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remainders of 4 and 6

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remainders of 4 and 6

by 4meonly » Sun Jan 11, 2009 12:10 am
If a is a positive integer and if remainders of 4 and 6 are obtained when 89 and 125, respectively, are divided by a, then a =
A. 7
B. 9
C. 15
D. 17
E. 19

OA D. Anybody with quick approach / shortcut?
I divided 85 and 119 by each answer :x
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by DanaJ » Sun Jan 11, 2009 12:17 am
Really quick one :P :
So you have the remainder 4 when you divide 89 by a. That means that (89-4)=85 is divisible by a. Well, 85=5*17, where both 5 and 17 are prime numbers. So a can only be 5 or 17. But since 125 is a multiple of 5, your answer is 17.

Answer: D
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by cramya » Sun Jan 11, 2009 12:20 am
I am sure there may be other approaches but IMO what u did is very short too.

All u had to do was divide 89 by each choice only D) will give 4. U can forget the 125 part.



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Cramya
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by cramya » Sun Jan 11, 2009 12:22 am
Dana's approach is a good one also!

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by acecoolan » Sun Jan 11, 2009 12:46 am
89 = ka + 4 -----1
125 = ma + 6 ----2

2 - 1

(m-k)a = 34

Check answer choices - only D) fits the bill

-A
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by 4meonly » Sun Jan 11, 2009 12:59 am
acecoolan wrote:89 = ka + 4 -----1
125 = ma + 6 ----2

2 - 1

(m-k)a = 34

Check answer choices - only D) fits the bill

-A
WOW! Interesting! Simple and fast! Just to check factors of 34! Thank you!
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