In first 1000 natural numbers, how many numbers exists such that when divided by 7 leaves a remainder of 6?
142
141
143
146
The numbers required will be 6, 13, 20, 27, 34, 41, ....., 1000
6 + (n-1) 7 = 1000
n = 143
In first 1000 natural numbers, how many numbers exists such that when divided by 11 leaves a remainder of 8?
How many integers that lie between 0 and 1000 which when divided by 2 or 4 leave a remainder of 1 and by 7 leaves a remainder of 2 ?