<p>Blue book Math question Practice test 3, section 2, question 20, page 523</p>
<p>20) When 15 is divided by the positive integer k, the remainder is 3. For how many different values of k is this true?
a) one
b) two
c) three
d) four
e) five</p>
<p>(answer is c)</p>
<p>Can you guys explain how to get the answer in detail please. Thanks.</p>
<p>Because the remainder is 3, you know the number has to be a factor of 12 so that 3 comes out as a remainder. So all the factors of 12 are 1,2,3,4,6,12 and you can test those accordingly. 1,2 and 3 you can eliminate instantly. It turns out that when 15 is divided by 4,6, and 12, the remainder is 3. So the answer is C.</p>
<p>Try all numbers 1-15 (obviously not 1 or 15 or the factors of 15). That leaves 2, 4, 6, 7, 8, 9, 10, 11, 12, 13, and 14. Using common sense, rule out 2 and 14. This method will probably work if you're fast at using your calculator :)</p>
<p>spazattack,
ur method is a bit long winded. I usually try tht very same method as well though:) But the thing at the end of the exam, we are short on time and if anyone can come up with a shorter way of doin this math, it will be helpful.</p>
<p>yep. k should fulfill these two requirements:
is a factor of 15-3=12
larger than the remainder 3
so k=4,6,12</p>
<p>DanQ, tht's a fast technique. Thanks for sharing!</p>