Difference between revisions of "1968 AHSME Problems/Problem 15"
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== Solution == | == Solution == | ||
| − | <math>\fbox{}</math> | + | 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> | ||
| + | *In first step we could also apply modular arthmetic and get the same answer. | ||
== See also == | == See also == | ||
| − | {{AHSME box|year=1968|num-b=14|num-a=16}} | + | {{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 
be the product of any three consecutive positive odd integers. The largest integer dividing all such is:
Solution
Product can be written as 
,
,
. Because is defined as a "3 consecutive odd integer" product impies that 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.
- 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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
