Question

What is the least number which when divided by 7,9 and 12, leaves the same remainder 1 in each case?
A. 253
B. 352
C. 505
D. 523

Answer 1

Hint: To find the least number which when divided by 7,9 and 12, leaves the same remainder 1 in each case, we have to the LCM of 7, 9, and 12. We will get this as 252. It is given that the least number divided by 7, 9 and 12, should leave a remainder 1. So we add 1 to 252. The result will be the required answer.

Complete step-by-step solution:
We have to find the least number which when divided by 7,9 and 12, leaves the same remainder 1 in each case.
Let us find the LCM of 7, 9 and 12.
$$\begin{array}{rl}& 2|\underset{―}{7,9,12}\\ & 2|\underset{―}{7,9,6}\\ & 3|\underset{―}{7,9,3}\\ & 3|\underset{―}{7,3,1}\\ & 7|\underset{―}{7,1,1}\\ & 1,1,1\end{array}$$
Hence, LCM of 7, 9, 12 is given as $2\times 2\times 3\times 3\times 7=252$
It is given that the least number divided by 7, 9 and 12, should leave a remainder 1.
We found that the least number divided by 7, 9, and 12 that leaves a remainder 0 is 252. So, if we want a remainder 1, we will have to add 1 to 252.
Hence, the least number divided by 7, 9 and 12, that leaves a remainder 1 is
$252+1=253$
Hence, the correct option is A.

Note: We added 1 to 252 since we need the remainder 1. This can be also checked by doing the division. For example, when 253 is divided by 7, that is,
$$7\stackrel{{\displaystyle 36}}{)\phantom{1\begin{array}{c}253\\ -21\\ \mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\\ \begin{array}{cc}& 43\end{array}\\ \text{ - 42}\\ \text{\_\_\_\_\_\_}\\ \text{ 1}\end{array}}\overline{\phantom{1}\begin{array}{c}253\\ -21\\ \mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\\ \begin{array}{cc}& 43\end{array}\\ \text{ - 42}\\ \text{\_\_\_\_\_\_}\\ \text{ 1}\end{array}}}$$
We can see that the remainder is 1.
In the question, it is we have to find the least number. If we had to find the highest number, we have to find the HCF and perform similar operations.