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If a, b, and c are consecutive integers and 0 < a < b < c, is the prod
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27 Jul 2017, 07:18
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If a, b, and c are consecutive integers and 0 < a < b < c, is the prod
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27 Jul 2017, 07:31
carcass wrote: If a, b, and c are consecutive integers and 0 < a < b < c, is the product abc a multiple of 8 ?
(1) The product ac is even.
(2) The product bc is a multiple of 4. hi, If a, b, and c are consecutive integers and 0 < a < b < c, is the product abc a multiple of 8 TWO cases..A) a and c are even AND b is ODD.... abc will always be multiple of 8 or rather 24.. ans will be YES B) a and c are ODD and b is even.. if b is multiple of 8.. YES otherwise NOlets see the statements (1) The product ac is even. this MEANS a and c are even.. falls under category A above..always YES sufficient (2) The product bc is a multiple of 4either b is multiple of 4, ans can be YES or NO case BOR c is multiple of 4, ans will be YES.. case Adifferent answers possible insuff A
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Re: If a, b, and c are consecutive integers and 0 < a < b < c, is the prod
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27 Jul 2017, 07:34
I Think answer is A
Statement 1: ac is even > a and c are even and b is odd, minimum possible values of abc is 2*3*4. Sufficient
Statement 2: bc multiple of 4 > b can be 4.. c can be 5.... so a can be 3, which gives abc=60 not sufficient
Answer A



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Re: If a, b, and c are consecutive integers and 0 < a < b < c, is the prod
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01 Dec 2017, 14:58
I don't get why a and c are even AND b is ODD.... abc will always be multiple of 8 or rather 24, could you explain me please



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Re: If a, b, and c are consecutive integers and 0 < a < b < c, is the prod
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01 Dec 2017, 19:38
analuisa wrote: I don't get why a and c are even AND b is ODD.... abc will always be multiple of 8 or rather 24, could you explain me please Hi analuisaas the numbers are consecutive, so \(a\), \(b\), \(c\) can take two options  Case 1: Even, Odd, Even OR Case 2: Odd, Even, Odd Statement 1: as \(ac\) is even, so it has to be Case 1 Now let \(a=2k\) (even number) so \(b=2k+1\) and \(c=2k+2=2(k+1)\), where \(k\) is any number greater than \(0\) (because \(0<a<b<c\)) Hence \(ac=2k*2(k+1)=4k(k+1)\) so if \(k\) is odd then \(k+1=even\), hence \(4k(k+1)\) will be a multiple of \(8\) if \(k\) is even then \(4k\) will be a multiple of \(8\). Therefore \(abc\) is a multiple of \(8\) because \(ac\) is a multiple of \(8\). SufficientStatement 2: \(bc\) is a multiple of \(4\). so it could be multiple of \(8\) such as \(16,24,32\) etc. or it could not be a multiple of \(8\) such as \(20, 28\) etc.. InsufficientOption A



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Re: If a, b, and c are consecutive integers and 0 < a < b < c, is the prod
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01 Dec 2017, 20:29
carcass wrote: If a, b, and c are consecutive integers and 0 < a < b < c, is the product abc a multiple of 8 ?
(1) The product ac is even.
(2) The product bc is a multiple of 4. The Best would be to plug in by numbers. Let b = n, a = n1, c = n+1 Stmnt 1: ac = even, \(n^2\)  1 =even, Here n can be only a odd number starting from 3,5,7,9,11 etc.. So abc will be Divisible by 8. Stmnt 2: n(n+1) = 4m.. Here, n can be 3,4,7,8,9,11,13,14.. etc. when n = 3, abc Divisible by 8 but n= 4, abc not divisible by 8. Hope this Helps.
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Re: If a, b, and c are consecutive integers and 0 < a < b < c, is the prod
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01 Dec 2017, 20:48
analuisa wrote: I don't get why a and c are even AND b is ODD.... abc will always be multiple of 8 or rather 24, could you explain me please Hi... Three consecutive integers and two even means even, odd, even... Now since three are consecutive, one of them would surely be a MULTIPLE of 3. Two of them even means one will be MULTIPLE of 2 and other will be MULTIPLE of atleast 4 so product will be MULTIPLE of atleast 3*2*4=24
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Re: If a, b, and c are consecutive integers and 0 < a < b < c, is the prod
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23 Aug 2018, 19:16
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If a, b, and c are consecutive integers and 0 < a < b < c, is the prod
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23 Aug 2018, 22:36
Hi, Another way to solve this question is trying out numbers. Always try out few numbers and understand the pattern of the answers. Here in the question, it’s given, a, b and c are consecutive integers and also positive (0 < a < b < c), Question: Is product abc a multiple of 8 ?
To be a multiple of 8 , it has to have minimum of three 2’s because of 8 can be written in prime factorization as 2^3 Now, lets try out some numbers, a, b and c minimum has to be 1,2 and 3. Here the product abc is 6. Not a multiple of 8. If a, b and c is 2,3 and 4. Then the product abc is 24. Multiple of 8. Okay, then lets check the next set. If a,b and c is 3,4 and 5. Then the product abc is 60. Not a multiple of 8. If a, b and c is 4,5 and 6. Then the product abc is 120. Multiple of 8. Note 1: We can note from the above pattern, everytime if a and c is even, then the product abc is a multiple of 8(Infact 24). Note 2: Also, if a and c are odd, then the product abc need not be a multiple of 8 but it is an even number. Statement I is sufficient.
Given, the product ac is even.Since, a, b and c are consecutive integers. If ac is even, then both a and c has to be even. So, from the note 1, we can see that product is a multiple of 8. So sufficient. Statement II is insufficient.
Given, the product bc is a multiple of 4.bc is a multiple of 4 means, either b is a multiple of 4 and c is odd or b is odd and c is a multiple of 4. For example, If a,b and c is 3,4 and 5. Then the product abc is 60. Not a multiple of 8. But here 4*5 = 20 is a multiple of 4. If a, b and c is 2,3 and 4. Then the product abc is 24. Multiple of 8. Here 3*4= 12 also a multiple of 4. So it is insufficient. So the answer is A. I alone sufficient.
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Re: If a, b, and c are consecutive integers and 0 < a < b < c, is the prod
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25 Aug 2018, 20:08
carcass wrote: If a, b, and c are consecutive integers and 0 < a < b < c, is the product abc a multiple of 8 ?
(1) The product ac is even.
(2) The product bc is a multiple of 4. a, b, and c are consecutive integers and 0 < a < b < c > b=a+1; c=a+2 Question: Is \(a(a+1)(a+2)\) a multiple of 8? (1) The product \(ac\) is even. > \(a(a+2)\) is even. Since \(a\) and \(a+2\) are either both odd or both even, we can conclude that \(a\) and \(a+2\) are both even in this case. The product of two consecutive even numbers is a multiple of 8, so \(a(a+2)\) is a multiple of 8. > \(a(a+1)(a+2)\) is a multiple of 8. > Sufficient. (2) The product \(bc\) is a multiple of 4 > (a+1)(a+2) is multiple of 4. If a=2 then a(a+1)(a+2)=2*3*4 > YES If a=3 then a(a+1)(a+2)=3*4*5 > NO > Statement 2 alone is not sufficient. Answer A
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