Solution:How many 3-digit numbers leave remainder 1 when divided by 7
First such number is 105+1 = 106
Last such number is 994 + 1 = 995
AP with d = 7 , a(First term) = 106,
I (last term) = 995, Then n = ?
So, 106 + (n - 1) × 7 = 995
n = 889/7 + 1 = 128