Question

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From the question, we can write that

$a$ + $100$ = $b_{2}$$a$ + $168$ = $c_{2}$

=> $c_{2}−b_{2}$ = 68

=> $(c−b)(c+b)=1×68$

Now solve the equation for $(c−b)=1$ and $(c+b)=68$

We get $c=35$ and $b=34$. (here $c=34.5$ which is $35$ and $b=33.5$ which can be written as $34$ by ignoring the point )

$a+100=b_{2}$

$a=34_{2}−100$

$a=1156−100=1056$ hence this the required number.

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