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When positive integer n is divided by 13, the remainder is 2

Expert replies
by BTGmoderatorAT » Sat Dec 30, 2017 6:30 pm
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

OA: B

I'm confused how to set up the formulas here. Can any experts help?
Join the discussion
Source: — Problem Solving |

by GMATGuruNY » Sat Dec 30, 2017 10:53 pm
A quick lesson on remainders:

When x is divided by 5, the remainder is 3.
In other words, x is 3 more than a multiple of 5:
x = 5a + 3.

When x is divided by 7, the remainder is 4.
In other words, x is 4 more than a multiple of 7:
x = 7b + 4.

Combined, the statements above imply that when x is divided by 35 -- the LOWEST COMMON MULTIPLE OF 5 AND 7 -- there will be a constant remainder R.
Put another way, x is R more than a multiple of 35:
x = 35c + R.

To determine the value of R:
Make a list of values that satisfy the first statement:
When x is divided by 5, the remainder is 3.
x = 5a + 3 = 3, 8, 13, 18...
Make a list of values that satisfy the second statement:
When x is divided by 7, the remainder is 4.
x = 7b + 4 = 4, 11, 18...
The value of R is the SMALLEST VALUE COMMON TO BOTH LISTS:
R = 18.

Putting it all together:
x = 35c + 18.

Another example:
When x is divided by 3, the remainder is 1.
x = 3a + 1 = 1, 4, 7, 10, 13...
When x is divided by 11, the remainder is 2.
x = 11b + 2 = 2, 13...

Thus, when x is divided by 33 -- the LCM of 3 and 11 -- the remainder will be 13 (the smallest value common to both lists).
x = 33c + 13 = 13, 46, 79...

Onto the problem at hand:
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
When n is divided by 13, the remainder is 2.
n = 13a + 2 = 2, 15, 28, 41, 54, 67, 80, 93...
When n is divided by 8, the remainder is 5.
n = 8b + 5 = 5, 13, 21, 29, 37, 45. 53, 61, 69, 77, 85, 93...

Thus, when n is divided by 104 -- the LCM of 13 and 8 -- the remainder will be 93 (the smallest value common to both lists).
n = 104c + 93 = 93, 197...
Of the list of options for n, only the first -- 93 -- is less than 180.

The correct answer is B.
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by Scott@TargetTestPrep » Sun Aug 18, 2019 6:22 pm
BTGmoderatorAT 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

OA: B

I'm confused how to set up the formulas here. Can any experts help?
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

Scott Woodbury-Stewart
Founder and CEO
[email protected]

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by Scott@TargetTestPrep » Sun Aug 18, 2019 6:22 pm
BTGmoderatorAT 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

OA: B

I'm confused how to set up the formulas here. Can any experts help?
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

Scott Woodbury-Stewart
Founder and CEO
[email protected]

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

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