product of three consecutive integers

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j_shreyans: Legendary Member; Posts: 510; Joined: Thu Aug 07, 2014 2:24 amFollowed by: 5 members; Thanked: 3 times

by j_shreyans » Mon Jun 08, 2015 7:32 am

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

OAB

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GMATGuruNY: GMAT Instructor; Posts: 15539; Joined: Tue May 25, 2010 12:04 pmFollowed by: 1906 members; Location: New York, NY; Thanked: 13060 times; GMAT Score:790

by GMATGuruNY » Mon Jun 08, 2015 7:45 am

j_shreyans wrote:The integer 6 is the product of two consecutive integers (6 = 2 Ã— 3) and the product of three consecutive integers (6 = 1 Ã— 2 Ã— 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

TEST options for the 3 consecutive integers:
2*3*4 = 24.
3*4*5 = 60.
4*5*6 = 120.
5*6*7 = 210.

Since 210 is among the answer choices, check whether 210 can also be expressed as the product of 2 consecutive integers:
210 = 2*3*5*7 = (2*7) * (3*5) = 14*15.
Success!

The correct answer is B.

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MartyMurray: Legendary Member; Posts: 2135; Joined: Mon Feb 03, 2014 9:26 amFollowed by: 140 members; Location: https://martymurraycoaching.com/; Thanked: 955 times; GMAT Score:800

by MartyMurray » Tue Jun 09, 2015 11:39 am

j_shreyans wrote:The integer 6 is the product of two consecutive integers (6 = 2 Ã— 3) and the product of three consecutive integers (6 = 1 Ã— 2 Ã— 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

OAB

Check each answer choice.

A) 153 is not possible because any two or three consecutive integers will always include at least one even integer. 153 is odd. So it cannot have any even factors. So eliminate A.

B) Since we are looking for a product of two consecutive integers, look for the nearest perfect square. 15Â² = 225. Ah, so 15 x 14 = 210! So that's two consecutive integers and the factors of them are 3, 5, 2 and 7. The factors can be rearranged into 5, 6, and 7. So we have our answer.

The question is finished at this point, but I'll do the rest for illustrative purposes.

C) Since we are looking for a product of two consecutive integers, look for the nearest perfect square. 16Â² = 256. Ah so 16 x 17 = 272. But 17 is prime. So we can't break down the factors further. So eliminate C.

D) Since we are looking for a product of two consecutive integers, I would look for the nearest perfect square, which is probably 18Â², but I am getting a sense of the pattern so I go with 18 x 20 = 360 and realize that 18 X 19 = 342, which is less than 18 away from 336. So there's no way to get two consecutive integers to multiply to 336, and D is out.

E) Since we are looking for a product of two consecutive integers, look for the nearest perfect square, which is 25Â² = 625. So 25 x 24 = 600. So the prime factors are 5, 5, 2, 2, 2 and 3. Immediately I see the two 5's and realize that as soon as they are split up there can't be three consecutive numbers, because any two unique numbers with 5 as a factor have to be at least 5 apart from each other. So E is out.

Choose B.

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nikhilgmat31: Legendary Member; Posts: 518; Joined: Tue May 12, 2015 8:25 pm; Thanked: 10 times

by nikhilgmat31 » Mon Jun 15, 2015 5:02 am

I think we can solve using factorization. But it is very important & difficult to strike the pair & triplet at same time.

B)210 7*6*5 or 15 * 14 - 7 * 3 * 2 * 5

Well its very good question.

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