AMC 10 Daily Practice Round 1

Complete problem set with solutions and individual problem pages

Problem 16 Hard

21!={{12}^{n}}\cdot M, where M is a positive integer. When n is the greatest positive integer that makes this equation valid, which of the following is the factor of M?

  • A.

    2

  • B.

    3

  • C.

    2 and 4

  • D.

    2 and 3

  • E.

    neither 2, nor 3

Answer:E

The number of factors of 2 in p:

\left[ 21\div 2 \right]+\left[ 21\div 4 \right]+\left[ 21\div 8 \right]+\left[ 21\div 16 \right]

=10+5+2+1

=18．

The number of factors of 3 in p:

\left[ 21\div 3 \right]+\left[ 21\div 9 \right]

=7+2

=9.

 

Since 12 is composed of 2^2 \times 3, each group of two 2's and one 3 forms a factor of 12. The number of complete groups we can form is:

 

\min\left(\frac{18}{2}, 9\right) = \min(9,9) = 9.

 

Thus, we can form 9 complete groups of 12, meaning:

 

n = 9.

 

Since all factors of 2 and 3 have been used to form 12, the remaining factor M does not contain 2 or 3, meaning M is not a multiple of 2 or 3.

 

Related Problems
Problem 14 AMC 10 Daily Practice Round 1
Problem 15 AMC 10 Daily Practice Round 1
Problem 17 AMC 10 Daily Practice Round 1
Problem 18 AMC 10 Daily Practice Round 1
Think Academy Powered by ChatGPT