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mina [271]
3 years ago
9

2. Suppose (220)_b and (251)_b are the squares of consecutive integers. Determine b.

Mathematics
1 answer:
Andrei [34K]3 years ago
3 0

Let x and x+1 be these consecutive integers. For some base b, we can write

(x_{10})^2=220_b=2b^2+2b

((x+1)_{10})^2=251_b=2b^2+5b+1

If

x^2=2b^2+2b

then

x^2+1=2b^2+2b+1

Now,

x^2+2x+1=2b^2+5b+1

2x+2b^2+2b+1=2b^2+5b+1

\implies2x=3b

Squaring both sides gives

4x^2=9b^2

and so

4(2b^2+2b)=9b^2

\implies8b^2+8b=9b^2

\implies b^2-8b=0\implies\boxed{b=8}

(because the base has to be non-zero)

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Explanation:
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   x + y = 140 ;
 
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