Difference between revisions of "2021 AMC 12A Problems/Problem 25"
Lopkiloinm (talk | contribs) (→Solution) |
Lopkiloinm (talk | contribs) (→Solution) |
||
| Line 2: | Line 2: | ||
These problems will not be posted until the 2021 AMC12A is released on Thursday, February 4, 2021. | These problems will not be posted until the 2021 AMC12A is released on Thursday, February 4, 2021. | ||
==Solution== | ==Solution== | ||
| − | Start off with the number 1. Multiply 1 by 2. When you multiply 1 by 2, you double the number of divisors and you divide it by <math>\sqrt[3]{2}</math>. Multiply 2 by 2. When you multiply 2 by 2, you triple and then halve the number of divisors. You must not forget to divide it by <math>\sqrt[3]{2}</math> again to get that function. Keep up multiplying by 2 until it starts to decrease. Keep on multiplying by 3 until it starts to decrease. Keep on multiplying by 5 until it starts to decrease. Keep on multiplying by 7 until it starts to decrease. You cannot multiply by 11 because first multiplication automatically causes a decrease. <math>1 \rightarrow 2 \rightarrow 2^2 \rightarrow 2^3 \rightarrow 2^33 \rightarrow 2^32^2 \rightarrow 2^33^25 \rightarrow2^33^257</math> | + | Start off with the number 1. Multiply 1 by 2. When you multiply 1 by 2, you double the number of divisors and you divide it by <math>\sqrt[3]{2}</math>. Multiply 2 by 2. When you multiply 2 by 2, you triple and then halve the number of divisors. You must not forget to divide it by <math>\sqrt[3]{2}</math> again to get that function. Keep up multiplying by 2 until it starts to decrease. Keep on multiplying by 3 until it starts to decrease. Keep on multiplying by 5 until it starts to decrease. Keep on multiplying by 7 until it starts to decrease. You cannot multiply by 11 because first multiplication automatically causes a decrease. <math>1 \rightarrow 2 \rightarrow 2^2 \rightarrow 2^3 \rightarrow 2^33 \rightarrow 2^32^2 \rightarrow 2^33^25 \rightarrow2^33^257</math> This is like the computer science algorithms. |
~Lopkiloinm | ~Lopkiloinm | ||
Revision as of 15:01, 11 February 2021
Contents
Problem
These problems will not be posted until the 2021 AMC12A is released on Thursday, February 4, 2021.
Solution
Start off with the number 1. Multiply 1 by 2. When you multiply 1 by 2, you double the number of divisors and you divide it by ![$\sqrt[3]{2}$](http://latex.artofproblemsolving.com/9/f/d/9fdfe572d1673f00f286d5ed753942d18f401cbb.png)
. Multiply 2 by 2. When you multiply 2 by 2, you triple and then halve the number of divisors. You must not forget to divide it by again to get that function. Keep up multiplying by 2 until it starts to decrease. Keep on multiplying by 3 until it starts to decrease. Keep on multiplying by 5 until it starts to decrease. Keep on multiplying by 7 until it starts to decrease. You cannot multiply by 11 because first multiplication automatically causes a decrease. 
This is like the computer science algorithms.
~Lopkiloinm
Note
See problem 1.
See also
| 2021 AMC 12A (Problems • Answer Key • Resources) | |
| Preceded by Problem 24 |
Followed by Last problem |
| 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 | |
| All AMC 12 Problems and Solutions | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
