What would be the smallest natural number which when divided either by 20 or by 42 or by 76 leaves a remainder
Q. What would be the smallest natural number which when divided either by 20 or by 42 or by 76 leaves a remainder of 7 in each case?
| (A) 3047 | (B) 6047 | (C) 7987 | (D) 63847 |
Ans: 7987
Sol:
Number is divided either by 20 or by 42 or by 76
= K × LCM (20, 42, 76) + constant difference
= 7980 K + 7 ( where K is natural number)
So, if we put K = 1 then,
Least number will be 7980 + 7 = 7987