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25 Sep 2023, 01:24
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C
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Difficulty:
95%
(hard)
Question Stats:
27% (02:21) correct
73%
(02:29)
wrong
based on 125
sessions
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If the units digit of a positive integer n is 2 and there are 4 factors of n that are greater than 1 and less than the square root of n, what is the total number of factors of n?
A. 8
B. 9
C. 10
D. 11
E. Cannot be determined
A. 8
B. 9
C. 10
D. 11
E. Cannot be determined
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26 Sep 2023, 00:15
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Bunuel
Factors exist in pairs: If \(p\) is a factor less than the \(\sqrt{n}\), then \(\frac{n}{p}\) is a corresponding factor greater or equal to the \(\sqrt{n}\).
\(\frac{n}{p} = p\) ➔ When n is a perfect square, for all other numbers \(\frac{n}{p}\) is greater than \(p\).
Applying this concept to this question. We know that "...there are 4 factors of n that are greater than 1 and less than the square root of n...", therefore the total number of factors can be (5*2) or (5*2)+1
- If n is a perfect square the total number of factors = 11
- If n is not a perfect square the total number of factors = 10.
As the number ends in 2, we can infer that the number is not a perfect square. Hence, the number of factors = 10.
Option C
General Discussion
25 Sep 2023, 22:38
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Bunuel
There are FIVE factors including 1 less than \(\sqrt{n}\).
Now, n ends in 2, so it cannot be a perfect square as squares end in 0,1,4,5,6, and 9.
For each factor less than \(\sqrt{n}\), there will be a factor greater than \(\sqrt{n}\)
Thus, 2*5 or 10.
