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

Question

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

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

farful · Accepted Answer

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.

PareshGmat · Answer

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

thefibonacci · Answer

Do it with options

210 = 6*7*5 and 14*15

bekerman · Answer

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.

PerfectScores · Answer

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

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

mike34170 · Answer

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.

andy233 · Answer

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

mbaprep2016 · Answer

factorize all options , we get 210 as 2*3*5*7
14*15 -- consecutive 
5*6*7 again consecutive

grichagupta · Answer

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

kannu44 · Answer

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

pra1785 · Answer

OE from Manhattan Prep

There’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).

ScottTargetTestPrep · Answer

The product of 3 consecutive integers must be divisible by 6.

We see that 210, 336, and 600 are all divisible by 6 (so we can eliminate 153 and 272). Let’s test 210.

Factoring 210, we get: 210 = 10 x 21 = 2 x 5 x 3 x 7

We see that 210 = (2 x 7) x (3 x 5) = 14 x 15 and 210 = 5 x (2 x 3) x 7 = 5 x 6 x 7. Therefore, 210 is the next greater integer that is both the product of two consecutive integers and the product of three consecutive integers.

Answer: B

kebab1990 · Answer

I used prime factorization to come up with the answer

Since we are looking for the closest number to 6 I started with option A:

A) 153 = 9*17= 3*3*17 --> no way to have even 2 consecutive numbers
B) 210 = 10*21 = 2*5*3*7=2*3*5*7--> 2*3 is 6, so I see 3 consecutive numbers 5,6,7. After noticing that I'm looking for 2 consecutive integers, out of 4 numbers: 2,3,5,7 we can multiply 2 and 7 to get 14 and 3 and 5 to get 15. BINGO! We've got two consecutive factors as well.

Hope it helps

Basshead · Answer

The product of three consecutive integers has to be divisible by 2 and 3, leaving B, D, and E.

Starting off with B:

210 = 2 * 3 * 5 * 7

We can form  5 * 6 * 7 . We can also form  14 * 15 .

Answer is B.

HarshG96 · Answer

Another way:

eliminate A and C since they are not divisible by 6.

Now coming to 210,336,600

Try finding closest perfect square and perfect cube near these numbers:
 Number | Nearest Square Root (a) | Nearest cube root (b) 
210 | 14 (14^2=196) | 6 (6^3=216)
336 | 18 (18^2=324) | 7 (7^3=343)
600 | 24 (24^2=576) | 8 (8^3=512)

a(a+1) | (b-1)b(b+1) 
14(14+1) =  210  | (6-1)(6)(6+1) =  210 
18(18+1) =  342  | (7-1)(7)(7+1) =  336 
24(24+1) =  600  | (8-1)(8)(8+1) =  504

Thus, B) 210 is the answer

nhutnguyen3009 · Answer

The product of n consecutive numbers is always divisible by the factorial of n. Therefore, for this problem, the number must be divisible by both 2! and 3!, which means it must be divisible by both 2 and 3. Option B meets these requirements (eliminate option D because we have to find the next number)

Vibhatu · Answer

The question is wrong!! You gave example for 1,2 and then 1,2,3... where 1,2 are same in both.. one might thought some number greater than 6 will be product of XY and these consecutive number XY WILL Followed by Z !!! You have to mention teo consecutive and three consecutive number may not be same!! Bunuel check

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Bunuel · Answer

The question is not wrong. The example of 6 being the product of two consecutive integers (2 × 3) and three consecutive integers (1 × 2 × 3) does not suggest that this pattern must hold for all numbers. The next integer greater than 6 that satisfies both conditions doesn’t require the two consecutive integers to overlap with the three consecutive integers. The question simply asks for "the product of two consecutive integers and the product of three consecutive integers," without mentioning any overlap between them.

In fact, the two sets of consecutive integers can be different, which is why the correct answer is found by considering different pairs and triples of consecutive integers. There is no requirement for the consecutive numbers to overlap in both cases.

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

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

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