When positive integer n is divided by 13

This topic has expert replies
Moderator
Posts: 2058
Joined: Sun Oct 29, 2017 4:24 am
Thanked: 1 times
Followed by:5 members

When positive integer n is divided by 13

by M7MBA » Sat Apr 28, 2018 1:20 am

Timer

00:00

Your Answer

A

B

C

D

E

Global Stats

When positive integer n is divided by 13, the remainder is 2. When n is divided by 8, the remainder is 5. How many such values are less than 180?

(A) 0
(B) 1
(C) 2
(D) 3
(E) 4

The OA is B.

How can I solve this PS question? I am confused. Help, please.

Legendary Member
Posts: 2898
Joined: Thu Sep 07, 2017 2:49 pm
Thanked: 6 times
Followed by:5 members

Answer

by Vincen » Sat Apr 28, 2018 5:15 am
Hello.

This is the way I solved it. I don't know if it is the best way, but it works. <i class="em em-stuck_out_tongue_winking_eye"></i>

When positive integer n is divided by 13, the remainder is 2 implies that $$n=13\cdot a+2\ ,\ where\ a\in\mathbb{N}.$$ When n is divided by 8, the remainder is 5 implies that $$n=8\cdot b+5\ ,\ where\ b\in\mathbb{N}.$$ Now, let's plug values for a: $$If\ \ a=0\ \ \ then\ \ \ n=2.\ \ Therefore\ b=-\frac{3}{8}\notin\mathbb{N}.\ \ NOT\ POSSIBLE.$$ $$If\ \ a=1\ \ \ then\ \ \ n=15.\ \ Therefore\ b=\frac{10}{8}\notin\mathbb{N}.\ \ NOT\ POSSIBLE.$$ $$If\ \ a=2\ \ \ then\ \ \ n=28.\ \ Therefore\ b=\frac{23}{8}\notin\mathbb{N}.\ \ NOT\ POSSIBLE.$$ $$If\ \ a=3\ \ \ then\ \ \ n=41.\ \ Therefore\ b=\frac{36}{8}\notin\mathbb{N}.\ \ NOT\ POSSIBLE.$$ $$If\ \ a=4\ \ \ then\ \ \ n=54.\ \ Therefore\ b=\frac{49}{8}\notin\mathbb{N}.\ \ NOT\ POSSIBLE.$$ $$If\ \ a=5\ \ \ then\ \ \ n=67.\ \ Therefore\ b=\frac{62}{8}\notin\mathbb{N}.\ \ NOT\ POSSIBLE.$$ $$If\ \ a=6\ \ \ then\ \ \ n=80.\ \ Therefore\ b=\frac{75}{8}\notin\mathbb{N}.\ \ NOT\ POSSIBLE.$$ $$If\ \ a=7\ \ \ then\ \ \ n=93.\ \ Therefore\ b=\frac{88}{8}=11\in\mathbb{N}.\ \ POSSIBLE.$$ $$If\ \ a=8\ \ \ then\ \ \ n=106.\ \ Therefore\ b=\frac{101}{8}\notin\mathbb{N}.\ \ NOT\ POSSIBLE.$$ $$If\ \ a=9\ \ \ then\ \ \ n=119.\ \ Therefore\ b=\frac{114}{8}\notin\mathbb{N}.\ \ NOT\ POSSIBLE.$$ We can continue like this and the next value that is possible is n=197.

Therefore, the unique value that holds the condition and is less than 180 is n=93.

The correct answer is [spoiler]B=1[/spoiler].

I hope it helps.

GMAT/MBA Expert

User avatar
GMAT Instructor
Posts: 1462
Joined: Thu Apr 09, 2015 9:34 am
Location: New York, NY
Thanked: 39 times
Followed by:22 members

by Jeff@TargetTestPrep » Wed May 02, 2018 9:33 am
M7MBA wrote:When positive integer n is divided by 13, the remainder is 2. When n is divided by 8, the remainder is 5. How many such values are less than 180?

(A) 0
(B) 1
(C) 2
(D) 3
(E) 4
We can create the equation:

n = 13Q + 2

So n can be 2, 15, 28, 41, 54, 67, 80, 93, ...

and

n = 8Q + 5

So n can be 5, 13, 21, 29, 37, 45, 53, 61, 69, 77, 85, 93, ...

We see that the first number that satisfies both conditions is 93. To find the other numbers, we can keep adding the LCM of 13 and 8, which is 13 x 8 = 104. Therefore, the next value that satisfies both conditions is 93 + 104 = 197. However, 197 is already greater than 180, so we only have one value, namely 93, that is less than 180 and satisfies both conditions.

Answer: B

Jeffrey Miller
Head of GMAT Instruction
[email protected]

Image

See why Target Test Prep is rated 5 out of 5 stars on BEAT the GMAT. Read our reviews

Legendary Member
Posts: 2248
Joined: Sun Oct 29, 2017 2:04 pm
Followed by:6 members

by swerve » Wed May 02, 2018 10:50 am
n=13Q1+2
n=8Q2+5

13Q1+2=8Q2+5
13Q1=8Q2+3

When we list the multiples of 13, only when n=91 can we have a remainder of 3 when n is divided by 8.
OR
We will get a multiple of 13 only when Q2=11. Option B.

Regards!