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K is divided by 1602

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by Brent@GMATPrepNow » Sun Jan 11, 2009 8:53 am
When positive integer K is divided by 1602, the remainder is 102. What is the remainder when K is divided by 89?
(A) 0
(B) 13
(C) 18
(D) 34
(E) 51
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by DanaJ » Sun Jan 11, 2009 10:44 am
well you have k = 1602*n+102.
All you have to notice, IMHO, is that 1602 = 2 *801 = 2(890-89)=2*9*89. So k = 2*9*89*n + 102 = 89(2*9*n+89) +13. That means that the remainder will be 13.
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Re: K is divided by 1602

by Ian Stewart » Sun Jan 11, 2009 11:01 am
Dana has a great solution above that proves that the remainder will always be 13. Her method is the one pro number theorists would use. Because this is a GMAT question, there is a shortcut:
Brent Hanneson wrote:When positive integer K is divided by 1602, the remainder is 102. What is the remainder when K is divided by 89?
(A) 0
(B) 13
(C) 18
(D) 34
(E) 51
Well, if the remainder is 102 when K is divided by 1602, K could certainly be equal to 102, in which case the remainder will be 13 when K is divided by 89. And, because the answer choices are all numbers (we don't have the answer 'cannot be determined exactly') if the remainder is 13 for one suitable value of K, it must be 13 for any suitable value of K, since this is a GMAT question; one of the answers must be correct.
For online GMAT math tutoring, or to buy my higher-level Quant books and problem sets, contact me at ianstewartgmat at gmail.com

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by Brent@GMATPrepNow » Sun Jan 11, 2009 11:33 am
Well, if the remainder is 102 when K is divided by 1602, K could certainly be equal to 102
Ian raises a very important point about finding numbers to plug into questions involving remainders.
Often, when students are given information about remainders, they begin examining numbers that are larger than the divisor. For example, if n is divided by 6, the remainder is 5, most students recognize that n could equal 11, 17, 23, etc, but they often fail to recognize that n might also equal 5.
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by logitech » Sun Jan 11, 2009 12:27 pm
Brent Hanneson wrote:
Well, if the remainder is 102 when K is divided by 1602, K could certainly be equal to 102
Ian raises a very important point about finding numbers to plug into questions involving remainders.
Often, when students are given information about remainders, they begin examining numbers that are larger than the divisor. For example, if n is divided by 6, the remainder is 5, most students recognize that n could equal 11, 17, 23, etc, but they often fail to recognize that n might also equal 5.
Great thread. Invaluable information.

K=1602X+102
K/89 = 18x + 89/89 + 13/89
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by woo » Mon Jan 12, 2009 4:05 am
DanaJ wrote:well you have k = 1602*n+102.
All you have to notice, IMHO, is that 1602 = 2 *801 = 2(890-89)=2*9*89. So k = 2*9*89*n + 102 = 89(2*9*n+1) +13. That means that the remainder will be 13.
There is an error in the explanation. Still a very good explanation
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by DanaJ » Mon Jan 12, 2009 4:10 am
True... Missed it! Thanks
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by welcome » Mon Jan 12, 2009 4:54 am
Good input Ian.

102/89 = 1+(13/89)
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by Mr2Bits » Mon Jan 12, 2009 12:23 pm
Easiest way I found to solve this is of course to solve for K to start:

K = 1602 +102

k = 1704 (check your answer as this number divided by 1602 leaves a remainder of 102..check)

Question now askes to solve the remainder(R) when K is divided by 89

r = 1704/89 .....Largest rounded down even multiple is 19. 89x19 = 1691 is the largest multiple and leaves a difference/remainder of 13

B is your answer
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