arrive at the final answer

Case 1 :  D=4  -->  S=2  -->  M=3  -->  A=8  -->  U=5  (as  A  is even and addition  3D  in Ten’s digit will give only  1  as carry forward. Since  S , in the thousands place, has to be  2  so  A  has to be  8 , the largest even digit possible to have a carry forward of  2 )

So in Case 1, we got all our distinct variables. We can stop here or we can test for  D=5,6,7,8,9  and match the subsequent conditions. For any other value of  D , one or the other condition will not satisfy. Hence our answer  U=5

Option  D

Luckisnoexcuse · Answer

ADD + ADD + ADD = SUMS

Addition of three 3-digit numbers giving sum of 4-digit number can yield S as only 1 or 2

multiplication of 3 * D = Last digit 1 ...D=7
multiplication of 3 * D = Last digit 2 ...D=4

CASE 1: S = 1 and D = 7

now A is even can take 2,4,6,8,0

we know
multiplication of 3 * A + 2 = 1U
0 and 2 not possible (no carry over to form 4-digit) (0*3 = 0, 2*3 = 6)
Putting A= 8 gives U= 6 but S= 2 ..not possible
Putting A= 6 gives U= 0 but S= 2 ..not possible
Putting A= 4 gives U= 4 and S= 1 ..not possible (unique digits)

CASE 2: S = 2 and D = 4
we know
multiplication of 3 * A + 2 = 1U
now A is even can take 2,4,6,8,0
A cannot take 2,4 (already taken-unique digits)
Putting A= 6 gives U= 9 but S= 1 ..not possible
Putting A= 8 gives U= 5 and S= 2 ..possible
Hence A = 8, D = 4, S = 2 and U = 5
D

Jerry1982 · Answer

Ans D ) u =5

D cannot be 1 , 2, 3 or 5 because then m= s which is not possible as given in the question.
3D = S and 3D = M and all digits are different which implies that M = 3D+ Carry from 3D = S. Also from the question it can be deduced that since 3A = SU then S can at max be digit 2 if we take the max value of A which is A= 8.

Let say s= 2 then D = 4 m=3 and u =5 only.
Plug in the answer and ADD = 844

Mohammad Ali Khan · Answer

very time consuming question. how to arrive to the conclusion fast??
I cant find other approach other than basic number sense for max. value of fourth digit. Rest was hit and trial

ruchik · Answer

Official explanation:
When the GMAT asks abstract calculation problems, your job is to make that abstraction more concrete using two methods:

1) Limit the number of options by adhering to and proving rules about the situation provided.

2) Test numbers via trial-and-error to eliminate options and to learn more about the situation.

For example here, you can start by limiting the options for A: since A must be even and it cannot be 0 (if it were, then the number ADD would just be DD), you only have four options for A: 2, 4, 6, and 8.

But if you quickly try 2, you'll see that even with the greatest possible D, you won't have large enough numbers to produce a thousands digit S in the sum. 299 + 299 + 299 is 897, and you need a number that's 1000 or greater. So you can limit your options for A to 4, 6, or 8.

Next, consider the sum SUMS. Since you're adding three three-digit numbers to produce SUMS, the S has a limit to it also. Even if you added the three greatest three-digit numbers possible, 999 + 999 + 999, you'd end up with a number less than 3000. So S can only be 1 or 2.

Also consider that in SUMS M and S must be different digits, meaning that adding three Ds must sum to something greater than 10 so that the operation forces you to carry a tens digit and make M different from S. (For example, 411 + 411 + 411 would give you 1233 with the same M and S. You need D to be large enough that you don't have repeat digits in the tens and units places in the sum).

So from quick trial and error and some application of the rules provided, you know three things:

-A can only be 4, 6, or 8 -S can only be 1 or 2 -D must be greater than 3

From here you can use some units digit rules along with trial-and-error to arrive at SUMS. In the units place, 3D (the sum of D + D + D) can only be 1 or 2. In order for it to be 1, D would have to be 7. So you might try:

477 + 477 + 477 = 1431

But note that in this situation U and D are each 4, which violates the situation that they must be different values. So this cannot work. The next possible value ending in 77 would be 677, but at that point the sum would begin with a 2 (677 + 677 + 677 = 2031, or you could just know that 667 is 1/3 of 2000 and so three 677s would be greater than that). And that doesn't work because the S values in SUMS would be different.

So S cannot be 1, meaning that it must be 2. In order for that to be the case, you'd need D to be 4 (since 4x3 = 12). Here you can try again: if A cannot be 4 (that would be a repeat value), then you could try 6, but recognize again that you'd need something greater than 666 to reach a thousands digit of 2, since 2000 divided by 3 is 666.67. So your only choice is 844 + 844 + 844 = 2532. This then means that the correct answer is 5.

ScottTargetTestPrep · Answer

First of all, we can simplify the expression as 3(ADD) = SUMS. Therefore, instead of looking 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

KritiG · Answer

Why cant ADD be 477, A is even, SUMS=1431, U=4

vs224 · Answer

U is 4 and A is already 4. Needs to be different digits.

Kinshook · Answer

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

Asked: What is the value of the digit U?

Let us take D = 1; S = 3 & M = 3; Since S & M are different digits D =1 is not possible
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

IMO D

samarpan.g28 · Answer

A is even so A can max be 8. Let the number ADD be 8xx. Now 8xx*3=2abc which means S cannot be anything but 2.
SUMS: 2pq2. For D, we have to come up with such a one-digit number which when multiplied by 3 (because ADD+ADD+ADD means ADD*3) gives 2. 
Only 4*3 yields a unit digit of 2. Therefore D=4. So, 844*3=2532. Here all of A, D, S, U, and M represent a different digit. So, U is 5(D).

Adarsh_24 · Answer

Another method 
S can be either 1 or 2. Because 1000*3=3000. Here only 3 three digit.
So last digit either 1 or 2.
3D last digit 1 or 2. So D is either 4 or 7.
When you do A77+A77+A77 we will get 1U31
So U=2+3A
none of the options will work.
So D must be 4=>A44. Then rest can be found.

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In the addition problem ADD + ADD + ADD = SUMS, each of A, D, S, U, an

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