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

Do it with options

210 = 6*7*5 and 14*15

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.

[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

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.

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

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

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

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

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).

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

The integer 6 is the product of two consecutive integers (6 = 2 × 3) and the product of three consecutive integers (6 = 1 × 2 × 3).

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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

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.

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

Why should it be divisible by 6?