Question

X is an integer such that it leaves a remainder of 2 when divided by 3, leaves a remainder of 3 when divided by 5, leaves a remainder of 5 when divided by 7. What could be a possible value of X from among the following options?
A. 53
B. 68
C. 74
D. 83

Answer 1

Hint: We have to find the particular constant where we can find the number having remainder of 2 when divided by 3, leaves a remainder of 3 when divided by 5, leaves a remainder of 5 when divided by 7. We use the trial-and-error method to find the right number from the given options.

Complete step by step solution:
We will try to use the given options. There are four given options out of which one leaves a remainder of 2 when divided by 3, leaves a remainder of 3 when divided by 5, leaves a remainder of 5 when divided by 7.
We take 53. When we divide 53 by 7.
$7\overset{7}{\overline{\left){\begin{align}
  & 53 \\
 & \underline{49} \\
 & 4 \\
\end{align}}\right.}}$
The remainder is 4. Therefore, this one is not applicable.
Now we take 74. When we divide 74 by 7.
$7\overset{10}{\overline{\left){\begin{align}
  & 74 \\
 & \underline{70} \\
 & 4 \\
\end{align}}\right.}}$
The remainder is 4. Therefore, this one is not applicable.
Now we take 83. When we divide 83 by 7.
$7\overset{11}{\overline{\left){\begin{align}
  & 83 \\
 & \underline{77} \\
 & 6 \\
\end{align}}\right.}}$
The remainder is 6. Therefore, this one is not applicable.
For 68, it leaves a remainder of 2 when divided by 3, leaves a remainder of 3 when divided by 5, leaves a remainder of 5 when divided by 7.
$3\overset{22}{\overline{\left){\begin{align}
  & 68 \\
 & \underline{66} \\
 & 2 \\
\end{align}}\right.}}$
$5\overset{13}{\overline{\left){\begin{align}
  & 68 \\
 & \underline{65} \\
 & 3 \\
\end{align}}\right.}}$
$7\overset{9}{\overline{\left){\begin{align}
  & 68 \\
 & \underline{63} \\
 & 5 \\
\end{align}}\right.}}$

So, the correct answer is “Option B”.

Note: It can also be explained as the remainder theorem where we find the remainder in the form of the dividend, divisor and the quotient.
If we take $y,x,q$ as dividend, divisor and the quotient and $r$ as remainder, then the remainder theorem tells us that $y=xq+r$ where $0\le r < x$.