60ab57377 is divisible by 99 .Find a and b?

In the given question we know that the number 60ab57377 is divisible by 99. We need to find the numbers a and b.

Here, use the fact that a number which is divisible by 99 is also divisible by 9 and 11 separately.

Now, the number 60ab57377 is divisible by 9.
Using the divisibility test, we know that if a number is divisible by 9, then the sum of all the digits of the number should be a multiple of 9. So,

$$\begin{gathered} 6+0+a+b+5+7+3+7+7=multiple of 9 \\ a+b+35=multiple of 9 \end{gathered}$$

Now, as a and b are single digit numbers, so they cannot take values greater than 9. So,

$$\begin{gathered} a+b+35=36 \\ a+b=36-35 \end{gathered}$$ 
$a+b=1$             ... (1)

or

$$\begin{gathered} a+b+35=45 \\ a+b=45-35 \end{gathered}$$
$a+b=10$         ... (2)

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Similarly the number 60ab57377 is divisible by 11.
We know that if a number is divisible by 11, then the sum of the alternate digits minus the sum of the remaining digits is a multiple of 11. So,

$$\begin{gathered} \left(6+a+5+3+7\right)-\left(0+b+7+7\right)=multiple of 11 \\ \left(a+21\right)-\left(b+14\right)=multiple of 11 \\ a+21-b-14=multiple of 11 \\ a-b+7=multiple of 11 \end{gathered}$$

Now, as a and b are single digit numbers, so they cannot take values greater than 9. So,

$$\begin{gathered} a-b+7=11 \\ a-b=11-7 \end{gathered}$$
$a-b=4$  ... (3)

or

$a-b+7=0$
$a-b=-7$    ... (4)

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Adding equation (2) and (3), we get

$$\begin{gathered} a+b+a-b=10+4 \\ 2a=14 \\ a=\frac{14}{2} \\ a=7 \end{gathered}$$

Substituting $a=7$ in equation (2), we get

$$\begin{gathered} a+b=10 \\ 7+b=10 \\ b=10-7 \\ b=3 \end{gathered}$$

Therefore, $\mathit{a}\mathbf{=}\mathbf{7}\mathbf{ }\mathit{a}\mathit{n}\mathit{d}\mathbf{ }\mathit{b}\mathbf{=}\mathbf{3}$