How many 3 digit positive integers exist that when divided by 7 leave a remainder of 5?
A) 128
B) 129
C) 141
D) 142
E) 143
A) 128
B) 129
C) 141
D) 142
E) 143
The number closest to 100 that is divisible by 7 is 98. So if 5 is added to 98 and the resultant number is divided by 7 the remainder would obviously be 5. So 98+5=103 is the first 3 digit number which when divided by 7 leaves a remainder of 5. Not the greatest 3 digit number divisible by 7 is 999
Now, all the numbers which when divided by 7 leave a remainder of 5 are an evenly spaced set with an increment of 7 i.e. the difference between any two consecutive numbers in this set is 7.
For and evenly spaced set,
The number of elements=[(Biggest-Smallest)/Increment]+1=(999-103)/7+1=(896/7)+1=128+1=129
Hence B
