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# If the fraction 98/(23*89) is written in the form a + b/23 + c/89, wil

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Joined: 02 Sep 2009
Posts: 60727
If the fraction 98/(23*89) is written in the form a + b/23 + c/89, wil  [#permalink]

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05 Dec 2019, 01:37
10
00:00

Difficulty:

55% (hard)

Question Stats:

24% (01:44) correct 76% (03:01) wrong based on 17 sessions

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If the fraction $$\frac{98}{23*89}$$ is written in the form $$a + \frac{b}{23} + \frac{c}{89}$$, with a, b and c are integers such that:

$$1 \leq b < 23$$
$$1 \leq c< 89$$

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

A. 30
B. 31
C. 32
D. 33
E. 34

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Math Expert
Joined: 02 Aug 2009
Posts: 8327
Re: If the fraction 98/(23*89) is written in the form a + b/23 + c/89, wil  [#permalink]

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05 Dec 2019, 06:06
2
Bunuel wrote:
If the fraction $$\frac{98}{23*89}$$ is written in the form $$a + \frac{b}{23} + \frac{c}{89}$$, with a, b and c are integers such that:

$$1 \leq b < 23$$
$$1 \leq c< 89$$

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

A. 30
B. 31
C. 32
D. 33
E. 34

Are You Up For the Challenge: 700 Level Questions

$$a + \frac{b}{23} + \frac{c}{89}=$$$$\frac{98}{23*89}$$..
So, now $$\frac{98}{23*89}$$<1, so a will surely not be >1.
Can it be 0? If a=0, then $$0 + \frac{b}{23} + \frac{c}{89}=$$$$\frac{98}{23*89}...89b+23c=98$$.... b and c are at least 1, so a cannot be 0.
Can it be -1? If a is -2, then (-2)+(0 to 1)+(0 to 1)=(-2)+(0 to 2) = -something. Thus a can only be -1

Now $$a + \frac{b}{23} + \frac{c}{89}=$$$$\frac{98}{23*89}....23*89a+89b+23c=98....89b+23c=98+23*89=9+89+23*89=9+24*89$$..
$$24*89-b*89=23c-9.....89(24-b)=23c-9$$
Now, 89 is a prime number, so 23c-9 has to be a multiple of 89...23c-9=89x
when x=1...23c-9=89...23c=98...c is NOT an integer
when x=2...23c-9=89*2...23c=198+9...c is NOT an integer
when x=3...23c-9=89*3...23c=267+9=276...c =12
so 24-b=x=3...b=21

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

C
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Re: If the fraction 98/(23*89) is written in the form a + b/23 + c/89, wil   [#permalink] 05 Dec 2019, 06:06
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