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# The product of 3 consecutive positive multiples of 4 must be divisible

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Math Expert
Joined: 02 Sep 2009
Posts: 57155
The product of 3 consecutive positive multiples of 4 must be divisible  [#permalink]

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09 Aug 2019, 06:27
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Difficulty:

45% (medium)

Question Stats:

70% (01:29) correct 30% (01:32) wrong based on 54 sessions

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The product of 3 consecutive positive multiples of 4 must be divisible by each of the following EXCEPT:

(A) 16
(B) 36
(C) 48
(D) 128
(E) 192

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Joined: 20 Nov 2018
Posts: 18
Re: The product of 3 consecutive positive multiples of 4 must be divisible  [#permalink]

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09 Aug 2019, 07:22
1
The product of 3 consecutive positive multiples of 4 must be divisible by each of the following EXCEPT:

Solution :-

The Least Consecutive Multiples of 4 are (4, 8, 12).
Prime Factorization of product of least multiples of 4 is :- $$2^7*3$$

So for the product to be divisible by a number in the answer choices, the answer choices available to us must have only 7 powers of 2's and only 1 power of 3 in their prime factorization. Lets check 'em out.

(A) 16 -- $$2^4$$

(B) 36 -- $$2^2*3^2$$ (2 powers of 3....Huhhh... )

(C) 48 -- $$2^4*3$$

(D) 128 -- $$2^7$$

(E) 192-- $$2^6*3$$

All the other Answer choices are with in our constraints, except B. So.....

IMO B

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Re: The product of 3 consecutive positive multiples of 4 must be divisible  [#permalink]

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09 Aug 2019, 07:31
(b)36

considering smallest consecutive positive multiples of 4 (4,8,12), we get 2^7 x 3^1.
All options except (b)36 have 1 or 0 3's in it. 36 has 2 3's in it hence it may or may not be a factor of the progression.
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Location: India
Re: The product of 3 consecutive positive multiples of 4 must be divisible  [#permalink]

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09 Aug 2019, 08:54
Bunuel wrote:
The product of 3 consecutive positive multiples of 4 must be divisible by each of the following EXCEPT:

(A) 16
(B) 36
(C) 48
(D) 128
(E) 192

Asked: The product of 3 consecutive positive multiples of 4 must be divisible by each of the following EXCEPT:

Product of 3 consecutive positive multiples of 4 is divisible by 4^3.3 = 64*2*3 = 2^7*3

(A) 16 = 2^4
(B) 36 =2^2*3^2
(C) 48 = 2^4*3
(D) 128 = 2^7
(E) 192 = 2^6*3

(B)36=3^2*2^2 require 3^2 NOT FEASIBLE

IMO B
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Re: The product of 3 consecutive positive multiples of 4 must be divisible  [#permalink]

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09 Aug 2019, 20:23
Hi All,

We're asked to consider the PRODUCT of 3 CONSECUTIVE positive multiples of 4. We're asked to find the number that is NOT necessarily divisible into that product. This question can be solved rather easily by TESTing VALUES.

IF... the three consecutive multiples of 4 are 4, 8 and 12, then the PRODUCT of those 3 numbers is (4)(8)(12) = 384. Four of the answer choices will divide evenly into 384; which one will NOT....? As a built-in shortcut, once you determine which answer does not divide evenly in, then that is your answer...

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Re: The product of 3 consecutive positive multiples of 4 must be divisible  [#permalink]

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12 Aug 2019, 19:04
1
Bunuel wrote:
The product of 3 consecutive positive multiples of 4 must be divisible by each of the following EXCEPT:

(A) 16
(B) 36
(C) 48
(D) 128
(E) 192

We can use the 3 smallest positive multiples of 4, so we have:

4 x 8 x 12

Factoring the above, we have:

2^2 x 2^3 x 2^2 x 3 = 2^7 x 3

Since 36 = 3^2 x 2^2, and since the above product contains only one factor of 3, 36 is the answer.

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Re: The product of 3 consecutive positive multiples of 4 must be divisible   [#permalink] 12 Aug 2019, 19:04
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