francoisph wrote:How many 3 digit positive integers exist that when divided by 7 leave a remainder of 5?
how to find quickly? that the smallest div by 7 and remainder 5 is 103
largest is 999
thks
The 3 digit integers range from 100 to 999. There are 900 of them.
900/7 = 128rem4. So, the answer must be either 128 or 129, depending on where 100 and 999 fall in the cycle of multiples of 7.
Here's where answer choices would be very useful (please always post the choices and the source of your question): if only one of those two appeared among the choices, we'd have our sure point and be finished.
Assuming that they both appear (which normally happens on the GMAT), we'd look at the biggest number under 100 divisible by 7: 98. So, the smallest three-digit number that will give a remainder of 5 is 103.
Since 103 is the 4th number in our sequence (starting at 100), and since we have a remainder of 4, we now know that "remainder 4" occurs 1 extra time, leading to our final answer:
128 + 1 = 129
