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

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

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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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Re: In the addition problem ADD + ADD + ADD = SUMS, each of A, D, S, U, an [#permalink]

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New post 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


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

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New post 11 Nov 2017, 06:24
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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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Re: In the addition problem ADD + ADD + ADD = SUMS, each of A, D, S, U, an [#permalink]

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New post 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.
Plug in the answer and ADD = 844

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Re: In the addition problem ADD + ADD + ADD = SUMS, each of A, D, S, U, an   [#permalink] 13 Nov 2017, 20:04
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