boomtangboy wrote:
There are between 100 and 110 cards in a collection of cards. If they are counted out 3 at a time, there are 2 left over, but if they are counted out 4 at a time, there is 1 left over. How many cards are in the collection?
(A) 101
(B) 103
(C) 106
(D) 107
(E) 109
So first we can find the set of numbers between 100 and 110 which,when divided by 3, have a reminder of 2.
The set includes: 101, 104, and 107.
Next we can find the set of numbers between 100 and 110 which, when divided by 4, have a reminder of 1.
Teh set includes: 101, 105, and 109.
The number common to both sets provides the answer. Thus, there must be 101 cards in the collection.
Alternate Solution:
Let x represent the number of cards in this collection. Since 2 cards are left over when the cards are counted 3 at a time, the remainder when x is divided by 3 is 2 or, in other words, x = 3s + 2 for some integer s. Since one card is left over when the cards are counted 4 at a time, the remainder when x is divided by 4 is 1 or, in other words, x = 4k + 1 for some integer k.
Notice that if we add 7 to x, we obtain x + 7 = 3s + 9 = 4k + 8. Notice also that x + 7 is divisible by both 3 and 4; therefore it is divisible by 12. Thus, the smallest possible value of x + 7 is 12; which means the smallest possible value of x is 5. Every other possible value of x can be found by adding multiples of 12 to 5; such as 17, 29, 41 etc. If we add 12 x 8 = 96 to 5, we obtain 96 + 5 = 101.
Answer: A
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