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AB + CD = AAA, where AB and CD are two-digit numbers and AA [#permalink]

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06 Aug 2012, 09:57

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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?

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).

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

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16 Aug 2012, 10:11

Hi Bunuel,

Doesn't 82+19=111. But you say no two digit number with 1st digit as 1 can be added to a number less than 90 to get a sum of 111. Can you elaborate??

One more thing is only 111 fits into this scenario as it is the sum of two digit numbers which cannot exceed 188 right??
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How did u check that there is only one such number as AAA when AB and CD are added together..Please can you explain the logic behind it?

AAA is a 3-digit number with all 3 digits alike, so it could be: 111, 222, 333, ..., 999. But the sum of any two 2-digit numbers cannot be more than 99+99=198, so AB+CD can only be 111.
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Re: AB + CD = AAA, where AB and CD are two-digit numbers and AA [#permalink]

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06 Oct 2015, 02:19

the greatest 2 digit number is 99 so 99+99 =198... thats shows that AAA can not be 222 or 333 or ...... 999, it has to be 111.... now that aaa is 111,ab+cd=aaa ... a bit of logic, a=1 and c=9...
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