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19 Nov 2020, 02:45
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Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
45%
(medium)
Question Stats:
70% (02:24) correct
30%
(02:15)
wrong
based on 185
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Not Attempted Yet
The sum of seven consecutive integers is 1,617. How many of them are prime?
A. 0
B. 1
C. 2
D. 3
E. 4
A. 0
B. 1
C. 2
D. 3
E. 4
19 Nov 2020, 06:47
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We have an equally spaced list (seven consecutive integers) and the median of an equally spaced list is equal to the average. We have 7 numbers that sum to 1617, so the average is 1617/7 = 231, and that's the median of our list. So our numbers are
228, 229, 230, 231, 232, 233, 234
Some of these are clearly not prime -- the even numbers are divisible by 2, and by summing digits, we see 231 is divisible by 3. That leaves only two potential primes, 229 and 233. The only way to prove conclusively that these are prime numbers is by trying to divide them by all of the primes less than √233 (if a number has a nontrivial divisor, it must have a prime divisor less than or equal to its square root), so by all primes less than about 15. This kind of prime testing is tedious and time consuming, and I've never seen a real GMAT question that asks test takers to do this. But we can: we know 229 and 233 are not divisible by 2, 3 or 5. The multiples of 7 that are nearby are 210, 217, 224, 231, etc, so 229 and 233 aren't divisible by 7. The multiples of 11 that are in that vicinity are 220, 231, 242, etc, so 229 and 233 are not multiples of 11. And the multiples of 13 that are nearby are (starting from 260 and subtracting 13s) 260, 247, 234, 221, etc, so 229 and 233 are not divisible by 13 either, and are therefore both prime numbers.
And the comment at the end of the post above is correct, but I fear it might be misinterpreted: it's true that any prime (besides 2 and 3) is in the form 6k + 1 or 6k - 1 (that's true because primes greater than 3 are odd, and are not divisible by 3, so they can't be in the form 6k + 3). But that's no guarantee a number is prime (e.g. 35 and 49 and 77 and 143 are all in the form 6k-1 or 6k+1, and none of them is prime). To establish that 229 and 233 are prime numbers, we need to do a lot more work than just to notice they're in the form 6k + 1 or 6k - 1.
228, 229, 230, 231, 232, 233, 234
Some of these are clearly not prime -- the even numbers are divisible by 2, and by summing digits, we see 231 is divisible by 3. That leaves only two potential primes, 229 and 233. The only way to prove conclusively that these are prime numbers is by trying to divide them by all of the primes less than √233 (if a number has a nontrivial divisor, it must have a prime divisor less than or equal to its square root), so by all primes less than about 15. This kind of prime testing is tedious and time consuming, and I've never seen a real GMAT question that asks test takers to do this. But we can: we know 229 and 233 are not divisible by 2, 3 or 5. The multiples of 7 that are nearby are 210, 217, 224, 231, etc, so 229 and 233 aren't divisible by 7. The multiples of 11 that are in that vicinity are 220, 231, 242, etc, so 229 and 233 are not multiples of 11. And the multiples of 13 that are nearby are (starting from 260 and subtracting 13s) 260, 247, 234, 221, etc, so 229 and 233 are not divisible by 13 either, and are therefore both prime numbers.
And the comment at the end of the post above is correct, but I fear it might be misinterpreted: it's true that any prime (besides 2 and 3) is in the form 6k + 1 or 6k - 1 (that's true because primes greater than 3 are odd, and are not divisible by 3, so they can't be in the form 6k + 3). But that's no guarantee a number is prime (e.g. 35 and 49 and 77 and 143 are all in the form 6k-1 or 6k+1, and none of them is prime). To establish that 229 and 233 are prime numbers, we need to do a lot more work than just to notice they're in the form 6k + 1 or 6k - 1.
General Discussion
19 Nov 2020, 02:59
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Explanation:
let the middle number is x.
(x-3) + (x-2) + (x-1)+ x + (x+1) + (x+2) + (x+3) = 1617
7x = 1617
x = 1617/7
x = 231
So, Numbers are = 228, 229, 230, 231, 232, 233, 234
Out of these prime number are : 229, 233 i.e. 2 numbers
IMO-C
Note: All prime numbers can be written in form of 6n+1 or 6n-1 (where n is a natural number); except 2 and 3
let the middle number is x.
(x-3) + (x-2) + (x-1)+ x + (x+1) + (x+2) + (x+3) = 1617
7x = 1617
x = 1617/7
x = 231
So, Numbers are = 228, 229, 230, 231, 232, 233, 234
Out of these prime number are : 229, 233 i.e. 2 numbers
IMO-C
Note: All prime numbers can be written in form of 6n+1 or 6n-1 (where n is a natural number); except 2 and 3
