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B
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Difficulty:
35%
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Question Stats:
49% (00:46) correct
51%
(00:50)
wrong
based on 65
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S is a list of consecutive integers. Does S contain a multiple of 7?
(1) 8 and 13 are in set S.
(2) Set S has 8 terms.
(1) 8 and 13 are in set S.
(2) Set S has 8 terms.
The official answer is B.
But I can think of an option where you have 8 terms and it does not contain a multiple of 7.
For example:
-1,0,1,2,3,4,5,6
I thought it was C.
But I can think of an option where you have 8 terms and it does not contain a multiple of 7.
For example:
-1,0,1,2,3,4,5,6
I thought it was C.
15 Oct 2023, 01:46
Kudos
Bookmarks
S is a list of consecutive integers. Does S contain a multiple of 7?
(1) 8 and 13 are in set S.
This does not specify the full range of the set, only a part of it. There could be more numbers before 8 or after 13 in the set S, and without a defined range, we cannot be sure whether or not a multiple of 7 is included. Not sufficient.
(2) Set S has 8 terms.
Given that there are 8 consecutive integers in S, at least one of those integers will be a multiple of 7. This is because the longest stretch of numbers you can have without a multiple of 7 is 6 (after which the next number will be a multiple of 7). Sufficient.
Generally, in any group of n consecutive integers, one will always be a multiple of n. For instance, in a set of three consecutive numbers, one will always be divisible by 3.
To address your doubt: your list includes a multiple of 7, specifically 0.
(1) 8 and 13 are in set S.
This does not specify the full range of the set, only a part of it. There could be more numbers before 8 or after 13 in the set S, and without a defined range, we cannot be sure whether or not a multiple of 7 is included. Not sufficient.
(2) Set S has 8 terms.
Given that there are 8 consecutive integers in S, at least one of those integers will be a multiple of 7. This is because the longest stretch of numbers you can have without a multiple of 7 is 6 (after which the next number will be a multiple of 7). Sufficient.
Generally, in any group of n consecutive integers, one will always be a multiple of n. For instance, in a set of three consecutive numbers, one will always be divisible by 3.
Galdamari
To address your doubt: your list includes a multiple of 7, specifically 0.
ZERO:
1. Zero is an INTEGER.
2. Zero is an EVEN integer. An even number is an integer that is "evenly divisible" by 2, i.e., divisible by 2 without a remainder and as zero is evenly divisible by 2 then it must be even.
3. Zero is neither positive nor negative (the only one of this kind).
4. Zero is divisible by EVERY integer except 0 itself (\(\frac{x}{0} = 0\), so 0 is a divisible by every number, x).
5. Zero is a multiple of EVERY integer (\(x*0 = 0\), so 0 is a multiple of any number, x).
6. Zero is NOT a prime number (neither is 1 by the way; the smallest prime number is 2).
7. Division by zero is NOT allowed: anything/0 is undefined.
8. Any non-zero number to the power of 0 equals 1 (\(x^0 = 1\))
9. \(0^0\) case is NOT tested on the GMAT.
10. If the exponent n is positive (n > 0), \(0^n = 0\).
11. If the exponent n is negative (n < 0), \(0^n\) is undefined, because \(0^{negative}=0^n=\frac{1}{0^{(-n)}} = \frac{1}{0}\), which is undefined. You CANNOT take 0 to the negative power.
12. \(0! = 1! = 1\).
1. Zero is an INTEGER.
2. Zero is an EVEN integer. An even number is an integer that is "evenly divisible" by 2, i.e., divisible by 2 without a remainder and as zero is evenly divisible by 2 then it must be even.
3. Zero is neither positive nor negative (the only one of this kind).
4. Zero is divisible by EVERY integer except 0 itself (\(\frac{x}{0} = 0\), so 0 is a divisible by every number, x).
5. Zero is a multiple of EVERY integer (\(x*0 = 0\), so 0 is a multiple of any number, x).
6. Zero is NOT a prime number (neither is 1 by the way; the smallest prime number is 2).
7. Division by zero is NOT allowed: anything/0 is undefined.
8. Any non-zero number to the power of 0 equals 1 (\(x^0 = 1\))
9. \(0^0\) case is NOT tested on the GMAT.
10. If the exponent n is positive (n > 0), \(0^n = 0\).
11. If the exponent n is negative (n < 0), \(0^n\) is undefined, because \(0^{negative}=0^n=\frac{1}{0^{(-n)}} = \frac{1}{0}\), which is undefined. You CANNOT take 0 to the negative power.
12. \(0! = 1! = 1\).
15 Oct 2023, 07:39
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Galdamari
Given: S is a list of consecutive integers.
Asked: Does S contain a multiple of 7?
(1) 8 and 13 are in set S.
If S = {8,9,10,11,12,13}: No
If S = {7,8,9,10,11,12,13,14}: Yes
NOT SUFFICIENT
(2) Set S has 8 terms.
Every seven consecutive integers will have one multiple of 7.
Since Set S has 8 terms, it contains a multiple of 7.
SUFFICIENT
IMO B
Given: S is a list of consecutive integers.
Asked: Does S contain a multiple of 7?
(1) 8 and 13 are in set S.
If S = {8,9,10,11,12,13}: No
If S = {7,8,9,10,11,12,13,14}: Yes
NOT SUFFICIENT
(2) Set S has 8 terms.
Every seven consecutive integers will have one multiple of 7.
Since Set S has 8 terms, it contains a multiple of 7.
SUFFICIENT
IMO B
