Problem Solving

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.

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.

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!