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VP  P
Joined: 07 Dec 2014
Posts: 1224
A four-digit number with consecutive positive integers is a product of  [#permalink]

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Difficulty:   65% (hard)

Question Stats: 57% (02:41) correct 43% (02:35) wrong based on 77 sessions

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A four-digit number with consecutive positive integers is a product of 2x and x^2. What is the sum of these two factors?

A. 80
B. 99
C. 168
D. 224
E. 288
Math Expert V
Joined: 02 Aug 2009
Posts: 7984
A four-digit number with consecutive positive integers is a product of  [#permalink]

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gracie wrote:
A four-digit number with consecutive positive integers is a product of 2x and x^2. What is the sum of these two factors?

A. 80
B. 99
C. 168
D. 224
E. 288

Number is 4-digit and =2x*x^2=2x^3, thus an even number..
Cases
1) 1234..
$$2x^3=1234....x^3=617... Between 8^3--9^3$$ , so no integer value possible
2) 3456..
$$2x^3=3456......x^3=1728......x=12$$, so we have a possible value
We are looking for $$2x+x^2=2*12+12^2=24+144=168$$

C
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Senior Manager  S
Joined: 12 Sep 2017
Posts: 302
Re: A four-digit number with consecutive positive integers is a product of  [#permalink]

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chetan2u wrote:
gracie wrote:
A four-digit number with consecutive positive integers is a product of 2x and x^2. What is the sum of these two factors?

A. 80
B. 99
C. 168
D. 224
E. 288

Number is 4-digit and =2x*x^2=2x^3, thus an even number..
Cases
1) 1234..
$$2x^3=1234....x^3=617... Between 8^3--9^3$$ , so no integer value possible
2) 3456..
$$2x^3=3456......x^3=1728......x=12$$, so we have a possible value
We are looking for $$2x+x^2=2*12+12^2=24+144=168$$

C

Hello!

Do you have another approach?

Kind regards! Re: A four-digit number with consecutive positive integers is a product of   [#permalink] 05 Jan 2019, 17:42
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