How many numbers of 7 consecutive positive integers are divisible by 6?
(1) Their average is divisible by 6
(2) Their median is divisible by 12
The OA is D.
Experts, I don't know how can I solve this DS question. I would appreciate your help. Thanks.
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How many numbers of 7 consecutive positive . . .
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ErikaPrepScholar
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Let's think about our list of 7 numbers like this:
x-3, x-2, x-1, x, x+1, x+2, x+3
Statement 1
The average here is
$$\frac{\left(x-3\right)\ +\ \left(x-2\right)+\left(x-1\right)+x+\left(x+1\right)+\left(x+2\right)+\left(x+3\right)}{7}=\ \frac{7x}{7}=x$$
So the average of all seven numbers is also the median (middle) number in the list. If the middle number is divisible by 6, and the largest and smallest numbers in the list are only 3 integers away from the middle number, none of the other numbers in the list are divisible by 6. So one number in the list is divisible by 6. Sufficient.
Statement 2
If a number is divisible by 12, it is also divisible by 6. So this statement tells us that the median number is divisible by 6. This is exactly what Statement 1 tells us, so this statement must be sufficient as well.
x-3, x-2, x-1, x, x+1, x+2, x+3
Statement 1
The average here is
$$\frac{\left(x-3\right)\ +\ \left(x-2\right)+\left(x-1\right)+x+\left(x+1\right)+\left(x+2\right)+\left(x+3\right)}{7}=\ \frac{7x}{7}=x$$
So the average of all seven numbers is also the median (middle) number in the list. If the middle number is divisible by 6, and the largest and smallest numbers in the list are only 3 integers away from the middle number, none of the other numbers in the list are divisible by 6. So one number in the list is divisible by 6. Sufficient.
Statement 2
If a number is divisible by 12, it is also divisible by 6. So this statement tells us that the median number is divisible by 6. This is exactly what Statement 1 tells us, so this statement must be sufficient as well.
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Hi M7MBA,
We're told that we have 7 CONSECUTIVE positive integers. We're asked how many of them are divisible by 6. To start, with 7 consecutive integers, only 1 or 2 of those integers will be divisible by 6. Once you have a number that is divisible by 6, the 'next' number that is divisible by 6 will either be '6 more' or '6 less' than the original number. For example: 0, 6, 12, 18, 24, etc.
1) Their average is divisible by 6.
With 7 consecutive integers, the AVERAGE will equal the "middle" number (in this case, the 4th number), so we can use a bit of 'brute force' to determine whether a pattern exists here or not.
IF... the numbers are...
3, 4, 5, 6, 7, 8, 9 .... there's 1 number that is divisible by 6.
9, 10, 11, 12, 13, 14, 15 .... there's 1 number that is divisible by 6.
15, 16, 17, 18, 19, 20, 21 .... there's 1 number that is divisible by 6.
Etc.
There will ALWAYS be just 1 number that is divisible by 6.
Fact 1 is SUFFICIENT
2) Their median is divisible by 12
With 7 consecutive integers, the AVERAGE = MEDIAN and both values will equal the "middle" number (in this case, the 4th number), so we can use a bit of 'brute force' here too. The difference is that the 4th value will have to be a multiple of 12.
IF... the numbers are....
9, 10, 11, 12, 13, 14, 15 .... there's 1 number that is divisible by 6.
21, 22, 23, 24, 25, 26, 27 .... there's 1 number that is divisible by 6.
Etc.
There will ALWAYS be just 1 number that is divisible by 6.
Fact 2 is SUFFICIENT
Final Answer: D
GMAT assassins aren't born, they're made,
Rich
We're told that we have 7 CONSECUTIVE positive integers. We're asked how many of them are divisible by 6. To start, with 7 consecutive integers, only 1 or 2 of those integers will be divisible by 6. Once you have a number that is divisible by 6, the 'next' number that is divisible by 6 will either be '6 more' or '6 less' than the original number. For example: 0, 6, 12, 18, 24, etc.
1) Their average is divisible by 6.
With 7 consecutive integers, the AVERAGE will equal the "middle" number (in this case, the 4th number), so we can use a bit of 'brute force' to determine whether a pattern exists here or not.
IF... the numbers are...
3, 4, 5, 6, 7, 8, 9 .... there's 1 number that is divisible by 6.
9, 10, 11, 12, 13, 14, 15 .... there's 1 number that is divisible by 6.
15, 16, 17, 18, 19, 20, 21 .... there's 1 number that is divisible by 6.
Etc.
There will ALWAYS be just 1 number that is divisible by 6.
Fact 1 is SUFFICIENT
2) Their median is divisible by 12
With 7 consecutive integers, the AVERAGE = MEDIAN and both values will equal the "middle" number (in this case, the 4th number), so we can use a bit of 'brute force' here too. The difference is that the 4th value will have to be a multiple of 12.
IF... the numbers are....
9, 10, 11, 12, 13, 14, 15 .... there's 1 number that is divisible by 6.
21, 22, 23, 24, 25, 26, 27 .... there's 1 number that is divisible by 6.
Etc.
There will ALWAYS be just 1 number that is divisible by 6.
Fact 2 is SUFFICIENT
Final Answer: D
GMAT assassins aren't born, they're made,
Rich
