First, you can use any 3 different prime numbers, so plug in using 3 simple prime numbers (one way to solve).
Let's use 3, 5, and 7. If we multiply these three together we get 105. Now, the first two are 1 and the number 105. You know 3, 5, and 7 must be factors because that's how you got 105. 105/3 = 35 105/5 = 21 and 105/7 = 15
In all, we have 8 factors that include 1, 3, 5, 7, 15, 21, 35, and 105. You can see this if you use a factor tree.
105
/ \
35 3
/ \
7 5
Any combination will be a factor of 105.
2007-07-17 07:20:13
·
answer #1
·
answered by aldorf17 2
·
0⤊
0⤋
This question is more simple than it looks. Let's say p, r and s equals to 3, 5 and 7 respectively. n would therefore be equal to 105. Now, try to find all the factors of 105. The factors are 1,3,5,7,15,21,35 and 105. Notice that the numbers 3, 5 and 7 come up again.
Therefore since we have three numbers (p, r and s) , we can come to the conclusion that there are six factors excluding 1 and n. If we include 1 and n, we have a total of eight factors.
2007-07-17 07:23:04
·
answer #2
·
answered by Anonymous
·
0⤊
0⤋
howdy! I took it right now to boot, yet i'm a junior :). i presumed it replaced into undesirable too; which evaluation e book did you utilize to attain your prepare examination? i did no longer take any prepare tests earlier I took the try yet I regarded on the Princeton evaluation scoring instruction manual, and consistent with which you will leave out fifteen and get a 800! i do no longer comprehend how precise that's in spite of the undeniable fact that. i'm uncertain what you will get, yet on the grounds which you skipped lots you will no longer get factors counted off for those .. possibly you scored interior the six hundred-630s. NO, you do no longer ought to deliver the scores!! :)
2017-01-21 07:05:50
·
answer #3
·
answered by ? 3
·
0⤊
0⤋
Let's just write them down:
The divisors of prs
are
1, p, r, s, pr, ps, rs, prs
or 8 in all.
2007-07-17 07:15:09
·
answer #4
·
answered by steiner1745 7
·
1⤊
0⤋