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# What is the sum of four consecutive odd integers? 1. Product

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What is the sum of four consecutive odd integers? 1. Product [#permalink]

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13 Nov 2007, 05:51
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What is the sum of four consecutive odd integers?

1. Product of four integers is 105
2. Median of four integers is 4

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13 Nov 2007, 06:33
four consecutive odd integer x, x+2, x+4, x+6

statement 1 product of x*(x+2)*(x+4)*(x+6)=105; therefore, 1*3*5*7=105, 1+3+5+7=16 sufficient

statement 2 median of four integers is 4, median is [(x+2)+(x+4)]/2=4; therefore, x=1 1+3+5+7=16 sufficient

CEO
Joined: 21 Jan 2007
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Kudos [?]: 861 [0], given: 4

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15 Nov 2007, 03:12
eileen1017 wrote:
four consecutive odd integer x, x+2, x+4, x+6

statement 1 product of x*(x+2)*(x+4)*(x+6)=105; therefore, 1*3*5*7=105, 1+3+5+7=16 sufficient

statement 2 median of four integers is 4, median is [(x+2)+(x+4)]/2=4; therefore, x=1 1+3+5+7=16 sufficient

OA is B.

Statement 1:
We know 105 is a multiple of 5, so at least 1 odd integer will be a multiple of 5. See what the remaining product is:
105/5=21, we know prime factors of 21 are 3 and 7, so now we have all 4 integers: 1, 3, 5, 7.

However, since there is an even number of numbers being multiplied, it'll also work if they're all negative (-7, -5, -3, -1). Thus, we can't determine which of the two sets is the one we're looking for, since one sum will be positive and the other negative. Insufficient.

Statement 2:
4 consecutive odd integers with mean of the two middle ones being 4, we know that can only be 3 and 5, thus the other two are 1 and 7.
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Joined: 06 Mar 2006
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Kudos [?]: 179 [0], given: 1

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15 Nov 2007, 06:19
bmwhype2 wrote:
eileen1017 wrote:
four consecutive odd integer x, x+2, x+4, x+6

statement 1 product of x*(x+2)*(x+4)*(x+6)=105; therefore, 1*3*5*7=105, 1+3+5+7=16 sufficient

statement 2 median of four integers is 4, median is [(x+2)+(x+4)]/2=4; therefore, x=1 1+3+5+7=16 sufficient

OA is B.

Statement 1:
We know 105 is a multiple of 5, so at least 1 odd integer will be a multiple of 5. See what the remaining product is:
105/5=21, we know prime factors of 21 are 3 and 7, so now we have all 4 integers: 1, 3, 5, 7.

However, since there is an even number of numbers being multiplied, it'll also work if they're all negative (-7, -5, -3, -1). Thus, we can't determine which of the two sets is the one we're looking for, since one sum will be positive and the other negative. Insufficient.

Statement 2:
4 consecutive odd integers with mean of the two middle ones being 4, we know that can only be 3 and 5, thus the other two are 1 and 7.

Thanks. I guess I can't assume the integers being all positive without the condition stated explicitly.
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Joined: 21 Jan 2007
Posts: 2756
Location: New York City
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Kudos [?]: 861 [0], given: 4

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15 Nov 2007, 06:43
eileen1017 wrote:
bmwhype2 wrote:
eileen1017 wrote:
four consecutive odd integer x, x+2, x+4, x+6

statement 1 product of x*(x+2)*(x+4)*(x+6)=105; therefore, 1*3*5*7=105, 1+3+5+7=16 sufficient

statement 2 median of four integers is 4, median is [(x+2)+(x+4)]/2=4; therefore, x=1 1+3+5+7=16 sufficient

OA is B.

Statement 1:
We know 105 is a multiple of 5, so at least 1 odd integer will be a multiple of 5. See what the remaining product is:
105/5=21, we know prime factors of 21 are 3 and 7, so now we have all 4 integers: 1, 3, 5, 7.

However, since there is an even number of numbers being multiplied, it'll also work if they're all negative (-7, -5, -3, -1). Thus, we can't determine which of the two sets is the one we're looking for, since one sum will be positive and the other negative. Insufficient.

Statement 2:
4 consecutive odd integers with mean of the two middle ones being 4, we know that can only be 3 and 5, thus the other two are 1 and 7.

Thanks. I guess I can't assume the integers being all positive without the condition stated explicitly.

yep. everytime i approach a DS question, i ask myself is it a positive integer fraction -1, 0, 1.
15 Nov 2007, 06:43
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