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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