Difference between revisions of "2021 AMC 12A Problems/Problem 25"

(Solution)
(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==
Randomly guess that the answer is 2520 and it adds up to 9.
+
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. $1 \rightarrow 2 \rightarrow 2^2 \rightarrow 2^3 \rightarrow 2^33 \rightarrow 2^32^2 \rightarrow 2^33^25 \rightarrow2^33^257
 
~Lopkiloinm
 
~Lopkiloinm
  

Revision as of 15:00, 11 February 2021

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 $\sqrt[3]{2}$. 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 $\sqrt[3]{2}$ 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. $1 \rightarrow 2 \rightarrow 2^2 \rightarrow 2^3 \rightarrow 2^33 \rightarrow 2^32^2 \rightarrow 2^33^25 \rightarrow2^33^257 ~Lopkiloinm

Note

See problem 1.

See also

2021 AMC 12A (ProblemsAnswer KeyResources)
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. AMC logo.png