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Inequalities + Fractions + Integers! Pleaaaase!!!!!

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Inequalities + Fractions + Integers! Pleaaaase!!!!! [#permalink] New post 22 Nov 2010, 06:09
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Can anybody help me with this? Is giving me a hard time....

If the fraction 98/(23*89) is written in the form a + b/23 + c/89 , with a,b anc c integer numbers such that:

1<= b < 23
1<= c< 89

Then the sum a + b + c is equal to?

a) 30
b) 31
c) 32
d) 33
e) 34


Thanks a lot!
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Re: Inequalities + Fractions + Integers! Pleaaaase!!!!! [#permalink] New post 22 Nov 2010, 21:24
mariellary wrote:
Can anybody help me with this? Is giving me a hard time....

If the fraction 98/(23*89) is written in the form a + b/23 + c/89 , with a,b anc c integer numbers such that:

1<= b < 23
1<= c< 89

Then the sum a + b + c is equal to?

a) 30
b) 31
c) 32
d) 33
e) 34


Thanks a lot!


a+b/23+c/89 = 98/(23*89)
(89*23a+89b+23c)/(23*89) = (98/23*89)

Hence, 89*23a+89b+23c=98
23*(89a+c)=98-89b

We need a b between 1 and 22 such that 98-89b is a multiple of 23
Lets look at the pattern of remainders with 23 that 98-89b leaves for different b :
b=1 ... remainder=9
b=2 ... remainder=12
b=3 ... remainder=15
...
b=5 ... remainder=21
b=6 ... remainder=1
...
b=13 ... remainder=22
b=14 ... remainder=2
...
b=21 ... remainder=0

98-89*21 = -77*23

So b=21 & 89a+c=-77
c=-77-89a
We know c is between 1 and 88, so only valid value of a is -1
with a=-1, c=12

a+b+c=-1+21+12=32

Answer is (c)
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Re: Inequalities + Fractions + Integers! Pleaaaase!!!!! [#permalink] New post 23 Nov 2010, 00:25
This question is really complicated. Thanks for sharing
Re: Inequalities + Fractions + Integers! Pleaaaase!!!!!   [#permalink] 23 Nov 2010, 00:25
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