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# If 2a – b = 3c, where a, b, and c are non-zero integers

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If 2a – b = 3c, where a, b, and c are non-zero integers  [#permalink]

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Updated on: 09 Nov 2012, 02:25
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Question Stats:

66% (02:27) correct 34% (02:57) wrong based on 343 sessions

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If 2a – b = 3c, where a, b, and c are non-zero integers, which of the following could be the average (arithmetic mean) of a and b, if the average must itself be an integer?

A. -2
B. -1
C. 1
D. 10
E. 12

Originally posted by zisis on 04 Sep 2010, 11:57.
Last edited by Bunuel on 09 Nov 2012, 02:25, edited 1 time in total.
Renamed the topic and edited the question.
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Re: Puzzling question  [#permalink]

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04 Sep 2010, 12:25
15
6
zisis wrote:
If 2a – b = 3c, where a, b, and c are integers, which of the following could be the average (arithmetic mean) of a and b, if the average must itself be an integer?

Choices
A -2
B -1
C 1
D 10
E 12

spent a fair amount of time on it and m stuck....please help

$$2a-b=3c$$ --> $$b=2a-3c$$.

Now, $$average(a,b)=\frac{a+b}{2}=\frac{a+(2a-3c)}{2}=3*\frac{a-c}{2}=integer$$, so average is a multiple of 3 (as given that average and all unknowns are integers). Only multiple of 3 among answers is 12.

Hope it's clear.
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Re: Puzzling question  [#permalink]

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04 Sep 2010, 14:40
Hi Bunuel,

I have a different approach to this question, Please let me know if It is correct...

Lets say Average of a & b is c

a+b/2=c
a+b= 2c

2a-b=3c
a+b=2c
--------
3a = 5c

a=5c/3

we can plug in all answer choices for average - " c " and see which one gives us an integer as "a" is also an integer.

5(-2)/3 = -10/3
5(-1)/3 = -5/3
5(1)/3 = 5/3
5(10)/3 = 50/3
5(12)/3 = 60/3= 20 ( Hence E- 12 is the correct answer)
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Re: Puzzling question  [#permalink]

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06 Sep 2010, 05:47
I also got E by using substitution approach. However Bunuel suggested a good approach. Thanks
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Re: Puzzling question  [#permalink]

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08 Sep 2010, 09:22
1
1
My solution:

Code:
2a - b = 3c
2a + 2b - 2b - b = 3c
2(a + b) = 3(b + c)
(a + b)/2 = average = (3/4) (b + c)

For average to be an integer, it must be a multiple of 12. Hence E
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Re: try this one  [#permalink]

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06 Feb 2011, 15:05
Merging similar topics.

bhandariavi wrote:
If 2a – b = 3c, where a, b, and c are non-zero integers, which of the following could be the average (arithmetic mean) of a and b, if the average must itself be an integer?
A) 12 B) 3 c) -1 d) 1 E) 4

For the question above there are two correct answer A (for example a=9, b=15, c=1) and E (for example a=3, b=3, c=1), as both are multiple of 3.
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Re: Puzzling question  [#permalink]

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19 May 2011, 01:04
2a-b/3 = c integer.

a+b = 2 * average

average = 12
a+b = 24

3(a-8)/3 = integer.

hence E
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Re: Puzzling question  [#permalink]

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19 May 2011, 21:11
E through substitution though its time consuming. Bunuel has the best approach
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Re: If 2a – b = 3c, where a, b, and c are non-zero integers, whi  [#permalink]

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08 Nov 2012, 23:48
2013gmat wrote:
If 2a – b = 3c, where a, b, and c are non-zero integers, which of the following could be the average (arithmetic mean) of a and b, if the average must itself be an integer?
A=-2
B=-1
C=1
D=10
E=12

checking options....

a+b = 24/20/2/-2/-4

-b = a - 24/20/2/-2/-4

Now, 2a - b = 3 c

only 24 gives us a multiple of 3 that can be taken out.

so E.
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Re: If 2a – b = 3c, where a, b, and c are non-zero integers  [#permalink]

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04 Jun 2013, 21:12
2
zisis wrote:
If 2a – b = 3c, where a, b, and c are non-zero integers, which of the following could be the average (arithmetic mean) of a and b, if the average must itself be an integer?

A. -2
B. -1
C. 1
D. 10
E. 12

Responding to a pm:

Bunuel has already given the algebraic approach which is quite simple and clear. I am guessing that since you are looking for another approach, you want me to solve it without using algebra.

I instinctively jumped to the options in this question. They are the average of a and b so sum of a and b will be twice of the average so depending on which option we pick, the sum (a + b) will be -4 or -2 or 2 or 20 or 24. Note that a and b will be either both even or both odd since their sum must be even.

The question says, which of the following could be the average i.e. there are probably many numbers that could be the average and one of them in included in this list.

We also know that 2a – b = 3c
Since right hand side has a 3, we know that 2a - b is divisible by 3. The easiest way to make it divisible by 3 is to make both a and b divisible by 3 which makes their sum divisible by 3 as well. Of the given options, only 24 is divisible by 3 hence (E) must be the answer.
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Re: If 2a – b = 3c, where a, b, and c are non-zero integers  [#permalink]

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22 Dec 2013, 18:59
1
zisis wrote:
If 2a – b = 3c, where a, b, and c are non-zero integers, which of the following could be the average (arithmetic mean) of a and b, if the average must itself be an integer?

A. -2
B. -1
C. 1
D. 10
E. 12

My way of solving

2a = 3c + b

a = (3c + b)/2

So average (a + b)/2

We replace a and we get 3c+b/2, finally we get 6c + 3b = 3(2c+b) as the average

So average must be a multiple of 3 and higher than 9 since numbers must be non zero integers

Therefore only 12 fits the bill

Hope it helps!
Let me see that Kudos rain!!!

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Re: If 2a – b = 3c, where a, b, and c are non-zero integers  [#permalink]

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02 May 2014, 09:44
1
Here's actually another way of solving. First since all integers are >=0 it will be impossible to get a negative number so A,B are out right off the bat

Now, we have that 2a - b = 3c and we need to find a+b / 2

So let's begin with answer choice E

We have that a+b = 24

2a - b = 3c
a + b = 24

We have 3a = 24 + 3c

Now 24 + 3c is always a multiple of 3 so this one stays

Let's try with D

2a - b = 3c
a+b = 10

3a = 3c+10

Now 3c+10 won't ever be a multiple of 3 so OUT

One final try, C

2a -b = 3c
a+b = 2

3a = 3c+2

Again, 3c + 2 will NEVER be a multiple of 3.

Therefore only answer choice that is valid is E

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Re: If 2a – b = 3c, where a, b, and c are non-zero integers  [#permalink]

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19 Jul 2014, 00:19
Let N be the average
so a+b = 2N
2a -b = 3c
Adding both the equation
3a = 2N + 3c

As a,N,c are all integers, N has to be multiple of 3 to have the equation balanced. Hence 12 - Option E)
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Re: If 2a – b = 3c, where a, b, and c are non-zero integers  [#permalink]

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19 Jul 2015, 07:32
$$2a - b = 3c$$

$$2a - b - 3c = 0$$

$$3a + 3b + 3c = a + 4b + 6c$$

$$\frac{(a+b+c)}{3} = a + 4b + 6c$$

$$(a+b+c) = 3(a + 4b + 6c)$$

So, the average is a multiple of 3, hence Ans (E).
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Re: If 2a – b = 3c, where a, b, and c are non-zero integers  [#permalink]

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19 Jul 2015, 07:51
zisis wrote:
If 2a – b = 3c, where a, b, and c are non-zero integers, which of the following could be the average (arithmetic mean) of a and b, if the average must itself be an integer?

A. -2
B. -1
C. 1
D. 10
E. 12

We can always solve such questions by taking certain values for a, b and c keeping in mind that if average of a and b must be integers then both a and b must be either even or both must be odd
Also 2a - b must be a multiple of 3

Trying with Even Numbers first

Let, a=8, b=4 i.e. 2a-b = 16-4 = 12 = 3c i.e. c=4 i.e. Average of a and b = (8+4)/2 = 6

Obtained Average Relates with 12 so doubling every number in previous step

i.e. Let, a=16, b=8 i.e. 2a-b = 32-8 = 24 = 3c i.e. c=8 i.e. Average of a and b = (16+8)/2 = 12 SUCCESS

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Re: If 2a – b = 3c, where a, b, and c are non-zero integers  [#permalink]

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26 Feb 2018, 11:19
zisis wrote:
If 2a – b = 3c, where a, b, and c are non-zero integers, which of the following could be the average (arithmetic mean) of a and b, if the average must itself be an integer?

A. -2
B. -1
C. 1
D. 10
E. 12

We can manipulate the first equation to read:

2a - 3c = b

Now let’s set up an expression for the average of a and b:

(a + b)/2 = ?

(a + 2a - 3c)/2 = ?

(3a - 3c)/2 = ?

3(a - c)/2 = ?

Since 3 is not divisible by 2, so a - c must be divisible by 2. Therefore, since (a - c)/2 must be an integer and 3(a - c)/2 must be a multiple of 3. The only multiple of 3 in the answer choices is E, 12; thus, E is the correct answer.

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Re: If 2a – b = 3c, where a, b, and c are non-zero integers  [#permalink]

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13 Oct 2019, 17:46
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Re: If 2a – b = 3c, where a, b, and c are non-zero integers   [#permalink] 13 Oct 2019, 17:46
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