Question

Which is the greatest number that will divide \[37,50\] and $123$ leaving the remainder $1,2$ and $3$ respectively?
A. $9$
B. $10$
C. $15$
D. $12$

Answer 1

Hint: Here we are asked to find the greatest number which will leave the remainder $1,2$ and $3$ on dividing the numbers \[37,50\] and $123$. This problem can be solved by finding the highest common factor of the given numbers after subtracting the remainder from it. Remainder is nothing but the extra part which is in the given number so removing them will give us the part which is divisible by the required number.

Complete step-by-step answer:
It is given that we need to find the greatest number which will divide the numbers \[37,50\] and $123$ leaving the remainder $1,2$ and $3$ respectively.
The greatest number that will divide the given three numbers \[37,50\] and $123$ can be found by finding the highest common factor for these three numbers. But it is given that when we divide these numbers by the required number it is leaving the remainder $1,2$ and $3$ respectively. Thus, we first need to find the part of the given three numbers that is totally divisible by the required number. This can be found by subtracting the respective remainder from that number.
For the number $37$, the remainder is $1$.
$37 - 1 = 36$
Thus, thirty-six is the part that is totally divisible by the required number.
For the number $50$, the remainder is $2$.
$50 - 2 = 48$
Thus, forty-eight is the part that is totally divisible by the required number.
For the number $123$, the remainder is $3$.
$123 - 3 = 120$
Thus, one hundred and twenty is the part that is totally divisible by the required number.
Now let us find the prime factors of these numbers that is $36,48$ and $120$.
$
2\left| \!{\underline {\,
{36} \,}} \right. \\
2\left| \!{\underline {\,
{18} \,}} \right. \\
3\left| \!{\underline {\,
9 \,}} \right. \\
3\left| \!{\underline {\,
3 \,}} \right. \\
$
Therefore, $36 = 2 \times 2 \times 3 \times 3 \times 3$
$
2\left| \!{\underline {\,
{48} \,}} \right. \\
2\left| \!{\underline {\,
{24} \,}} \right. \\
2\left| \!{\underline {\,
{12} \,}} \right. \\
2\left| \!{\underline {\,
6 \,}} \right. \\
3\left| \!{\underline {\,
3 \,}} \right. \\
$
Thus, $48 = 2 \times 2 \times 2 \times 2 \times 3 \times 3$
$
2\left| \!{\underline {\,
{120} \,}} \right. \\
2\left| \!{\underline {\,
{60} \,}} \right. \\
2\left| \!{\underline {\,
{30} \,}} \right. \\
5\left| \!{\underline {\,
{15} \,}} \right. \\
3\left| \!{\underline {\,
3 \,}} \right. \\
$
Thus, $120 = 2 \times 2 \times 2 \times 5 \times 3$
From these factors the highest common factor of these three numbers is $2 \times 2 \times 3 = 12$.
Thus, the greatest required number is twelve which divides the numbers \[37,50\] and $123$ leaving the remainder $1,2$ and $3$ respectively.
Hence, option d) $12$ is the correct option.

So, the correct answer is “Option d)”.

Note: We can also check whether the answer we got in the above calculation is correct or not by dividing and see whether we get the same remainder as given in the question.
$37 \div 12$ will give you $3$ as quotient $\left( {12 \times 3 = 36} \right)$ and $1$ as remainder.
$50 \div 12$ will give you $48$ as quotient $\left( {12 \times 4 = 48} \right)$ and $2$ as remainder.
$123 \div 12$ will give you $10$ as quotient $\left( {12 \times 10 = 120} \right)$ and $3$ as remainder.
Thus, our answer was correct as we got the same remainders as given in the question.