gettinit wrote:
when 777 is divided by positive integer n, the remainder is 77. How many possibilities are there for n?
a. 2
b. 3
c. 4
d. 5
e. 6
Let me your thought process on this. I know we have to put this in remainder format of n(y) + 77= 777 so n(y)=700 but from here I would find the number of factors for 700 to be 18. Is there a quicker way of figuring out the number of factors greater than 77?
Positive integer
a divided by positive integer
d yields a reminder of
r can always be expressed as
a=qd+r, where
q is called a quotient and
r is called a remainder, note here that
0\leq{r}<d (remainder is non-negative integer and always less than divisor).
So we'd have:
777=qn+77, where
remainder=77<n=divisor -->
qn=700=2^2*5^2*7 --> as
n must be more than 77 then
n could take only 5 values: 100, 140, 175, 350, and 700 (the least factor more than 77 is
2^2*5^2=100, now if you
substitute 2, 2^2, and 5 by 7 you'll get the factors more than 100 plus if you
include 7 you'll get one more factor, 700 itself, so total 1+3+1=5).
Answer: D.
Discussed here:
remainder-101074.html?hilit=possibilities#p782422Hope it's clear.
THanks for the reference Bunuel. So I am guessing its best to write out the factors in these types of questions? I didn't quite understand your methodology of replacing 2, 2^2, and 5 by 7 etc. I understand how it is applied here but if this were a different question I'd be lost. I think the key is just to mulitply the different factors to arrive at numbers larger than 77.
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