What is the smallest natural number which, when divided by 7, 9 and 11, leaves remainders 1, 3 and 5 respectively?
687
Note that in each case, divisor minus remainder is the same: 7-1=6, 9-3=6, 11-5=6. So the required number = LCM(7,9,11) - 6. Since 7, 9, 11 are pairwise coprime, LCM = 7×9×11 = 693. Required number = 693 - 6 = 687. Hence option (D) is correct.
Important Questions from Number System
Consider the following statements :
1. (25)! + 1 is divisible by 26
2. (6)! + 1 is divisible by 7
Which of the above statements is/are correct ?
If the sum S is divided by 8, what is the remainder ?
If the sum S is divided by 60, what is the remainder ?
Find the sum of squares of the greatest value and the smallest value of K in the number so that the number 45082K is divisible by 3.
How many composite numbers are there from 53 to 97 ?
