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AB + CD = AAA, where AB and CD are 2 digit numbers and AAA [#permalink]

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08 Jun 2005, 11:32

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AB + CD = AAA, where AB and CD are two-digit numbers and AAA is a three digit number; A, B, C, and D are distinct positive integers. In the addition problem above, what is the value of C?

Yep, thats right. Can you please explain how you got that? These kinda questions always stumps me.

AB +
CD =
AAA

We know that A must be 1 since A + C (+1 maybe)= AA
(+1 maybe) means that an extra unit can be carried over from summing B and D.
The largest number AA can be cannot exceed 19.

Now since A=1 1+C = 11, C cannot be 10, so we need one unit to be carried over from B+D so C equals 9.

p.s. there are multiple combinations for B and D as long as B+D = 11

AB + CD = AAA, where AB and CD are 2 digit numbers and AAA is a 3 digit number. A,B,C and D are distinct positive numbers. In the above addition problem, what is the value of C?

from AB+CD=AAA:
B+D=10+A
A+C+1=10A+A -> C+1=10A
A has to be 1 as if the sum of two 2-digit numbers equals a 3-digit number, the max that the 3-digit number can be is 198...
if A=1, C=9

AB + CD = AAA, where AB and CD are 2 digit numbers and AAA is a 3 digit number. A,B,C and D are distinct positive numbers. In the above addition problem, what is the value of C?

The sum of two 2 digit numbers cannot be more than 200. Hence A had to be 1.

Re: AB + CD = AAA, where AB and CD are 2 digit numbers and AAA [#permalink]

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24 Feb 2014, 09:49

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AB + CD = AAA, where AB and CD are two-digit numbers and AAA is a three digit number; A, B, C, and D are distinct positive integers. In the addition problem above, what is the value of C?

(A) 1 (B) 3 (C) 7 (D) 9 (E) Cannot be determined

Since AB and CD are two-digit integers, their sum can give us only one three digit integer of a kind of AAA: 111.

So, A=1 and we have 1B+CD=111

Now, C can not be less than 9, because no two-digit integer with first digit 1 (1B<20) can be added to two-digit integer less than 90, so that to have the sum 111 (if CD<90, so if C<9, CD+1B<111).

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