Question

how many three digit number are divisible by 7

Answer 1

15*7=105 is the first three digit number divisible by 7 and the last number divisible by 7 is 142*7=994. Using Arithmetic progression, Formula for last term is a+(n-1)d
a=105 d=7 l=994
994=105+(n-1)7
994=105+7n-7
994=98+7n
7n=994-98
7n=896
n=896/7=128

Answer 2

1st three digit no. divisible by 7=105(7a)
2nd(7b)=994
Let the number of terms be n . Then tn = 994.
994 = 105 + 7(n -1)
994-105=7(n-1)
889 = 7(n-1)
889/7=n-1
n -1 = 127
n = 128.
so therefore, 128 three digit numbers which are divisible by 7.

Hope this helps guys