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Cards Problem

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by raunakrajan » Wed Jun 16, 2010 9:32 pm
there are between 100 and 110 cards in a collection of cards. If they are counted 3 at a time there are 2 left over but if they are counted 4 at a time there is one left over. How many cards are there in the collection?
(A) 101
(B) 103
(C) 106
(D) 107
(E) 109
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by Testluv » Wed Jun 16, 2010 10:19 pm
If they are counted 3 at a time, there are 2 left over. In other words, the number of cards when divided by 3, leaves a remainder of 2. The largest multiple of 3 under 100 is 99. So, from this constraint, the number of cards can be 101, 104, 107 (and 110). Only 101 and 107 appear among the answer choices. Eliminate B, C, and E.

The number of cards when divided by 4 leaves a remainder of 1. We know that 100 is divisible by 4. Thus, the number of cards must be 101.

Choose A.
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by amising6 » Wed Jun 16, 2010 10:22 pm
so basivcally we can say when n is divide by 3 =you get 2 remainder so n=3k+2
when n is divided by 4 you get remainder 1 =4p+1

so take option
a)if we take first option it gives remainder 2 with 3 and 1 with 4
b)it gives remainder 1 with 3 so not possible
c)106 gives remainder 2 with 4
d)gives remainder 3 with when divided by 4
e) gives remainder 1 when divided by 3
so answer is A
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by Testluv » Wed Jun 16, 2010 10:30 pm
Right! But you don't have to check all the choices against both rules. You only need to check the choices against 1 rule at a time so that you can eliminate them more quickly. So, once we say that n = 3k + 2, given that n is between 100 and 110, we can instantly eliminate three choices without having to check them against the other constraint.

OR: as soon as you see that choice A satisfies both rules, there is NO NEED to check the other choices--there can only be one correct answer.
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