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

But the given answer is "C". Can anybody explain, please?
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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Re: The product of 3 consecutive positive even integers x+1, x+3 and x+5 [#permalink]
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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".

But the given answer is "C". Can anybody explain, please?
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]
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".

But the given answer is "C". Can anybody explain, please?
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]
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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Re: The product of 3 consecutive positive even integers x+1, x+3 and x+5 [#permalink]
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


To make p/60 an integer, p has to be at least 60 (the sum of 18, 20 and 22). When p =120 (38, 40, 42) or 180 (58, 60, 62), p =240 (78, 80, 82). From this, we can get x has to be an odd integer which lowest value can be 17 and from that it can take any value which is greater by 20 from it. The series for x will be like 17, 37, 57, 77, 97, 117....

1) 20 has 5 factors: 1, 4, 5, 10, 20. According to the question x + 3 is an even integer. But if x + 3 is either 4 or 10, p/60 will not be an integer. So, not sufficient.
2) Since, x has to be an odd integer as per the question, we can take any values of x from the odd multiples of 5, e.g 5, 15, 25, 35. Since, the unit value will always be 5, whereas we need an integer x with unit value 7 to make p a multiple of 60. So from this statement we can understand p/ 60 can never be an integer. Sufficient.

B is the answer.
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Re: The product of 3 consecutive positive even integers x+1, x+3 and x+5 [#permalink]
Bunuel wrote:
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.


Answer: B.



Bunuel :- Aren't the 2 statements contradicting?

i) x+3 = 4,10 or 20 => x = 1, 7 ,17

ii) x is divisible by 5 => x = 5, 10, 15

We don't have any numbers matching in i) and ii)
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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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Re: The product of 3 consecutive positive even integers x+1, x+3 and x+5 [#permalink]
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