question_answer

Find the least multiple of 23, which when divided by 18, 21 and 24 leaves the remainder 7, 10 and 13, respectively?

A)  3013

B)  3024   

C)  3002                           

D)  3036

Correct Answer: A

Solution :

LCM of 18, 21 and 24 LCM\[=2\times 3\times 3\times 7\times 4=504\] Now, compare the divisors with their respective remainders. We observe that in all the assets the remainder is just 11 less than their respective divisor. So the number can be given by \[504\,\,K-11,\]where K is an positive integer. Since, \[23\times 21=483\] We can write \[=504\,\,K-11=(483+21)\,\,K-11\] \[=483\,\,K+(21\,\,K-11)\] 483 K is multiple of 23, since 483 is divisible by 23, So, for \[(504K-11)\]to be multiple of 23, the remainder\[(21K-11)\] must be divisible by 23. Put the value of K = 1, 2, 3, 4, 5, 6 ... and so on successively. We find that the minimum value of K for which \[(21\,\,K-11)\]divisible by 23, is 6, \[(21\times 16-11)=115\]is which is divisible by 23. Therefore, the required least number \[=504\times 6-11=3013\]