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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.
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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
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
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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