\(a + \frac{b}{23} + \frac{c}{89}=\)\(\frac{98}{23*89}\)..
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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