A certain number when successively divided by $8$ and $11$ leaves remainders of $3$ and $7$, respectively. What will be the remainder when the number is divided by the product of $8$ and $11$, $88$?
[Math] A certain number when successively divided by $8$ and $11$ leaves remainders of $3$ and $7$, respectively.
elementary-number-theory
Best Answer
$n=8a+3=11b+7$. So 8 divides $11b+4$, so $b=4\bmod 8$. Hence $n=11(8k+4)+7=51\bmod88$. Check: $51=6\cdot8+3=4\cdot11+7$.