60ab57377 is divisible by 99 .Find 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,

$6+0+a+b+5+7+3+7+7=multipleof9\phantom{\rule{0ex}{0ex}}a+b+35=multipleof9$

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

$a+b+35=36\phantom{\rule{0ex}{0ex}}a+b=36-35$

$a+b=1$ ... (1)

or

$a+b+35=45\phantom{\rule{0ex}{0ex}}a+b=45-35$

$a+b=10$ ... (2)

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,

$\left(6+a+5+3+7\right)-\left(0+b+7+7\right)=multipleof11\phantom{\rule{0ex}{0ex}}\left(a+21\right)-\left(b+14\right)=multipleof11\phantom{\rule{0ex}{0ex}}a+21-b-14=multipleof11\phantom{\rule{0ex}{0ex}}a-b+7=multipleof11$

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

$a-b+7=11\phantom{\rule{0ex}{0ex}}a-b=11-7$

$a-b=4$ ... (3)

or

$a-b+7=0$

$a-b=-7$ ... (4)

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

$a+b+a-b=10+4\phantom{\rule{0ex}{0ex}}2a=14\phantom{\rule{0ex}{0ex}}a=\frac{14}{2}\phantom{\rule{0ex}{0ex}}a=7$

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

$a+b=10\phantom{\rule{0ex}{0ex}}7+b=10\phantom{\rule{0ex}{0ex}}b=10-7\phantom{\rule{0ex}{0ex}}b=3$

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

**
**