2000 PMWC Problems/Problem I15: Difference between revisions
| Line 3: | Line 3: | ||
==Solution== | ==Solution== | ||
Try 5: 2*3*4*5*6 = 720 | Try <cmath>5: 2*3*4*5*6 = 720</cmath> | ||
Try 4: 3*4*6*7 = 504 | Try <cmath>4: 3*4*6*7 = 504</cmath> | ||
Try 3 | Try 3, then realize it's going to be smaller. | ||
6 numbers is too many | 6 numbers is too many, since <cmath>1+2+3+4+5+6=21</cmath> | ||
Thus the answer is 720 | |||
==See Also== | ==See Also== | ||
Latest revision as of 13:26, 31 January 2017
Problem
The sum of several positive integers is 
. Find the largest product that can be formed by these integers.
Solution
Try ![\[5: 2*3*4*5*6 = 720\]](http://latex.artofproblemsolving.com/1/1/c/11c22eb0174181ecd2c30a1d2d27b88d1639ddac.png)
Try ![\[4: 3*4*6*7 = 504\]](http://latex.artofproblemsolving.com/4/5/3/453667a07be111ae92326dca9b4f7a0f081da41d.png)
Try 3, then realize it's going to be smaller.
6 numbers is too many, since ![\[1+2+3+4+5+6=21\]](http://latex.artofproblemsolving.com/7/c/e/7cea75cd91286e850ef4923b19782b463dc49be2.png)
Thus the answer is 720
