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The expression x#y denotes the product of the consecutive multiples of

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The expression x#y denotes the product of the consecutive multiples of  [#permalink]

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New post 24 Apr 2015, 03:10
2
6
00:00
A
B
C
D
E

Difficulty:

  45% (medium)

Question Stats:

70% (02:55) correct 30% (02:33) wrong based on 167 sessions

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Re: The expression x#y denotes the product of the consecutive multiples of  [#permalink]

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New post 24 Apr 2015, 10:36
1
1
21#42
= 21*24*27*30*33*36*39*42
= 3^11 * 7^2 * 2^7 * 5^1 * 11^1 * 13^1

so total is = 11+2+7+1+1+1=23

Answer A
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Re: The expression x#y denotes the product of the consecutive multiples of  [#permalink]

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New post 24 Apr 2015, 11:31
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1
Bunuel wrote:
The expression x#y denotes the product of the consecutive multiples of 3 between x and y, inclusive. What is the sum of the exponents in the prime factorization of 21#42?

A. 23
B. 24
C. 25
D. 26
E. 27


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21#42 = 21*24*27*30*33*36*39*42 => 3(7*8*9*10*11*12*13*14) => \(3^{11} * {2^7}* {7^2}*5*11*13\) => 11+7+2+1+1+1 = 23

Hence, answer is A
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Re: The expression x#y denotes the product of the consecutive multiples of  [#permalink]

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New post 25 Apr 2015, 12:45
1
The question is basically asking for the total number of factors between 21 and 42, inclusive.
21, 24, 27, 30, 33, 36, 39, 42

2^7 = 7
3^11= 11
5^1= 1
7^2= 2
11^1= 1
13^1= 1

23 is the total

Answer: A
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Re: The expression x#y denotes the product of the consecutive multiples of  [#permalink]

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New post 25 Apr 2015, 19:43
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The product of the consecutive multiples of 3 between 21 and 42 inclusive
= 21 * 24 * 27 * 30 * 33 * 36 * 39 * 42

Factorize them separately and we get,

= (3 * 7) * (3 * \(2^3\)) * (\(3^3\)) * (3 * 2 * 5) * (3 * 11) * (\(3^2\) * \(2^2\)) * (3 * 13) * (3 * 2 * 7)

= (\(3^{11}\)) * (\(2^7\)) * (\(5^1\)) * (\(7^2\)) *(\(11^1\)) * (\(13^1\))

Sum of exponents = 11 + 7 + 1 + 2 + 1 + 1 = 23

Correct answer is (A)
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Re: The expression x#y denotes the product of the consecutive multiples of  [#permalink]

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New post 27 Apr 2015, 01:54
Bunuel wrote:
The expression x#y denotes the product of the consecutive multiples of 3 between x and y, inclusive. What is the sum of the exponents in the prime factorization of 21#42?

A. 23
B. 24
C. 25
D. 26
E. 27


Kudos for a correct solution.


MANHATTAN GMAT OFFICIAL SOLUTION:

First, translate the expression 21#42, using the definition given:

21#42 = 21×24×27×30×33×36×39×42

You need the prime factorization of this enormous product.

Since these are consecutive multiples of 3, a good move is to factor out that 3 from each multiple. You have 8 multiples, all multiplied together, so you get 3^8:

21#42 = 3^8(7×8×9×10×11×12×13×14)

Now replace each consecutive integer with its prime factorization:

21#42 = 3^8(7×2^3×3^2×(2×5)×11×(2^2×3)×13×(2×7))

Group up the prime bases:

21#42 = 2^7×3^11×5×7^2×11×13

Don’t forget to put in an understood 1 for the primes lacking explicit exponents:

21#42 = 2^7×3^11×5^1×7^2×11^1×13^1

Finally, add up the exponents:

7 + 11 + 1 + 2 + 1 + 1 = 23

The correct answer is A.
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Re: The expression x#y denotes the product of the consecutive multiples of  [#permalink]

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New post 06 May 2016, 13:49
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LMAO this should be on the verbal section!!!

Took me 6 minutes to figure out what "product of the consecutive multiples of 3 between x and y" and "sum of the exponents in the prime factorization" meant. Another 3 to actually solve it, but at least I figured it out by myself in the end. I guess the right part of my brain kept spaming messages to the left asking "Why would you wanna do this? Why would you wanna do that? Whats the context? This makes no sense!!".
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Re: The expression x#y denotes the product of the consecutive multiples of  [#permalink]

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Re: The expression x#y denotes the product of the consecutive multiples of   [#permalink] 11 Apr 2019, 20:07
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