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

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BTGmoderatorDC: Moderator; Posts: 7187; Joined: Thu Sep 07, 2017 4:43 pmFollowed by: 23 members

by BTGmoderatorDC » Fri Feb 15, 2019 6:21 pm

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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

OA A

Source: Veritas Prep

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GMATGuruNY: GMAT Instructor; Posts: 15539; Joined: Tue May 25, 2010 12:04 pmFollowed by: 1906 members; Location: New York, NY; Thanked: 13060 times; GMAT Score:790

by GMATGuruNY » Sat Feb 16, 2019 5:31 am

BTGmoderatorDC wrote: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

MULTIPLE OF X + MULTIPLE OF X = MULTIPLE OF X.
MULTIPLE OF X + NON-MULTIPLE OF X = NON-MULTIPLE OF X.

Since the prime-factorization of 15! includes 13, we get:
a = 15! + 13 = multiple of 13 + multiple of 13 = multiple of 13.
Since a is a multiple of 13, option I is a factor of a.
Eliminate any answer that does not include option I.
Eliminate E.

Since the prime-factorization of 15! includes 2*7=14, we get:
a = 15! + 13 = multiple of 14 + non-multiple of 14 = non-multiple of 14.
Since a is not a multiple of 14, option II is a NOT factor of a.
Eliminate any remaining answer that includes option II.
Eliminate B and D.

Since the prime-factorization of 15! includes 2*13=26, we get:
a = 15! + 13 = multiple of 26 + non-multiple of 26 = non-multiple of 26.
Since a is not a multiple of 26, option III is a NOT factor of a.
Eliminate any remaining answer that includes option III.
Eliminate C.

The correct answer is A.

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Brent@GMATPrepNow: GMAT Instructor; Posts: 16207; Joined: Mon Dec 08, 2008 6:26 pmFollowed by: 1268 members; Location: Vancouver, BC; Thanked: 5254 times; GMAT Score:770

by Brent@GMATPrepNow » Sat Feb 16, 2019 6:55 am

Here's a similar question to practice with: https://www.beatthegmat.com/divisibility-t111432.html

Cheers,
Brent

Brent Hanneson - Creator of GMATPrepNow.com

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Scott@TargetTestPrep: GMAT Instructor; Posts: 8157; Joined: Sat Apr 25, 2015 10:56 amFollowed by: 29 members; Location: Los Angeles, CA; Thanked: 43 times

by Scott@TargetTestPrep » Fri Feb 22, 2019 3:17 pm

BTGmoderatorDC wrote: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

OA A

Source: Veritas Prep

Let m, n, and p be positive integers. If m is divisible by k and n is divisible by k, then m + n is divisible by k. On the other hand, if m is divisible by k and n is NOT divisible by k, then m + n is NOT divisible by k.

Keeping the above fact in mind, let's analyze each Roman numeral choice.

I. 13

Since 15! is divisible by 13 and 13 is divisible by 13, then 15! + 13 is divisible by 13. In other words, 13 is a factor of 15! + 13.

II. 14

Since 15! is divisible by 14 and 13 is NOT divisible by 14, then 15! + 13 is NOT divisible by 14. In other words, 14 is NOT a factor of 15! + 13.

III. 26

Since 15! is divisible by 26 (15! has 2 and 13 as factors) and 13 is NOT divisible by 26, then 15! + 13 is NOT divisible by 26. In other words, 26 is NOT a factor of 15! + 13. Alternatively, we can also notice that 15! + 13 is odd (since 15! is even and 13 is odd) and 26 is even; thus 26 cannot be a factor of 15! + 13 since 15! + 13 has no factors of 2.

Answer: A

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deloitte247: Legendary Member; Posts: 2214; Joined: Fri Mar 02, 2018 2:22 pmFollowed by: 5 members

by deloitte247 » Sat Feb 23, 2019 5:15 am

$$a=15!+13$$
$$=13\left(\left(15\cdot14\cdot12!\right)+1\right)$$
For first option 13 ;
$$a=13\left(\frac{\left(15\cdot14\cdot12!\right)+1}{13}\right)=Odd\ number$$
Hence 13 is a factor of a,
$$Option\ I\ is\ correct$$

For second option 14 ;
$$a=13\left(\frac{\left(15\cdot14\cdot12!\right)+1}{14}\right)=even\ +\ remainder$$
Option II is wrong because a=odd and 14 is even ; even can't factor out odd without remainder.

For third option 26 ;
$$a=13\left(\frac{\left(15\cdot14\cdot12!\right)+1}{26}\right)=even+remainder$$
Option III is also wrong because a=odd and 26=even
even can factor out odd without remainder.
$$answer\ is\ Option\ A;\ I\ only\ $$

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