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The product of 3 consecutive positive even integers x+1, x+3 and x+5
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Updated on: 07 Jan 2019, 22:30
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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? (1) x+3 is a factor of 20 (2) x is divisible by 5
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Originally posted by akurathi12 on 06 Jan 2019, 08:46.
Last edited by Bunuel on 07 Jan 2019, 22:30, 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
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06 Jan 2019, 20:33
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.



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Re: The product of 3 consecutive positive even integers x+1, x+3 and x+5
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06 Jan 2019, 21:36
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
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07 Jan 2019, 22:08
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
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07 Jan 2019, 22:30
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.
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Re: The product of 3 consecutive positive even integers x+1, x+3 and x+5
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07 Jan 2019, 22:46
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 nitesh50x 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
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08 Jan 2019, 00:53
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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