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The integer 6 is the product of two consecutive integers (6 [#permalink]
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22 Jan 2014, 05:23
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The integer 6 is the product of two consecutive integers (6 = 2 × 3) and the product of three consecutive integers (6 = 1 × 2 × 3). What is the next integer greater than 6 that is both the product of two consecutive integers and the product of three consecutive integers? A) 153 B) 210 C) 272 D) 336 E) 600 B) 210 210 = 2*5*3*7 = 14*15 = 5*6*7 Is there any algebraic way to solve this? Plugging in would consume too much time in my opinion... Thanks in advance
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Re: The integer 6 is the product of two consecutive integers (6 [#permalink]
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22 Jan 2014, 11:02
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TheZezet wrote: The integer 6 is the product of two consecutive integers (6 = 2 × 3) and the product of three consecutive integers (6 = 1 × 2 × 3). What is the next integer greater than 6 that is both the product of two consecutive integers and the product of three consecutive integers? A) 153 B) 210 C) 272 D) 336 E) 600 B) 210 210 = 2*5*3*7 = 14*15 = 5*6*7 Is there any algebraic way to solve this? Plugging in would consume too much time in my opinion... Thanks in advance The OA actually has a pretty nifty solution that I didn't think of. The OA is suggesting you factor each of the answers and see if you can get two consecutive and three consecutive integers. Personally, I would have multiplied three digit numbers from the beginning, until an answer choice showed up. 1x2x3 = 6 2x3x4 = 24 3x4x5 = 60 4x5x6 = 120 5x6x7 = 210 6x7x8 = 336 210 is the first that shows up, and then figuring out that it's also 14x15 can easily be done by estimating its square root or even by trial and error.
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Re: The integer 6 is the product of two consecutive integers (6 [#permalink]
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22 Jan 2014, 11:05
TheZezet wrote: The integer 6 is the product of two consecutive integers (6 = 2 × 3) and the product of three consecutive integers (6 = 1 × 2 × 3). What is the next integer greater than 6 that is both the product of two consecutive integers and the product of three consecutive integers? A) 153 B) 210 C) 272 D) 336 E) 600 B) 210 210 = 2*5*3*7 = 14*15 = 5*6*7 Is there any algebraic way to solve this? Plugging in would consume too much time in my opinion... Thanks in advance Do it with options 210 = 6*7*5 and 14*15
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Re: The integer 6 is the product of two consecutive integers (6 [#permalink]
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29 Mar 2014, 19:54
I too would like to see an algebraic solution to this  when I did the problem, I prime factorized each answer choice and just missed the right answer, spending way too much time in the process too.
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Re: The integer 6 is the product of two consecutive integers (6 [#permalink]
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30 Mar 2014, 01:32
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[quote="christianbze"]The integer 6 is the product of two consecutive integers (6 = 2 × 3) and the product of three consecutive integers (6 = 1 × 2 × 3). What is the next integer greater than 6 that is both the product of two consecutive integers and the product of three consecutive integers? A) 153 B) 210 C) 272 D) 336 E) 600 Since the question asks the lowest possible so let us start with 153: Any number divisible by 6 would satisfy that. 153  17 x 9 210  divisible by 6 so it will have the pair 1 x 2 x 3
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Re: The integer 6 is the product of two consecutive integers (6 [#permalink]
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03 Apr 2014, 04:37
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Filtering the available options Multiplication of 3 consecutive numbers is always divisible by 2 AND 3 i.e 6 So option A & C can be discarded Let the consecutive integers be a, a+1 & a+2 Consider option B = 210 Setting up quadratic equation \(a^2 + a  210 = 0\) \(a^2 + 15a  14a  210 = 0\) a = 14 ; a+1 = 15 Taking 2 out from 14 & 3 out from 15 gives 5 , 6 & 7 As answer is obtained from option B, other options not calculated Answer = B Had 210 been in other options, we can still setup quadratic equations as below & try to find the values of a For example: \(a^2 + a  336 = 0\) \(a^2 + a  600 = 0\)
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Re: The integer 6 is the product of two consecutive integers (6 [#permalink]
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24 Jun 2015, 01:04
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Hello,
The first thing we can do is to check very quickly the numbers that are both divisible by 2 and 3 ( we are talking about the product of three consecutive integers).
Only values that remain are 210, 336 and 600.
Then you prime factorize those numbers and see wath happens.



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Re: The integer 6 is the product of two consecutive integers (6 [#permalink]
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25 Jun 2015, 23:01
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One thing which can be done is that the number must be divisible by 6, out of the 5 numbers 210, 336 and 600 are divisible by 6.
when you divide 210/6 = 35 (5,7) 336/6 = 56 (6,7,8) 600/6 = 100 (no consecutive #'s)
So between 336 and 210 336 ( the factors have to be less than or equal 19 as 19 x 19 = 361, 18 * 19 = 342, 17 * 18 = 306) so this is not the one
210 ( the factors have to be less than or equal to 15 as 15 * 15 = 225) 15 * 14 =210



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Re: The integer 6 is the product of two consecutive integers (6 [#permalink]
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13 Jul 2016, 23:09
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factorize all options , we get 210 as 2*3*5*7 14*15  consecutive 5*6*7 again consecutive



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Re: The integer 6 is the product of two consecutive integers (6 [#permalink]
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04 May 2017, 09:21
Since the question asks for the next integer greater than 6, my first inclination was to check if 153 satisfied the given criteria. Since, it does not, I checked 210 by factorizing it. 210 = 2*3*5*7 = 5 *6 *7 .... product of three consecutive integers = (2*7) * (3*5) = 14 * 15 .... product of two consecutive integers
So 210 is our no and no need to check for other numbers.
Ans B



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Re: The integer 6 is the product of two consecutive integers (6 [#permalink]
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04 May 2017, 21:25
a(a+1)(a+2) and b(b+1) should result in any one of the option given . Here a and b are the integer a(a+1)(a+2) checked with 5 = 210
checked whether for any integer b(b+1) =210 solved b =14 so 210 is the answer



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Re: The integer 6 is the product of two consecutive integers (6 [#permalink]
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01 Nov 2017, 06:35
OE from Manhattan PrepThere’s no obvious way to set this problem up algebraically, so work backwards from the answer choices. The problem asks for the next integer that satisfies the requirements—in other words, the smallest of the given choices—so begin by testing the smallest value, 153, and then work upward. First, break the number into its prime components: 153 = (51)(3) = (17)(3)(3). Next, can these primes be combined so that two consecutive integers will produce 153? No. (3)(3) = 9, but this is not consecutive with 17. (17)(3) would be way too big. This can’t be the right answer. Try the next smallest answer. First, break 210 into primes: 210 = (10)(21) = (2)(5)(3)(7). (2)(3) = 6, so 210 is the product of 3 consecutive integers (5, 6, and 7). Is there a way to combine the primes into two consecutive integers? Yes! (2)(7) = 14 and (3)(5) = 15. The number 210 fits the requirements, and it is the smallest of the answer choices, so it must be the next largest integer (after 6) that can be written as the product of two consecutive integers and as the product of three consecutive integers. The correct answer is (B).




Re: The integer 6 is the product of two consecutive integers (6
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