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The product of three consecutive positive integers is 8 times their su

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The product of three consecutive positive integers is 8 times their su  [#permalink]

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New post 19 Mar 2019, 00:20
00:00
A
B
C
D
E

Difficulty:

  45% (medium)

Question Stats:

63% (01:48) correct 37% (02:49) wrong based on 35 sessions

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Re: The product of three consecutive positive integers is 8 times their su  [#permalink]

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New post 19 Mar 2019, 01:18

Solution


Given:
    • The product of 3 consecutive positive integers is 8 times their sum.

To find:
    • The sum of the squares of the 3 integers.

Approach and Working:
Let us assume the 3 consecutive numbers are n – 1, n and n + 1.

As per the given condition,
    • (n – 1) n (n + 1) = 8 (n – 1 + n + n + 1)
    Or, \(n (n^2 – 1) = 24n\)
    Or, \(n^2 – 1 = 24\) (as n ≠ 0)
    Or, \(n^2 = 25\)
    Or, \(n = 5\) (as n is positive)

Hence, the integers are 4, 5 and 6.

Therefore, the sum of their squares = \(4^2 + 5^2 + 6^2 = 16 + 25 + 36 = 77\)

Hence, the correct answer is option B.

Answer: B

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Re: The product of three consecutive positive integers is 8 times their su  [#permalink]

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New post 19 Mar 2019, 04:39
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Bunuel wrote:
The product of three consecutive positive integers is 8 times their sum. What is the sum of their squares?

(A) 50
(B) 77
(C) 110
(D) 149
(E) 194


given
a-1 * a * a+1 = 8* 3a
a^2-1= 24a
(a+1)*(a-1) = 24a
a=5
so
4^2+5^2+6^2= 16+25+36 ; 77
IMO B
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Re: The product of three consecutive positive integers is 8 times their su  [#permalink]

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New post 21 Mar 2019, 06:22
I tried the same approach with x, x+1, x+2 but it seems more complicated.

How did you know that we should use x-1,x,x+1?
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Re: The product of three consecutive positive integers is 8 times their su  [#permalink]

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New post 21 Mar 2019, 08:33
Berlin92 wrote:
I tried the same approach with x, x+1, x+2 but it seems more complicated.

How did you know that we should use x-1,x,x+1?


Berlin92
well you would get the answer by using x, x+1, x+2 as well but it would be tedious and solving<2 mins would be a challenge
so use alternate option x-1,x,x+1 , saves time :cool:
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Re: The product of three consecutive positive integers is 8 times their su  [#permalink]

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New post 21 Mar 2019, 08:59
Is it generally easier to use n-1,n,n+1 for three consecutive integers or is it just for this case?
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Re: The product of three consecutive positive integers is 8 times their su  [#permalink]

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New post 21 Mar 2019, 09:05
Berlin92 wrote:
Is it generally easier to use n-1,n,n+1 for three consecutive integers or is it just for this case?


well trick is to use the most convienent and time saving method , which is why the method n-1*n*n+1 has been used..
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Re: The product of three consecutive positive integers is 8 times their su  [#permalink]

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New post 24 Mar 2019, 18:32
Bunuel wrote:
The product of three consecutive positive integers is 8 times their sum. What is the sum of their squares?

(A) 50
(B) 77
(C) 110
(D) 149
(E) 194


We let n = the middle integer; thus, we can create the equation:

(n - 1)(n)(n + 1) = 8(n - 1 + n + n + 1)

n(n^2 - 1) = 8(3n)

n^3 - n = 24n

n^3 - 25n = 0

n(n^2 - 25) = 0

n(n - 5)(n + 5) = 0

n = 0 or n = 5 or n = -5

Since we are told that the integers are positive, then n = 5 only. Therefore, the three integers are 4, 5 and 6, and the sum of their squares is 4^2 + 5^2 + 6^2 = 16 + 25 + 36 = 77.

Alternate Solution:

If we let n be the middle integer, then the sum of the squares, in terms of n, is:

(n - 1)^2 + n^2 + (n + 1)^2 = n^2 -2n + 1 + n^2 + n^2 + 2n + 1 = 3n^2 + 2

From the information given to us, we can write the equation:

(n - 1)(n)(n + 1) = 8(n - 1 + n + n + 1)

n(n^2 - 1) = 8(3n)

Since all the integers are positive, we can divide each side by n:

n^2 - 1 = 24

n^2 = 25

Thus, the sum of their squares is 3n^2 + 2 = 3(25) + 2 = 77.

Answer: B
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Re: The product of three consecutive positive integers is 8 times their su   [#permalink] 24 Mar 2019, 18:32
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