Question

The greatest 4-digit number which when divided by 20, 24 and 45 leaves a remainder of 11 in each case is
A. 9999
B. 9998
C. 9997
D. 9731

Answer 1

Hint: First of all, find the L.C.M. of the given numbers by finding their prime factors. Then find the greatest four-digit number divisible by the L.C.M. of the given numbers. Then add the remainder to the obtained greatest four-digit number which is our required answer.

Complete step-by-step answer:
Given numbers are 20, 24 and 45
Consider the L.C.M. of 20, 24 and 45
Prime factors of 20 \[ = {2^2} \times 5\]
Prime factors of 24 \[ = {2^3} \times 3\]
Prime factors of 45 \[ = {3^2} \times 5\]
We know that L.C.M. is the product of each prime factor of highest power.
So, L.C.M. of 20, 24 and 45 \[ = {2^3} \times {3^2} \times 5 = 360\]
Now divide the greatest four-digit number i.e., 9999 with the obtained L.C.M. of 20, 24 and 45.
\[ \Rightarrow 9999 \div 360 = 27.775\]
So, the greater four-digit number perfectly divisible by the L.C.M is \[360 \times 27 = 9720\]
Now, the greatest four-digit number divisible by 360 and leaving a remainder of 11 is \[9720 + 11 = 9731\]
So, the greatest four-digit number divisible by 20, 24 and 45 leaves a remainder of 11 is 9731.
Thus, the correct option is D. 9731

Note: The L.C.M. (least common multiple) is the smallest positive integer that is divisible by all the given numbers. For verifying our answer, we can divide by all the numbers, and if we get 360 as quotient and 11 as remainder in all cases, the obtained answer is correct otherwise incorrect.