Question

How many three-digit numbers would you find, which when divided by 3, 4, 5, 6, 7 leave the remainders 1, 2, 3, 4, and 5 respectively? (A) 4 (B) 3 (C) 2 (D) 1​

Answer 1

Answer: Correct option is

D

1

Here (3-1) =2

(4-2) =2 , (5-3)= 2, (6-4) =2 and (7-5) =2

LCM of 3, 4, 5, 6, 7 is 420.

Consider three digit number 999

When 999 is divided by 420, the remainder is 159 with quotient 2.

Subtracting 159 from 999 gives 840.

again subtract 2 from 840 = 838. {here 2 is subtracted from 840 as the difference between the given divisors and the remainders is  2 in each case)

Now,when  838 is divided by 3, 4, 5, 6, 7 we get the remainders as 1, 2, 3, 4, 5 respectively.

Hence only one such number is possible.