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 Post subject: Arithmetic Problem #82, page 98Posted: Tue Aug 10, 2010 10:51 am

Joined: Mon Jun 14, 2010 9:07 am
Posts: 1
After read the explanation in the end of the Book II, I still don't understand how to do exercise 82.

"X is the product of all even numbers from 2 to 50, inclusive. The smallest prime factor of x+1 must be:

a) Between 1 and 10
b) Between 11 and 15
c) Between 15 and 20
d) Between 20 and 25
e) Greater than 25"

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 Post subject: Re: Arithmetic ProblemPosted: Wed Aug 11, 2010 1:59 pm

Joined: Thu Jul 03, 2008 2:13 pm
Posts: 117
Hey Guilherme,

This problem is one of my favorites - it's a huge number, so you'll probably want to test the concept with smaller numbers and then just extrapolate it to the larger one in question (a pretty good strategy on these big-number number properties problems in general).

Look at it this way: Every second number (2, 4, 6, 8, 10...) is divisible by 2. But if you add 1 to any of them, it's no longer divisible by 2 (3, 5, 7, etc.). You're off that "every second" cycle by adding 1.

Same thing for 3: every third number (3, 6, 9, 12...) is divisible by 3, but if you add 1 you're off that every third cycle (4, 7, 10, 13...).

Essentially, if you take a number and add 1 to it, it's no longer divisible by any of its previous factors (other than 1), as it's off of each of those cycles.

If you take the number 2*4*6*8...*50 and break it down to its prime factors, you know that it's divisible by every prime number up to 25. If you add 1 to it, it won't be divisible by those factors anymore. That's why the answer is E - it won't be divisible by any prime factors that are less than 25.

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 Post subject: Re: Arithmetic ProblemPosted: Wed Oct 27, 2010 6:17 am

Joined: Thu Oct 21, 2010 5:19 am
Posts: 16
Fantastic! Very clear! Thank you!

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 Post subject: Re: Arithmetic ProblemPosted: Fri Feb 18, 2011 11:24 am

Joined: Fri Feb 04, 2011 9:29 pm
Posts: 1

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 Post subject: Re: Arithmetic ProblemPosted: Thu Jun 02, 2011 6:09 pm

Joined: Mon Oct 04, 2010 6:09 pm
Posts: 2
My apologies, however I'm still a little unclear, as to how the smallest prime factor of x+1, is greater than 25, when X is the product of all even numbers from 2 to 50, inclusive.

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 Post subject: Re: Arithmetic ProblemPosted: Sat Jun 04, 2011 9:22 am

Joined: Thu Feb 12, 2009 6:32 pm
Posts: 497
x = 2*4*6 * ... 48*50

x = 2*1 * 2*2 * 2*3 * ... 2*24 * 2*25

x = 2^25 * 1*2*3* ... 24*25

so, all numbers from 2 to 25 are factors of x

if they are factors of x they can't be factors of a number that is one larger than x

so, the smallest prime factor must be larger than 25 since it can't be between 2 and 25

for example ... numbers that are factors of 65 can't be factors of 65, numbers that are factors of 100 can't be factors of 101, etc.

-Veritas Help

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