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OpenStudy (anonymous):
When 20 is divided by a positive integer n, the remainder is 2. How many possible values are there for n?
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OpenStudy (campbell_st):
well n has to be less than 20 and not a factor of 20. so that rules out 1, 2, 4, 5, 10 and 20
OpenStudy (anonymous):
a little confused still?
OpenStudy (anonymous):
i am confused
OpenStudy (campbell_st):
well if n is larger than 20 the remainder will always be 20....
e.g 20/21 remainder 20 20/35 r = 20....
if n is a factor there is no remainder
OpenStudy (anonymous):
we can try them all
20 divided by 18 leaves a remainder of 2
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OpenStudy (anonymous):
likewise 9
OpenStudy (anonymous):
also perhaps 6
OpenStudy (campbell_st):
likewise 3
OpenStudy (anonymous):
lol
OpenStudy (anonymous):
think we got them all?
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Directrix (directrix):
Let 20 = nk + 2 where k is an integer.
18= nk
18/n = k
So n would have to be a factor of 18.
Directrix (directrix):
n could be 1, 2, 3, 6, 9, 18 (total of 6) but should we exclude 18?
OpenStudy (anonymous):
Oh, I see where you guys are getting at. Thanks! :)
OpenStudy (anonymous):
20/n = x + 2/n = (nx + 2) /n
-> nx + 2 = 20
-> nx = 20 -2 = 18
The chance is: 1 * 18, 2 * 9, 3 * 6
Thus possible values of n = 1, 2, 3
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