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On a weekend, 6 college friends went skiing and evenly split the cost

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Joined: 02 Sep 2009
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On a weekend, 6 college friends went skiing and evenly split the cost  [#permalink]

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13 Apr 2016, 04:30
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On a weekend, 6 college friends went skiing and evenly split the cost of their cabin. If the amount paid by each person was an integer, which of the following could not have been the cost of the cabin?

A. \$126
B. \$156
C. \$218
D. \$648
E. \$1,728

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Joined: 02 Aug 2009
Posts: 8638
On a weekend, 6 college friends went skiing and evenly split the cost  [#permalink]

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13 Apr 2016, 19:49
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Bunuel wrote:
On a weekend, 6 college friends went skiing and evenly split the cost of their cabin. If the amount paid by each person was an integer, which of the following could not have been the cost of the cabin?

A. \$126
B. \$156
C. \$218
D. \$648
E. \$1,728

Evenly distributing within 6 friends MEANS the Total should be a multiple of 6..

All choices are EVEN, so lets check for Div by 3..

A. \$126 ... 1+2+6 = 9... YES
B. \$156 ... 1+5+6 = 12... YES
C. \$218 ... 2+1+8 = 11... NO ... ANSWER
D. \$648 ... 6+4+8 = 18... YES
E. \$1,728... 1+7+2+8 = 18... YES

ans C
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Re: On a weekend, 6 college friends went skiing and evenly split the cost  [#permalink]

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15 Apr 2020, 09:02
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Top Contributor
Bunuel wrote:
On a weekend, 6 college friends went skiing and evenly split the cost of their cabin. If the amount paid by each person was an integer, which of the following could not have been the cost of the cabin?

A. \$126
B. \$156
C. \$218
D. \$648
E. \$1,728

If the total cost is divisible by 6, then each of the 6 friends will pay an integer amount.

Take answer choice A (\$126) for example.
126 is divisible by 6
\$126/6 = \$31
So, each person pays an integer amount (\$31 each)
ELIMINATE A

The question is basically asking us to find the answer choice that is NOT divisible by 6

CONCEPT #1: A number is divisible by 6 if that number is divisible by 2 AND by 3
We can see that all 5 answer choices are divisible by 2.

CONCEPT #2: A number is divisible by 3 if the SUM OF THE DIGITS of that number is divisible by 3
Take answer choice B (\$156) for example.
The sum of the digits = 1 + 5 + 6 = 12
Since 12 is divisible by 3, we can conclude that 156 is divisible by 3
ELIMINATE B

The sum of the digits = 2 + 1 + 8 = 11
Since 11 is NOT divisible by 3, we can conclude that 218 is NOT divisible by 3
If 218 is NOT divisible by 3, then 218 is also NOT divisible by 6

Cheers,
Brent
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Re: On a weekend, 6 college friends went skiing and evenly split the cost  [#permalink]

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10 May 2016, 12:28
Condition states that amount paid by each of the 6 persons for the cost of the cabin is an integer amount
Clearly we have to check for divisibility by 6 in all five answer choices
126 = 6 * 21 Yes so eliminate
156 = 6 * 26 Yes so eliminate
218 = 6 * 36.3333 clearly not divisible so correct answer
648 = 6 * 108 Yes so eliminate
1728 = 6 * 288 Yes so eliminate
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Re: On a weekend, 6 college friends went skiing and evenly split the cost  [#permalink]

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17 Apr 2020, 03:28
Bunuel wrote:
On a weekend, 6 college friends went skiing and evenly split the cost of their cabin. If the amount paid by each person was an integer, which of the following could not have been the cost of the cabin?

A. \$126
B. \$156
C. \$218
D. \$648
E. \$1,728

We are looking for a number that is not divisible by 6, i.e., a number not divisible by both 2 and 3. Since all the given numbers are even, they are divisible by 2. So we are looking for a number that is not divisible by 3.

Recall that a number is divisible by 3 if the sum of its digits is divisible by 3. Looking at the choices, we see that the sum of the digits of 218 is 2 + 1 + 8 = 11, which is not divisible by 3. So 218 is not divisible by 6, either.

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Re: On a weekend, 6 college friends went skiing and evenly split the cost   [#permalink] 17 Apr 2020, 03:28