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Math Expert
Joined: 02 Sep 2009
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09 Nov 2017, 22:13
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29% (01:43) correct 71% (02:13) wrong based on 96 sessions

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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
[Reveal] Spoiler: OA

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Kudos [?]: 135614 [0], given: 12705

Director
Joined: 18 Aug 2016
Posts: 598

Kudos [?]: 181 [0], given: 138

GMAT 1: 630 Q47 V29
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09 Nov 2017, 22:34
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Bunuel wrote:
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

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
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Director
Joined: 25 Feb 2013
Posts: 619

Kudos [?]: 308 [0], given: 39

Location: India
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11 Nov 2017, 06:24
3
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Bunuel wrote:
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

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

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

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13 Nov 2017, 20:04
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.

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