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Re: challenging remainder problem! [#permalink]
24 Mar 2008, 15:23
I found the solution but thanks for the explanation. I have a more detail explanation for those who are interested.
The problem states that when x is divided by y the remainder is 6. In general, the divisor (y in this case) will always be greater than the remainder. To illustrate this concept, let's look at a few examples:
15/4 gives 3 remainder 3 (the divisor 4 is greater than the remainder 3) 25/3 gives 8 remainder 1 (the divisor 3 is greater than the remainder 1) 46/7 gives 6 remainder 4 (the divisor 7 is greater than the remainder 4)
In the case at hand, we can therefore conclude that y must be greater than 6.
The problem also states that when a is divided by b the remainder is 9. Therefore, we can conclude that b must be greater than 9.
If y > 6 and b > 9, then y + b > 6 + 9 > 15. Thus, 15 cannot be the sum of y and b.