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The sum of three integers is 40. The largest integer is 3

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The sum of three integers is 40. The largest integer is 3 [#permalink]

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The sum of three integers is 40. The largest integer is 3 times the middle integer, and the smallest integer is 23 less than the largest integer. What is the product of the three integers?

A) 1,104
B) 972
C) 672
D) 294
E) 192

Who can solve this in less than 60 seconds? And how?
[Reveal] Spoiler: OA

Last edited by Bunuel on 26 Feb 2013, 02:31, edited 1 time in total.
Renamed the topic.

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Re: Algebriac Translation [#permalink]

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New post 25 Feb 2013, 19:53
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mejia401 wrote:
62. The sum of three integers is 40. The largest integer is 3 times the middle integer, and the smallest integer is 23 less than the largest integer. What is the product of the three integers?

A) 1,104
B) 972
C) 672
D) 294
E) 192

Who can solve this in less than 60 seconds? And how?


Actually, from what I know of the people on this forum, many of them can solve it in under a minute.

Say middle integer is x. Largest is 3x. Smallest is 3x - 23.

3x - 23 + x + 3x = 40
x = 9
The middle integer is 9.

Now look for the option which is divisible by 9 (since the product of the 3 integers will be a multiple of 9). Sum of the digits of the number should be divisible by 9 for the number to be divisible by 9.

(B) 9+7+2 = 18 which is divisible by 9.
Only (B) works.
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Re: Algebriac Translation [#permalink]

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New post 25 Feb 2013, 20:23
VeritasPrepKarishma wrote:
mejia401 wrote:
62. The sum of three integers is 40. The largest integer is 3 times the middle integer, and the smallest integer is 23 less than the largest integer. What is the product of the three integers?

A) 1,104
B) 972
C) 672
D) 294
E) 192

Who can solve this in less than 60 seconds? And how?


Actually, from what I know of the people on this forum, many of them can solve it in under a minute.

Say middle integer is x. Largest is 3x. Smallest is 3x - 23.

3x - 23 + x + 3x = 40
x = 9
The middle integer is 9.

Now look for the option which is divisible by 9 (since the product of the 3 integers will be a multiple of 9). Sum of the digits of the number should be divisible by 9 for the number to be divisible by 9.

(B) 9+7+2 = 18 which is divisible by 9.
Only (B) works.


Yes, that's it...perfect. I see my mistake now. Thank you so much VeritasPrepKarishma!

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Re: Algebriac Translation [#permalink]

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New post 25 Feb 2013, 21:07
VeritasPrepKarishma wrote:
mejia401 wrote:
62. The sum of three integers is 40. The largest integer is 3 times the middle integer, and the smallest integer is 23 less than the largest integer. What is the product of the three integers?

A) 1,104
B) 972
C) 672
D) 294
E) 192



Let the middle integer be x
Then the largest integer will be 3x
The smallest integer will be 3x – 23
Sum = x + 3x + 3x – 23 = 40
Then 7x = 63 and x = 9
Hence the numbers are 9(27) (4), since the option must be divisible by 9 hence the answer is (2).
(To check whether a number is divisible by 9, check if the sum of digits is divisible by 9)
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Re: The sum of three integers is 40. The largest integer is 3 [#permalink]

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New post 06 Jul 2014, 10:37
A+B+C = 40
A = C-23
B = C/3
Solving these 3 equations => C = 27, B = 9, A = 4 => ABC = 972

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Re: The sum of three integers is 40. The largest integer is 3 [#permalink]

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New post 13 Nov 2016, 22:31
mejia401 wrote:
The sum of three integers is 40. The largest integer is 3 times the middle integer, and the smallest integer is 23 less than the largest integer. What is the product of the three integers?

A) 1,104
B) 972
C) 672
D) 294
E) 192

Who can solve this in less than 60 seconds? And how?


The three integers are

a + b + c = 40

3b - 23 + b + 3b = 40

So. 7b = 63

Or, b = 9

So, The three numbers are 4 , 9 & 27

Hence, product of three numbers is 4*9*27 = 972

Thus, the answer will be (B) 972

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Re: The sum of three integers is 40. The largest integer is 3 [#permalink]

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Re: The sum of three integers is 40. The largest integer is 3   [#permalink] 02 Dec 2017, 14:56
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