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The number 571+572+573 can be expressed as the products of 3

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The number 571+572+573 can be expressed as the products of 3 [#permalink]

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The number 571+572+573 can be expressed as the products of 3 consecutive integers. What is the sum of these 3 consecutive integers?

A. 32
B. 33
C. 34
D. 35
E. 36
[Reveal] Spoiler: OA

Last edited by Bunuel on 25 Mar 2014, 02:08, edited 1 time in total.
Added the OA.

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Re: The number 571+572+573 cab be expressed [#permalink]

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New post 26 Nov 2012, 05:19
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tinashine20 wrote:
The number 571+572+573 can be expressed as the products of 3 consecutive integers. What is the sum of these 3 consecutive integers?

A. 32
B. 33
C. 34
D. 35
E. 36


571+572+573 = 3*572 = 3*(2^2*11*13) = 11*12*13 --> 11+12+13 = 36.

Answer: E.

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Re: The number 571+572+573 can be expressed as the products of 3 [#permalink]

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New post 25 Mar 2014, 01:46
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Addition of any 3 consecutive numbers is always divisible by 3

So, options A, C , D are ruled out


Now focus on options 33 & 36

It can be either

10, 11, 12 (Gives zero at the end after multiplication; ignore this series)

or

11, 12, 13 (Matches fine)

Answer = E
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Re: The product of three consecutive integers is equal to 571+572+573. [#permalink]

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petrified17 wrote:
niks18 wrote:
petrified17 wrote:
:roll: The product of three consecutive integers is equal to 571+572+573.
What is the sum of thoseconsecutive integers?

A) 32 B) 33 C) 34 D) 35 E) 36


\(571+572+573 = 1716\)

Now, \(1716 = 11*12*13\)

So \(11+12+13=36\)

Option \(E\)


how did you derive 11*12*13 from 1716?


Hi petrified17

Factorize 1716, you will get 2*2*3*11*13. so it is evident that the three consecutive numbers has to be 11, 12 & 13

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Re: The number 571+572+573 can be expressed as the products of 3 [#permalink]

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tinashine20 wrote:
The number 571+572+573 can be expressed as the products of 3 consecutive integers. What is the sum of these 3 consecutive integers?

A. 32
B. 33
C. 34
D. 35
E. 36


\(=571+572+573=1716=(13)(11)(2)(2)(3)\)

after seeing 11 and 13, I knew the third should be 12 without calculating rem. factors.

\(=11+12+13= 36\)

Answer: E
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Re: The number 571+572+573 can be expressed as the products of 3 [#permalink]

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New post 24 Dec 2014, 06:33
571+572+573 = 1716 = 2*2*3*11*13 = 12*11*13

11+12+13 = 36

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Re: The number 571+572+573 can be expressed as the products of 3 [#permalink]

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New post 25 Dec 2014, 00:05
571 + 572 + 573 can be written as
572 -1 + 572 + 572 + 1 = 3*572 = 3 * 4 * 11 * 13 = 11 * 12 * 13...
Ans. E

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Re: The product of three consecutive integers is equal to 571+572+573. [#permalink]

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New post 20 Sep 2017, 00:27
petrified17 wrote:
:roll: The product of three consecutive integers is equal to 571+572+573.
What is the sum of thoseconsecutive integers?

A) 32 B) 33 C) 34 D) 35 E) 36


\(571+572+573 = 1716\)

Now, \(1716 = 11*12*13\)

So \(11+12+13=36\)

Option \(E\)

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Re: The product of three consecutive integers is equal to 571+572+573. [#permalink]

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New post 20 Sep 2017, 00:29
niks18 wrote:
petrified17 wrote:
:roll: The product of three consecutive integers is equal to 571+572+573.
What is the sum of thoseconsecutive integers?

A) 32 B) 33 C) 34 D) 35 E) 36


\(571+572+573 = 1716\)

Now, \(1716 = 11*12*13\)

So \(11+12+13=36\)

Option \(E\)


how did you derive 11*12*13 from 1716?

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Re: The number 571+572+573 can be expressed as the products of 3 [#permalink]

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New post 20 Sep 2017, 03:01
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Another method is to note that the sum of these three will end in 6.

If product of 3 consecutive integers ends in 6, the units digits of integers would be either 1, 2, 3 or 6, 7, 8.

The units digit of sum of the units digits 1, 2, 3 would be 6 giving only possible answer 36.
The units digit of sum of the units digits 6, 7, 8 would be 1 which is not there in the options.

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Re: The number 571+572+573 can be expressed as the products of 3   [#permalink] 20 Sep 2017, 03:01
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