Question

Let the least number of six digits, which when divided by 4, 6, 10 and 15, leaves in each case the same remainder of 2, be N . The sum of the digits in N is ?

Answer 1

Hint: Find the least number of six digits which when divided by 4, 6, 10 and 15 leaves 2 as a remainder . Use LCM of 4, 6, 10 and 15 to find that number and divide the least six digit number with that LCM. You’ll get a remainder and quotient. Now, subtract remainder from LCM (divisor) and add the least six digit number taken. You’ll get a number which satisfies the condition of the question. Then, add all terms of that number and you’ll get the final answer.

Complete step-by-step answer:

Least six digit number = 100000

Let us find the smallest number divisible by 4 , 6 , 10 , 15
i.e LCM of 4 , 6 , 10 , 15 = 60 ( using prime factorization method )
Let us divide 100000 by 60 to check the remainder and find the smallest six digit number divisible by 60.

$ \Rightarrow \dfrac{{100000}}{{60}}$ remainder = 40 , quotient = 1666

Now to find the number divisible by 60 we need to subtract 40 from 100000 but if we do that it becomes a five digit number . Therefore the next divisible number would be \[100000 + \left( {60 - 40} \right) = 100020\]

Which leaves the remainder 0 , but the least number of six digits (N) , which when divided by 4, 6, 10 and 15, leaves in each case the same remainder of 2 will be 100020+2 = 100022

Therefore the sum of digits = 1+0+0+0+2+2= 5

Note: Using LCM of numbers to find the number divisible by all of them has to be known. Remember that if we subtract the remainder the final number would not be of six digits .