Question

ICDS (1) Exam, 20151
What is the sum of digits of the least multiple of 13, which when divided by 6, 8 and 12 leaves
5.7 and 1l respectively as the remainders?​

Answer 1

$\large\underline{\sf{Solution-}}$

Given that,

A least multiple of 13, which when divided by 6, 8 and 12 leaves 5, 7 and 11 respectively as the remainders.

Let first find a number which when divided by 6, 8 and 12 leaves 5, 7 and 11 respectively as the remainders.

Let assume that, the required number is 'x' which when divided by 6, 8 and 12 leaves 5, 7 and 11 respectively as the remainders

If we check out the divisors and remainders, i.e 6 and 5, 8 and 7, 12 and 11, we find that the difference between divisor and remainder is 1.

So, Required number is x + 1 = multiple of LCM (6, 8, 12)

Now, LCM (6, 8, 12) = 24

So,

$\bf\implies \:x + 1 = 24m$

$\bf\implies \:x = 24m - 1$

So, 24m - 1 is a required number which when divided by 6, 8 and 12 leaves 5, 7 and 11 respectively as the remainders.

Now, we have to find the least multiple of 13 which when divided by 6, 8 and 12 leaves 5, 7 and 11 respectively as the remainders.

Thus,

We have to find value of m in such a way that x is divisible by 13 and it should be least.

So, using hit and trial method for different values of m

$\begin{gathered}\boxed{\begin{array}{c|c} \bf m & \bf x = 24m - 1 \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf 1 & \sf 23 \\ \\ \sf 2 & \sf 47\\ \\ \sf 3 & \sf 73\\ \\ \sf 4 & \sf 95\\ \\ \sf 5 & \sf 119\\ \\ \sf 6 & \sf 143 \end{array}} \\ \end{gathered}$

So, least multiple of 13 is 143 which when divided by 6, 8 and 12 leaves 5, 7 and 11 respectively as the remainders.

• So, sum of digits = 1 + 4 + 3 = 8.

Answer 2

Answer:

1 plus 4 plus 3 is equal to 8