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The product of 3 consecutive positive even integers x+1, x+3 and x+5  [#permalink]

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Difficulty:   95% (hard)

Question Stats: 28% (02:38) correct 72% (02:24) wrong based on 95 sessions

HideShow timer Statistics The product of 3 consecutive positive even integers x+1, x+3 and x+5 is p. Is p/60 an integer?

(1) x+3 is a factor of 20

(2) x is divisible by 5

Originally posted by akurathi12 on 06 Jan 2019, 09:46.
Last edited by Bunuel on 07 Jan 2019, 23:30, edited 1 time in total.
Edited the OA.
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Re: The product of 3 consecutive positive even integers x+1, x+3 and x+5  [#permalink]

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Quote:
(1) x+3 is a factor of 20

(2) x is divisible by 5

Statement 1: so x+3=20 or 10 or 5 or 4 or 2 or 1.
1 & 2 are not possible because x+1 can't be zero or negative. x+3 is positive even integer so 5 is not possible.
So x+3=20 or 10 or 4.
So x+1=18 or 8 or 2 and x+5=22 or 12 or 6.
For first two numbers answer is "Yes" for the last number answer is "No".
Not sufficient.

Statement 2: x=5 or 10 or 15 or 20 so on.
When x=5, x+1=6, x+3=8 & x+5=10 and answer is "Yes".
When x=10, x+1=11, x+3=13 & x+5=15 and answer is "No".
Not sufficient.

Taking 1&2 together we have no common value. So I have "E".

Thank you.
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Re: The product of 3 consecutive positive even integers x+1, x+3 and x+5  [#permalink]

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Mahfuz1469 wrote:
Quote:
(1) x+3 is a factor of 20

(2) x is divisible by 5

Statement 1: so x+3=20 or 10 or 5 or 4 or 2 or 1.
1 & 2 are not possible because x+1 can't be zero or negative. x+3 is positive even integer so 5 is not possible.
So x+3=20 or 10 or 4.
So x+1=18 or 8 or 2 and x+5=22 or 12 or 6.
For first two numbers answer is "Yes" for the last number answer is "No".
Not sufficient.

Statement 2: x=5 or 10 or 15 or 20 so on.
When x=5, x+1=6, x+3=8 & x+5=10 and answer is "Yes".
When x=10, x+1=11, x+3=13 & x+5=15 and answer is "No".
Not sufficient.

Taking 1&2 together we have no common value. So I have "E".

Thank you.

Statement (1), x can be 1,3,7, 17
when x =1, consecutive even integers will be, 2 *4 * 6 / 60, this will answer the question as No
when x =7, consecutive even integers will be , 8*10*12/ 60, this will answer the question as Yes

Statement (2), x can be 5,10
when x =5, consecutive even integers will be, 6 *8 * 10 / 60, this will answer the question as Yes
when x =10, consecutive even integers will be , 11*13*15/ 60, this will answer the question as No

I agree with your explanation about C, If there is no common value, will it make the overall question a No.(only then it can be a C)
Because there is nothing told about the magnitude of x, it can be -ive as well as +ive.

Only value to satisfy both the statements will be a negative one i.e. -5, but this value goes against the initial statement.

Would like to see an expert replying on this.
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1
KanishkM wrote:
Mahfuz1469 wrote:
Quote:
(1) x+3 is a factor of 20

(2) x is divisible by 5

Statement 1: so x+3=20 or 10 or 5 or 4 or 2 or 1.
1 & 2 are not possible because x+1 can't be zero or negative. x+3 is positive even integer so 5 is not possible.
So x+3=20 or 10 or 4.
So x+1=18 or 8 or 2 and x+5=22 or 12 or 6.
For first two numbers answer is "Yes" for the last number answer is "No".
Not sufficient.

Statement 2: x=5 or 10 or 15 or 20 so on.
When x=5, x+1=6, x+3=8 & x+5=10 and answer is "Yes".
When x=10, x+1=11, x+3=13 & x+5=15 and answer is "No".
Not sufficient.

Taking 1&2 together we have no common value. So I have "E".

Thank you.

Statement (1), x can be 1,3,7, 17
when x =1, consecutive even integers will be, 2 *4 * 6 / 60, this will answer the question as No
when x =7, consecutive even integers will be , 8*10*12/ 60, this will answer the question as Yes

Statement (2), x can be 5,10
when x =5, consecutive even integers will be, 6 *8 * 10 / 60, this will answer the question as Yes
when x =10, consecutive even integers will be , 11*13*15/ 60, this will answer the question as No

I agree with your explanation about C, If there is no common value, will it make the overall question a No.(only then it can be a C)
Because there is nothing told about the magnitude of x, it can be -ive as well as +ive.

Only value to satisfy both the statements will be a negative one i.e. -5, but this value goes against the initial statement.

Would like to see an expert replying on this.

Hi
I am no expert.
But I feel the answer should be B.

Statement 2:
when x =10, consecutive even integers will be , 11*13*15/ 60, this will answer the question as No

11 13 15 are no even integers. So this case is not valid.

Hence IMO the OA is incorrect.
It should be B
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Re: The product of 3 consecutive positive even integers x+1, x+3 and x+5  [#permalink]

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The product of 3 consecutive positive even integers x+1, x+3 and x+5 is p. Is p/60 an integer?

The question asks whether (x+1)(x+3)(x+5) is a multiple of $$60 = 2^2*3*5$$.

First of all notice that we are told that x+1, x+3 and x+5 are consecutive positive even integers.

Next notice that:
a. The product of any three consecutive positive even integers, will for sure be a multiple of 2^2.
b. One of the THREE consecutive even integers must be a multiple of 3.

So, (x+1)(x+3)(x+5) will be a multiple of 2^2*3 in ALL cases. We need to find whether this product is a multiple of 5, in order it to be a multiple of 60.

(1) x+3 is a factor of 20.

Factors of 20 are 1, 2, 4, 5, 10 and 20. Since we know that x+3 is even and positive, then x + 3 can be 4, 10 or 20. If x+3 is 10 and 20, then (x+1)(x+3)(x+5) will be a multiple of 5. If x+3 is 4, then the product is (x+1)(x+3)(x+5) = 2*4*6 and it's not a multiple of 5. Not sufficient.

(2) x is divisible by 5. If x is a multiple of 5, then so is x + 5, which means that (x+1)(x+3)(x+5) will be a multiple of 5. Sufficient.

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Re: The product of 3 consecutive positive even integers x+1, x+3 and x+5  [#permalink]

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nitesh50 wrote:
(2) x is divisible by 5

Statement 1: so x+3=20 or 10 or 5 or 4 or 2 or 1.
1 & 2 are not possible because x+1 can't be zero or negative. x+3 is positive even integer so 5 is not possible.
So x+3=20 or 10 or 4.
So x+1=18 or 8 or 2 and x+5=22 or 12 or 6.
For first two numbers answer is "Yes" for the last number answer is "No".
Not sufficient.

Statement 2: x=5 or 10 or 15 or 20 so on.
When x=5, x+1=6, x+3=8 & x+5=10 and answer is "Yes".
When x=10, x+1=11, x+3=13 & x+5=15 and answer is "No".
Not sufficient.

Taking 1&2 together we have no common value. So I have "E".

Thank you.

Statement (1), x can be 1,3,7, 17
when x =1, consecutive even integers will be, 2 *4 * 6 / 60, this will answer the question as No
when x =7, consecutive even integers will be , 8*10*12/ 60, this will answer the question as Yes

Statement (2), x can be 5,10
when x =5, consecutive even integers will be, 6 *8 * 10 / 60, this will answer the question as Yes
when x =10, consecutive even integers will be , 11*13*15/ 60, this will answer the question as No

I agree with your explanation about C, If there is no common value, will it make the overall question a No.(only then it can be a C)
Because there is nothing told about the magnitude of x, it can be -ive as well as +ive.

Only value to satisfy both the statements will be a negative one i.e. -5, but this value goes against the initial statement.

Would like to see an expert replying on this.

Hi
I am no expert.
But I feel the answer should be B.

Statement 2:
when x =10, consecutive even integers will be , 11*13*15/ 60, this will answer the question as No

11 13 15 are no even integers. So this case is not valid.

Hence IMO the OA is incorrect.
It should be B

I have the same thoughts as you nitesh50

x cannot be 10 or 20 or 30...it has to be an odd number since x+1 is even.

So x = 5 or 15 or 25... and so on...
And corresponding even nos will be 6, 8, 10....or 16, 18, 20 ...and so on....

IMO B.
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The product of 3 consecutive positive even integers x+1, x+3 and x+5  [#permalink]

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akurathi12 wrote:
The product of 3 consecutive positive even integers x+1, x+3 and x+5 is p. Is p/60 an integer?

(1) x+3 is a factor of 20

(2) x is divisible by 5

60 = 2^2 * 3* 5

so we need to find value of x such that x+1 * x+3*x+5 is divisible by 2^2 * 3* 5

#1:
x+3 factor of 20
so x= 1,15,17 many values not sufficeint
#2:
x is divisible by 5
sufficient for x=5,10,15,20 we would always get value of p/60 an integer
sufficient

IMO B
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