# Math problem

<p>How many positive integers less than 100 have a remainder of 3 when divided by 7?</p>

<p>I got 7. The Barrons book says 14 - I'm not sure how this is possible... maybe I am missing something? Please explain.</p>

<p>Any number with a remainder of 3 when divided by 7 is of the form 7k + 3. How many values of k can you have before the expression hits 100? </p>

<p>Set up the inequality: </p>

<p>7k + 3 < 100
7k < 97
k < 13.85</p>

<p>Since we discard fractional values, we may assume this is equivalent to saying
k <= 13; There are 13 nonzero integers plus zero that satisfy the inequality giving us *14 *.</p>

<p>^Wow, that will help me a lot since there always seems to be at least one of these types of questions, and I always end up spending up much more time than needed to solve them.</p>

<p>It's always useful to think in more general terms than to look for specific examples. If I had been trained to think that way back in my SAT days, I'd have finished the sections much faster :P</p>

<p>I got 13, not through arachnotron's method, but the same logic. How many times does 7 go into 100? 14. 7x14=98. 98+3=101, so that doesn't work. So I would say 13. Why would you include 0, Arachnotron? 0 is not a positive integer, nor does it have a remainder of 3 when divided by 7. Help me out here?</p>

<p>I think anachron meant 0 for the k in his equation. So the actual number would be 3 ([7* 0]+ 3).</p>

<p>I dont see how the answer can be 14. </p>

<p>If we use Arochnotrons equation where k<13, and the question asks for POSTIVE INTEGERS, 0 can not be included.</p>

<p>If u went the long method and logically worked it out, you would have arrived at 13 +ve numbers. </p>