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Difference between revisions of "2007 iTest Problems/Problem 43"

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She notes that two of them have exactly <math>8</math> positive divisors each. Find the common prime divisor of those two integers.
 
She notes that two of them have exactly <math>8</math> positive divisors each. Find the common prime divisor of those two integers.
  
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== Solution ==
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==Solution==
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Notice that each number can be written in form <math>n(n+12)(n+18) + 320</math>, where <math>n</math> is an integer from <math>700</math> to <math>799</math>.  Expanding the [[polynomial]] results in <math>n^3 + 30n^2 + 216n + 320</math>, and factoring that results in <math>(n+2)(n+8)(n+20)</math>.
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<br>
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If a number has eight positive divisors, then it’s [[prime factorization]] is in the form <math>a^7</math>, <math>a^3b</math>, or <math>abc</math>, where <math>a</math>, <math>b</math>, and <math>c</math> are primes.  The most likely option is three different [[prime|prime numbers]] being multiplied together.
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<br>
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After finding the prime numbers from <math>702</math> to <math>819</math>, the values of <math>n</math> that work are <math>731</math> and <math>749</math> (after finding one value, note that both share a factor to find the other value).  Thus, the prime factorizations of the two numbers are <math>733 \cdot 739 \cdot 751</math> and <math>751 \cdot 757 \cdot 769</math>, so the common prime divisor of the two is <math>\boxed{751}</math>.
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==See Also==
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{{iTest box|year=2007|num-b=42|num-a=44}}
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[[Category:Intermediate Number Theory Problems]]

Latest revision as of 04:22, 16 June 2018

Problem

Bored of working on her computational linguistics thesis, Erin enters some three-digit integers into a spreadsheet, then manipulates the cells a bit until her spreadsheet calculates each of the following one hundred $9$-digit integers:

\begin{align*}700\cdot 712\,\cdot\, &718+320,\\701\cdot 713\,\cdot\, &719+320,\\ 702\cdot 714\,\cdot\, &720+320,\\&\vdots\\798\cdot 810\,\cdot\, &816+320,\\799\cdot 811\,\cdot\, &817+320.\end{align*}

She notes that two of them have exactly $8$ positive divisors each. Find the common prime divisor of those two integers.

Solution

Notice that each number can be written in form $n(n+12)(n+18) + 320$, where $n$ is an integer from $700$ to $799$. Expanding the polynomial results in $n^3 + 30n^2 + 216n + 320$, and factoring that results in $(n+2)(n+8)(n+20)$.


If a number has eight positive divisors, then it’s prime factorization is in the form $a^7$, $a^3b$, or $abc$, where $a$, $b$, and $c$ are primes. The most likely option is three different prime numbers being multiplied together.


After finding the prime numbers from $702$ to $819$, the values of $n$ that work are $731$ and $749$ (after finding one value, note that both share a factor to find the other value). Thus, the prime factorizations of the two numbers are $733 \cdot 739 \cdot 751$ and $751 \cdot 757 \cdot 769$, so the common prime divisor of the two is $\boxed{751}$.

See Also

2007 iTest (Problems, Answer Key)
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Problem 42 		Followed by:
Problem 44
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