How many 3 digit positive integers exist that when divided by 7 leave a remainder of 5?
Options.
A) 128 B) 142 C) 143 D) 141 E) 129

Rationale: "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 e is the correct answer ."
Answer: E