Difference between revisions of "1968 AHSME Problems/Problem 15"
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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 | + | 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 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> | ||
Revision as of 03:18, 17 December 2020
Problem
Let 
be the product of any three consecutive positive odd integers. The largest integer dividing all such is:
Solution
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 E is ruled out because we have already proved that 3 is possible, and the question asks for greatest possible.
See also
| 1968 AHSME (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 | ||
| All AHSME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
