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Here is this week’s problem:

X is a three-digit positive integer in which each digit is either 1 or 2. Y has the same digits as X, but in reverse order. What is the remainder when X is divided by 3?

(1) The hundreds digit of XY is 6.
(2) The tens digit of XY is 4.

To see the answer choices, and to submit your answer, visit our Challenge Problem Showdown page on our site.

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#### Tom Williams

Tom Williams is a Marketing Associate at Manhattan Prep. His background and interests are primarily within New Media and Online Marketing. As such, he spends approximately 18 hours a day in front of a computer. When he’s not interacting with Manhattan Prep students on Facebook and Twitter, Tom can be found listening to records, rooting for the Mets and Jets, and reading history books.

### 11 responses to Challenge Problem Showdown – July 23rd, 2012

1. the only possibilities are 111, 222, 112, 121, 221 and their opposites (122 and 211 comes in the reverse order anyway)
Stmt (1) There are only 2 possibilities 121 * 121 = 14641
and 112 * 211 = 23632. here X can be 121, 112 or 211 but it is still sufficient as if you take any no. and divide by 3, the remainder is 1.
Strike of BCE, AD remains

Stmt (2) only one possibility 121 * 121 = 14641
sufficient.

Ans is D

2. The only possibilities are 111, 222, 112, 121, 221 and their opposites (122 and 211 comes in the reverse order anyway)

Stmt (1) There are only 2 possibilities 121 * 121 = 14641
and 112 * 211 = 23632. here X can be 121, 112 or 211 but it is still sufficient as if you take any no. and divide by 3, the remainder is 1.
Strike of BCE, AD remains

Stmt (2) only one possibility 121 * 121 = 14641
sufficient.

Ans is D

3. i m sorry rohit ,but my answer is A as in the second fact statement there is an another option that is 212 which staisfies the statement but gives the remainder 2

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