How many 3 digit positive integers exist that when divided by 7 leave a remainder of 5? Choose one.
A) 128	B) 142	C) 143	D) 141	E) 129
Explanatory Answer   Step 1: Find the first and last term of the series   The smallest 3-digit positive integer that leaves a remainder of 5 when divided by 7 is 103.   The largest 3-digit positive integer that leaves a remainder of 5 when divided by 7 is 999.   The series of numbers that satisfy the condition that the number should leave a remainder of 5 when divided by 7 is an A.P (arithmetic progression) with the first term being 103 and the last term being 999. The common difference of the sequence is 7.   Step 2: Compute the number of terms   In an A.P, the last term l = a + (n - 1) * d, where 'a' is the first term, 'n' is the number of terms of the series and 'd' is the common difference.   Therefore, 999 = 103 + (n - 1) * 7   Or 999 - 103 = (n - 1) * 7   Or 896 = (n - 1) * 7   So, n - 1 = 128 or n = 129   Choice 129 is the correct answer.
<result>E</result>
How many 3 digit positive integers exist that when divided by 7 leave a remainder of 5?