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If the product of 7 consecutive integers is equal to the med
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18 Dec 2017, 12:37
18 Dec 2017, 12:37
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62% (01:03) correct
37% (02:13) wrong based on 35 sessions
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If the product of 7 consecutive integers is equal to the median of the integers, what is the least of the 7 integers?
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Show: :: OA
\(-3\)
Re: If the product of 7 consecutive integers is equal to the med
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20 Dec 2017, 06:45
20 Dec 2017, 06:45
5
Carcass wrote:
If the product of 7 consecutive integers is equal to the median of the integers, what is the least of the 7 integers?
enter your value
enter your value
Show: :: OA
\(-3\)
Let x be the least integer then we have
x(x+1)(x+2)(x+3)(x+4)(x+5)(x+6) = x+3
or (x+3){x(x+1)(x+2)(x+4)(x+5)(x+6) - 1} =0
i,e. either x+3 = 0
x = -3
or x(x+1)(x+2)(x+4)(x+5)(x+6) - 1 =0, but this won't provide x = integer
Hence x = -3
Re: If the product of 7 consecutive integers is equal to the med
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28 Sep 2021, 07:44
28 Sep 2021, 07:44
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Let's say, seven consecutive integers \(= x - 3, x - 2, x - 1, x, x + 1, x + 2, x + 3\)
Product = the median of the integers
\(x (x^2-1)(x^2-4)(x^2-9) = x\)
Now, this can happen only if \(x = 0\)
So the least integer \(= -3\)
Product = the median of the integers
\(x (x^2-1)(x^2-4)(x^2-9) = x\)
Now, this can happen only if \(x = 0\)
So the least integer \(= -3\)
aishumurali wrote:
Can someone pls explain this?
