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The sum of the two 5-digit numbers AMC10 and AMC12 is 123422. What is

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The sum of the two 5-digit numbers AMC10 and AMC12 is 123422. What is  [#permalink]

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New post 19 Mar 2019, 02:23
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A
B
C
D
E

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Question Stats:

85% (02:12) correct 15% (01:42) wrong based on 34 sessions

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The sum of the two 5-digit numbers AMC10 and AMC12 is 123422. What is  [#permalink]

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New post 19 Mar 2019, 04:23
Bunuel wrote:
The sum of the two 5-digit numbers AMC10 and AMC12 is 123422. What is A + M + C?

(A) 10
(B) 11
(C) 12
(D) 13
(E) 14


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Answer: Option E
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Re: The sum of the two 5-digit numbers AMC10 and AMC12 is 123422. What is  [#permalink]

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New post 19 Mar 2019, 05:41
Bunuel wrote:
The sum of the two 5-digit numbers AMC10 and AMC12 is 123422. What is A + M + C?

(A) 10
(B) 11
(C) 12
(D) 13
(E) 14


solve for expression
we observe
c= 7, m =1 and a= 6
a+m+c = 14
IMO E
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Re: The sum of the two 5-digit numbers AMC10 and AMC12 is 123422. What is  [#permalink]

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New post 21 Mar 2019, 17:55
Bunuel wrote:
The sum of the two 5-digit numbers AMC10 and AMC12 is 123422. What is A + M + C?

(A) 10
(B) 11
(C) 12
(D) 13
(E) 14


We see that the last two digits of numbers do add up to 22 (10 + 12 = 22). Therefore, AMC + AMC = 1234 or 2(AMC) = 1234. So AMC = 1234/2 = 617. That is, A = 6, M = 1 and C = 7. So A + M + C = 6 + 1 + 7 = 14.

Alternate Solution:

Carrying out the addition, we notice that the units digit of C + C is 4; therefore C = 2 or C = 7. Next, we notice that the result from M + M is 3; this is only possible if the units digit of M + M is 2 and 1 carried over from the addition of C + C. Thus C = 7 and M = 1 or M = 6. If M were 6, then 1 would carry over to A + A and the first two digits of the sum would have been an odd number. Since the first two digits of the sum is 12, M can only be 1. Finally, since nothing carried over when we add M + M, A must be 6. Thus, A + M + C = 6 + 1 + 7 = 14.

Answer: E
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Re: The sum of the two 5-digit numbers AMC10 and AMC12 is 123422. What is   [#permalink] 21 Mar 2019, 17:55
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