Question

Find the least number which when divided separately by 15, 20, 48 and 36 will leave a remainder 9 in each case
(a) 189
(b) 349
(c) 683
(d) 729

Answer 1

Hint: We start solving the problem by assigning a variable ‘x’ for the required number. We then make use of the result dividend = $\left( \text{divisor}\times \text{quotient} \right)$ + remainder by assuming quotients when the given number is divided by 15, 20, 48, 36. We then find that $x-9$ is exactly divisible by 15, 20, 48, 36. We then find the L.C.M of 15, 20, 48, 36 and then add 9 to the obtained L.C.M to get the required answer.

Complete step by step answer:
According to the problem, we are asked to find the least number which when divided separately by 15, 20, 48 and 36 will leave a remainder 9 in each case.
Let us assume the value of the required number is ‘x’.
We know that dividend = $\left( \text{divisor}\times \text{quotient} \right)$ + remainder.
Let us assume the quotient when x is divided by 15 be ${{q}_{1}}$.
So, we have $x=\left( 15\times {{q}_{1}} \right)+9$.
$\Rightarrow x-9=15\times {{q}_{1}}$ ---(1).
Let us assume the quotient when x is divided by 20 be ${{q}_{2}}$.
So, we have $x=\left( 20\times {{q}_{2}} \right)+9$.
$\Rightarrow x-9=20\times {{q}_{2}}$ ---(2).
Let us assume the quotient when x is divided by 48 be ${{q}_{3}}$.
So, we have $x=\left( 48\times {{q}_{3}} \right)+9$.
$\Rightarrow x-9=48\times {{q}_{3}}$ ---(3).
Let us assume the quotient when x is divided by 36 be ${{q}_{4}}$.
So, we have $x=\left( 36\times {{q}_{4}} \right)+9$.
$\Rightarrow x-9=36\times {{q}_{4}}$ ---(4).
From equations (1), (2), (3) and (4), we can see that $x-9$ is clearly divisible by 15, 20, 48, 36.
We know that the least number that is divisible by two or more numbers is known as L.C.M (Least Common Multiple).
So, we get $x-9$ as the L.C.M of the numbers 15, 20, 48, 36. Now, let us find the L.C.M of the numbers 15, 20, 48 and 36 which is as shown below:
$\begin{align}
  & 2\left| \!{\underline {\,
  15,20,48,36 \,}} \right. \\
 & 2\left| \!{\underline {\,
  15,10,24,18 \,}} \right. \\
 & 5\left| \!{\underline {\,
  15,5,12,9 \,}} \right. \\
 & 3\left| \!{\underline {\,
  3,1,12,9 \,}} \right. \\
 & \left| \!{\underline {\,
  1,1,4,3 \,}} \right. \\
\end{align}$.
So, the L.C.M of the numbers 15, 20, 48, 36 is $2\times 2\times 5\times 3\times 4\times 3=720$.
Now, we have $x-9=720$.
$\Rightarrow x=729$.
We have found the value of the required number as 729.

So, the correct answer is “Option d”.

Note: Whenever we get this type of problems, we first try to find the relation between the L.C.M of the divisors and the given numbers which will lead us to the required answer. We should not forget to add 9 to the obtained L.C.M, as this is the most common mistake done by students. We should not make mistakes while finding the L.C.M of the given numbers. Similarly, we can expect problems to find the sum of the quotients when the number is divided by 15, 20, 48, 36.