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Difference between revisions of "2005 Cyprus Seniors TST/Day 1/Problem 4"

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== Solution ==
 
== Solution ==
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We first prove that there is a multiple of 2 and a multiple of 3 in three consecutive integers:
  
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n, n+1, n+2
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If n is odd, n+1 is even, and therefore a multiple of 2. If n is even, n is a multiple of 2.
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n can be a multiple of 3. If <math>n\equiv 1 \pmod 3</math>, then n+2 is a multiple of 3. If <math>n\equiv 2 \pmod 3</math>, n+1 is a multiple of 3.
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 +
Therefore, there is a multiple of 2 and a multiple of 3 in any three consecutive integers. Therefore, the product of three consecutive integers is a multiple of 6.
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Now we prove the second:
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<math>\nu=2x-1</math>
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<math>(10x-4)(10x-2)10x</math>
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10 is in there. The argument used on the last theorem can be used here to get that 3 is in here. Now we just need to find an 8 in here.
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We can use <math>(10x-4)(10x-2)</math>. We can divide out a 4:
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<math>4(5x-2)(5x-1)</math>
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We prove that there is a 2 in there, so there is an 8 in there.
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240 divides the product <math>(5\nu+1)(5\nu+3)(5\nu+5)</math>, where <math>\nu</math> is any odd number.
  
  

Latest revision as of 20:12, 9 October 2007

Problem

Prove that number 6 divides the product of every three consecutive integer numbers and then show that 240 divides the product $(5\nu+1)(5\nu+3)(5\nu+5)$, where $\nu$ is any odd number.

Solution

We first prove that there is a multiple of 2 and a multiple of 3 in three consecutive integers:

n, n+1, n+2

If n is odd, n+1 is even, and therefore a multiple of 2. If n is even, n is a multiple of 2.


n can be a multiple of 3. If $n\equiv 1 \pmod 3$, then n+2 is a multiple of 3. If $n\equiv 2 \pmod 3$, n+1 is a multiple of 3.

Therefore, there is a multiple of 2 and a multiple of 3 in any three consecutive integers. Therefore, the product of three consecutive integers is a multiple of 6.

Now we prove the second:

$\nu=2x-1$

$(10x-4)(10x-2)10x$

10 is in there. The argument used on the last theorem can be used here to get that 3 is in here. Now we just need to find an 8 in here.

We can use $(10x-4)(10x-2)$. We can divide out a 4:

$4(5x-2)(5x-1)$

We prove that there is a 2 in there, so there is an 8 in there.

240 divides the product $(5\nu+1)(5\nu+3)(5\nu+5)$, where $\nu$ is any odd number.


See also

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