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# If {x} is the product of all integer multiples of 5 from 1 to x, inclu

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Math Expert
Joined: 02 Sep 2009
Posts: 47037
If {x} is the product of all integer multiples of 5 from 1 to x, inclu [#permalink]

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02 Oct 2017, 00:59
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Difficulty:

55% (hard)

Question Stats:

52% (01:22) correct 48% (01:27) wrong based on 103 sessions

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If {x} is the product of all integer multiples of 5 from 1 to x, inclusive, what is the greatest prime number, y, such that ({36} + {41})/y is an integer?

A. 5
B. 11
C. 19
D. 23
E. 41

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Re: If {x} is the product of all integer multiples of 5 from 1 to x, inclu [#permalink]

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02 Oct 2017, 01:38
2
Bunuel wrote:
If {x} is the product of all integer multiples of 5 from 1 to x, inclusive, what is the greatest prime number, y, such that ({36} + {41})/y is an integer?

A. 5
B. 11
C. 19
D. 23
E. 41

{36} = 5*10*15*20*25*30*35 = 5^7 (7!)
{41} = 5*10*15*20*25*30*35*40 = 5^8 (8!)

({36} + {41}) = 5^7 (7!) (1+8*5) = 5^7 (7!) (41)

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Re: If {x} is the product of all integer multiples of 5 from 1 to x, inclu [#permalink]

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03 Oct 2017, 09:34
{x} is the product of all integer multiples of 5 from 1 to x.

So {36} = 5*10*15*20*25*30*35
So {41} = 5*10*15*20*25*30*35*40

{36} + {41} = (5*10*15*20*25*30*35) + (5*10*15*20*25*30*35*40)
= (5*10*15*20*25*30*35) (1 + 40)
= (5*10*15*20*25*30*35) * 41

Question asks us what is the greatest prime number, y, such that ({36} + {41})/y is an integer?

(5*10*15*20*25*30*35) * 41 --> 41 is the greatest prime number y can be so that the ({36} + {41})/y is still an integer
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Re: If {x} is the product of all integer multiples of 5 from 1 to x, inclu [#permalink]

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04 Oct 2017, 16:37
1
Bunuel wrote:
If {x} is the product of all integer multiples of 5 from 1 to x, inclusive, what is the greatest prime number, y, such that ({36} + {41})/y is an integer?

A. 5
B. 11
C. 19
D. 23
E. 41

Let’s simplify {36} and {41}:

{36} = 5 x 10 x 15 x 20 x 25 x 30 x 35

{41} = 5 x 10 x 15 x 20 x 25 x 30 x 35 x 40

Thus:

{36} + {41} = (5 x 10 x 15 x 20 x 25 x 30 x 35) + (5 x 10 x 15 x 20 x 25 x 30 x 35 x 40)

{36} + {41} = 5 x 10 x 15 x 20 x 25 x 30 x 35 x (1 + 40)

{36} + {41} = 5 x 10 x 15 x 20 x 25 x 30 x 35 x 41

Thus, the largest prime that will divide the above expression is 41.

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Joined: 26 Sep 2017
Posts: 14
Re: If {x} is the product of all integer multiples of 5 from 1 to x, inclu [#permalink]

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07 Oct 2017, 03:16
Hi!
I didn't understand that solution. Can anybody help me with that?
I can understand the following, but the rest I dont!
{36} = 5 x 10 x 15 x 20 x 25 x 30 x 35
{41} = 5 x 10 x 15 x 20 x 25 x 30 x 35 x 40
{36} + {41} = (5 x 10 x 15 x 20 x 25 x 30 x 35) + (5 x 10 x 15 x 20 x 25 x 30 x 35 x 40)
Math Expert
Joined: 02 Sep 2009
Posts: 47037
Re: If {x} is the product of all integer multiples of 5 from 1 to x, inclu [#permalink]

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07 Oct 2017, 03:24
lor12345 wrote:
If {x} is the product of all integer multiples of 5 from 1 to x, inclusive, what is the greatest prime number, y, such that ({36} + {41})/y is an integer?

A. 5
B. 11
C. 19
D. 23
E. 41

Hi!
I didn't understand that solution. Can anybody help me with that?
I can understand the following, but the rest I dont!
{36} = 5 x 10 x 15 x 20 x 25 x 30 x 35
{41} = 5 x 10 x 15 x 20 x 25 x 30 x 35 x 40
{36} + {41} = (5 x 10 x 15 x 20 x 25 x 30 x 35) + (5 x 10 x 15 x 20 x 25 x 30 x 35 x 40)

{36} + {41} = (5 x 10 x 15 x 20 x 25 x 30 x 35) + (5 x 10 x 15 x 20 x 25 x 30 x 35 x 40)

Factor out common part which is 5 x 10 x 15 x 20 x 25 x 30 x 35 to get 5 x 10 x 15 x 20 x 25 x 30 x 35 x (1 + 40);

5 x (2 x 5) x (3 x 5) x (2^2 x 5) x (5^2) x (2 x 3 x 5) x (5 x 7) x 41.

The largest prime that will divide the above expression is 41.

Hope it's clear.
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Joined: 26 Sep 2017
Posts: 14
Re: If {x} is the product of all integer multiples of 5 from 1 to x, inclu [#permalink]

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08 Oct 2017, 09:05
Bunuel wrote:
lor12345 wrote:
If {x} is the product of all integer multiples of 5 from 1 to x, inclusive, what is the greatest prime number, y, such that ({36} + {41})/y is an integer?

A. 5
B. 11
C. 19
D. 23
E. 41

Hi!
I didn't understand that solution. Can anybody help me with that?
I can understand the following, but the rest I dont!
{36} = 5 x 10 x 15 x 20 x 25 x 30 x 35
{41} = 5 x 10 x 15 x 20 x 25 x 30 x 35 x 40
{36} + {41} = (5 x 10 x 15 x 20 x 25 x 30 x 35) + (5 x 10 x 15 x 20 x 25 x 30 x 35 x 40)

{36} + {41} = (5 x 10 x 15 x 20 x 25 x 30 x 35) + (5 x 10 x 15 x 20 x 25 x 30 x 35 x 40)

Factor out common part which is 5 x 10 x 15 x 20 x 25 x 30 x 35 to get 5 x 10 x 15 x 20 x 25 x 30 x 35 x (1 + 40);

5 x (2 x 5) x (3 x 5) x (2^2 x 5) x (5^2) x (2 x 3 x 5) x (5 x 7) x 41.

The largest prime that will divide the above expression is 41.

Hope it's clear.

Thank you! Yes, now it's clear!
Re: If {x} is the product of all integer multiples of 5 from 1 to x, inclu   [#permalink] 08 Oct 2017, 09:05
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