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Re: If the product of 7 consecutive integers is equal to the med [#permalink]
Can someone pls explain this?
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If the product of 7 consecutive integers is equal to the med [#permalink]
aishumurali wrote:
Can someone pls explain this?


We have 7 consecutive integers in total, so we must have 3 integers before the Median and 3 integers after.
i.e. (M - 3), (M - 2), (M - 1), M, (M + 1), (M + 2), (M + 3)

Now, the product of 7 integers equals the Median
This could only happen if Median is 0

So, the integers are: -3, -2, -1, 0, 1, 2, 3
The least integer could be -3
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Re: If the product of 7 consecutive integers is equal to the med [#permalink]
pranab223 wrote:
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

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


hi, how did you get x+3=0? since it is in both sides shouldn't this be cancelled? could you please explain..
or x(x+1)(x+2)(x+4)(x+5)(x+6) - 1 =0, but this won't provide x = integer

Hence x = -3
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