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Bunuel
If a = 15! + 13, which of the following are factors of a?

I. 13
II. 14
III. 26

A. I only
B. I and II only
C. I and III only
D. I, II, and III only
E. None of the above

\(a = 15! + 13 = 13 * (1*2*3..12*14*15 + 1)\)

Note the two factors of a - "13" and "1 more than a number that has all other primes till 15"
Hence, this second factor will have none of these primes (2, 3, 5, 7, 11)

I. 13
13 is a factor of a.

II. 14
2 and 7 are not factors of a.

III. 26
2 is not a factor of a.
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KarishmaB
Bunuel
If a = 15! + 13, which of the following are factors of a?

I. 13
II. 14
III. 26

A. I only
B. I and II only
C. I and III only
D. I, II, and III only
E. None of the above

\(a = 15! + 13 = 13 * (1*2*3..12*14*15 + 1)\)

Note the two factors of a - "13" and "1 more than a number that has all other primes till 15"
Hence, this second factor will have none of these primes (2, 3, 5, 7, 11)

I. 13
13 is a factor of a.

II. 14
2 and 7 are not factors of a.

III. 26
2 is not a factor of a.

Check:
https://www.gmatclub.com/forum/veritas- ... c-or-math/
https://www.gmatclub.com/forum/veritas- ... h-part-ii/

Karishma in a simple way, we can also look at this problem. Because 15! will have 13 as a factor and N= 15!+13 so on the number line it will be a multiple of 13 + 13,, N+13 so the number will be only divisible by 13 and no other integer in options
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Ved22
KarishmaB
Bunuel
If a = 15! + 13, which of the following are factors of a?

I. 13
II. 14
III. 26

A. I only
B. I and II only
C. I and III only
D. I, II, and III only
E. None of the above

\(a = 15! + 13 = 13 * (1*2*3..12*14*15 + 1)\)

Note the two factors of a - "13" and "1 more than a number that has all other primes till 15"
Hence, this second factor will have none of these primes (2, 3, 5, 7, 11)

I. 13
13 is a factor of a.

II. 14
2 and 7 are not factors of a.

III. 26
2 is not a factor of a.

Check:
https://www.gmatclub.com/forum/veritas- ... c-or-math/
https://www.gmatclub.com/forum/veritas- ... h-part-ii/

Karishma in a simple way, we can also look at this problem. Because 15! will have 13 as a factor and N= 15!+13 so on the number line it will be a multiple of 13 + 13,, N+13 so the number will be only divisible by 13 and no other integer in options

Yes, it is easy to say why the number will be a multiple of 13. What does need to be noted though is why the number will not be a multiple of 26 which is just 2*13. So we need to notice why 2 is not a factor of a.
To do that we have multiple ways - Either notice that 15! will be even and if we add 13 to it, even + odd = odd so 2 will not be a factor.
Or note that here 13* (1*2...*12*14*15 + 1), the underlined will have 2 as a factor so (1*2...*12*14*15 + 1) will not.

P.S. - Request you to tag me if you would like me to give some input. I am unable to see all the replies.
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