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Let A, M, and C be digits with (100A + 10M + C)(A + M + C) = 2005. Wha

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Let A, M, and C be digits with (100A + 10M + C)(A + M + C) = 2005. Wha  [#permalink]

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New post 22 Mar 2019, 01:05
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A
B
C
D
E

Difficulty:

  55% (hard)

Question Stats:

46% (01:41) correct 54% (01:48) wrong based on 35 sessions

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Re: Let A, M, and C be digits with (100A + 10M + C)(A + M + C) = 2005. Wha  [#permalink]

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New post 22 Mar 2019, 02:32
1
On Prime factorising 2005
We get 5 × 401
Comparing with
(100A + 10M + C)(A + M + C) we get
(4×100+10×0+1 )*(4+0+1)
So A=4

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Re: Let A, M, and C be digits with (100A + 10M + C)(A + M + C) = 2005. Wha  [#permalink]

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New post 22 Mar 2019, 04:05
Bunuel wrote:
Let A, M, and C be digits with (100A + 10M + C)(A + M + C) = 2005. What is A ?

(A) 1
(B) 2
(C) 3
(D) 4
(E) 5



2005 = 5*401
so
A= 4,B=0 C= 1
IMO D
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Re: Let A, M, and C be digits with (100A + 10M + C)(A + M + C) = 2005. Wha  [#permalink]

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New post 24 Mar 2019, 18:29
Bunuel wrote:
Let A, M, and C be digits with (100A + 10M + C)(A + M + C) = 2005. What is A ?

(A) 1
(B) 2
(C) 3
(D) 4
(E) 5


We see that we have a 3-digit number, AMC, such that when it is multiplied by the sum of its digits, the product is 2005. Therefore, it must be a factor of 2005. Let’s factor 2005:

2005 = 401 x 5

We see that we’ve found the number. AMC = 401 (notice that the sum of its digits is 5). So A = 4.

Answer: D
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Re: Let A, M, and C be digits with (100A + 10M + C)(A + M + C) = 2005. Wha   [#permalink] 24 Mar 2019, 18:29
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