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Difference between revisions of "1968 AHSME Problems/Problem 15"

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== Solution ==
 
== Solution ==
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Notice that this product can be written as n-2,n,n+2. Because it is 3 consecutive odd integers we know that it must be divisible by at least 3. A is ruled out because factors of 5 only arise every 5 terms, if we were to take the 3 terms in the middle of the factors of 5 we wouldn't have a factor of 5. IE: 7,9,11. Obviously B is impossible because we are multiplying odd numbers and 2 would never become one of our prime factors.C is ruled out with the same logic as A. Lastly D is ruled out because we have already proved that 3 is possible, and the question asks for greatest possible.
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Product <math>P</math> can be written as <math>2n-1</math>,<math>2n+1</math>,<math>2n+3</math>. Because <math>P</math> is defined as a "3 consecutive odd integer" product impies that <math>P</math> must be divisible by at least 3. A is ruled out because factors of 5 only arise every 5 terms, if we were to take the 3 terms in the middle of the factors of 5 we wouldn't have a factor of 5. Obviously B is impossible because we are multiplying odd numbers and 2 would never become one of our prime factors. C is ruled out with the same logic as A. Lastly E is ruled out because we have already proved that 3 is possible, and the question asks for greatest possible.
 
<math>\fbox{D}</math>
 
<math>\fbox{D}</math>
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*In first step we could also apply modular arthmetic and get the same answer.
  
 
== See also ==
 
== See also ==
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{{AHSME box|year=1968|num-b=14|num-a=16}}   
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{{AHSME 35p box|year=1968|num-b=14|num-a=16}}   
  
 
[[Category: Introductory Number Theory Problems]]
 
[[Category: Introductory Number Theory Problems]]
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 08:24, 1 January 2024

Problem

Let $P$ be the product of any three consecutive positive odd integers. The largest integer dividing all such $P$ is:

$\text{(A) } 15\quad \text{(B) } 6\quad \text{(C) } 5\quad \text{(D) } 3\quad \text{(E) } 1$

Solution

Product $P$ can be written as $2n-1$,$2n+1$,$2n+3$. Because $P$ is defined as a "3 consecutive odd integer" product impies that $P$ must be divisible by at least 3. A is ruled out because factors of 5 only arise every 5 terms, if we were to take the 3 terms in the middle of the factors of 5 we wouldn't have a factor of 5. Obviously B is impossible because we are multiplying odd numbers and 2 would never become one of our prime factors. C is ruled out with the same logic as A. Lastly E is ruled out because we have already proved that 3 is possible, and the question asks for greatest possible. $\fbox{D}$

  • In first step we could also apply modular arthmetic and get the same answer.

See also

1968 AHSC (Problems • Answer Key • Resources)
Preceded by
Problem 14 		Followed by
Problem 16
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35
All AHSME Problems and Solutions

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