Divisibility/Remainder questions w/ solutions... : GMAT Quantitative Section
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# Divisibility/Remainder questions w/ solutions...

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Divisibility/Remainder questions w/ solutions... [#permalink]

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03 Dec 2010, 16:18
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Hey all...found some nice divisibility questions that will give you a bit of a workout:
Source: http://www.analyzemath.com/numbers/divisibility_questions.html

Keep in mind:
$$n = q m + r$$

Where $$n$$ is the dividend, $$q$$ is the quotient, $$m$$ is the divisor and $$r$$ is the remainder.

Good luck.

Question 1:
If a positive integer n is divided by 5, the remainder is 3. Which of the numbers below yields a remainder of 0 when it is divided by 5?
A) n + 3
B) n + 2
C) n - 1
D) n - 2
E) n + 1
[Reveal] Spoiler: Solution 1
n divided by 5 yields a remainder equal to 3 is written as follows
n = 5 k + 3 , where k is an integer.
add 2 to both sides of the above equation to obtain
n + 2 = 5 k + 5 = 5(k + 1)
The above suggests that n + 2 divided by 5 yields a remainder equal to zero. The answer is B.

Question 2:
If an integer n is divisible by 3, 5 and 12, what is the next larger integer divisible by all these numbers?
A) n + 3
B) n + 5
C) n + 12
D) n + 60
E) n + 15
[Reveal] Spoiler: Solution 2
If n is divisible by 3, 5 and 12 it must a multiple of the lcm of 3, 5 and 12 which is 60.
n = 60 k
n + 60 is also divisible by 60 since
n + 60 = 60 k + 60 = 60(k + 1)
The answer is D.

Question 3:
What is the smallest integer that is multiple of 5, 7 and 20?
A) 70
B) 35
C) 200
D) 280
E) 140
[Reveal] Spoiler: Solution 3
It is the lcm of 5, 7 and 20 which is 140.
The answer is E.

Question 4:
When the integer n is divided by 8, the remainder is 3. What is the remainder if 6n is divided by 8?
A) 0
B) 1
C) 2
D) 3
E) 4
[Reveal] Spoiler: Solution 4
When n is divided by 8, the remainder is 3 may be written as
n = 8 k + 3
multiply all terms by 6
6 n = 6(8 k + 3) = 8(6k) + 18
Write 18 as 16 + 2 since 16 = 8 * 2.
= 8(6k) + 16 + 2
Factor 8 out.
= 8(6k + 2) + 2
The above indicates that if 6n is divided by 8, the remainder is 2. The answer is C.

Question 5:
If n is an integer, when (2n + 2)^2 is divided by 4 the remainder is
A) 0
B) 1
C) 2
D) 3
E) 4
[Reveal] Spoiler: Solution 5
We first expand (2n + 2)2
(2n + 2)^2 = 4n^2 + 8n + 4
Factor 4 out.
= 4(n^2 + 2n + 1)
(2n + 2)2 is divisible by 4 and the remainder is equal to 0. The answer is A.

Question 6:
What is the smallest positive 2-digit whole number divisible by 3 and such that the sum of its digits is 9?
A) 27
B) 33
C) 72
D) 18
E) 90
[Reveal] Spoiler: Solution 6
Let xy be the whole number with x and y the two digits that make up the number. The number is divisible by 3 may be written as follows
10 x + y = 3 k
The sum of x and y is equal to 9.
x + y = 9
Solve the above equation for y
y = 9 - x Substitute y = 9 - x in the equation 10 x + y = 3 k to obtain.
10 x + 9 - x = 3 k
Solve for x
x = (k - 3) / 3
x is a positive integer smaller than 10
Let k = 1, 2, 3, ... and select the first value that gives x as an integer. k = 6 gives x = 1
Find y using the equation y = 9 - x = 8
The number we are looking for is 18 and the answer is D. It is divisible by 3 and the sum of its digits is equal to 9 and it is the smallest and positive whole number with such properties.

Question 7:
Which of these numbers is not divisible by 3?
A) 339
B) 342
C) 552
D) 1111
E) 672
[Reveal] Spoiler: Solution 7
One may answer this question using a calculator and test for divisibility by 3. However we can also test for divisibilty by adding the digits and if the result is divisible by3 then the number is divisible by 3.
3 + 3 + 9 = 15 , divisible by 3.
3 + 4 + 2 = 9 , divisible by 3.
5 + 5 + 2 = 12 , divisible by 3.
1 + 1 + 1 + 1 = 4 , not divisible by 3.
The number 1111 is not divisible by 3 the answer is D.

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Last edited by martie11 on 21 Feb 2011, 06:45, edited 1 time in total.
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Re: Divisibility/Remainder questions... [#permalink]

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16 Dec 2010, 07:35
thanks for the questions and explanations.. On Qs 4,5,6 I took the "plug numbers" approach..
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Re: Divisibility/Remainder questions... [#permalink]

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16 Dec 2010, 22:46
Yeah, plugin method will save time especially with Q6.
Re: Divisibility/Remainder questions...   [#permalink] 16 Dec 2010, 22:46
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# Divisibility/Remainder questions w/ solutions...

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