Find the least number which when divided separately by 15, 20, 48 and 36 will leave a remainder 9 in each case.

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189

349

683

729

answer is D.

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Detailed Solution

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 = (divisor×quotient) + remainder.
Let us assume the quotient when x is divided by 15 be q1.
So, we have x=(15×q1)+9.
⇒x−9=15×q1 ---(1).
Let us assume the quotient when x is divided by 20 be q2.
So, we have x=(20×q2)+9.
⇒x−9=20×q2 ---(2).
Let us assume the quotient when x is divided by 48 be q3.
So, we have x=(48×q3)+9.
⇒x−9=48×q3 ---(3).
Let us assume the quotient when x is divided by 36 be q4.
So, we have x=(36×q4)+9.
⇒x−9=36×q4 ---(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:
Question Image

So, the L.C.M of the numbers 15, 20, 48, 36 is 2×2×5×3×4×3=720.
Now, we have x−9=720.
⇒x=729.
We have found the value of the required number as 729.
So, the correct answer is “Option 4”.

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