Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2004 AMC 10B Problems/Problem 4"

(→‎Solution)
(→‎Solution 3)
 
(3 intermediate revisions by 2 users not shown)
Line 10: Line 10:
 
Clearly <math>L=60</math> and the answer is <math>720/60=\boxed{\mathrm{(B)}\ 12}</math>.
 
Clearly <math>L=60</math> and the answer is <math>720/60=\boxed{\mathrm{(B)}\ 12}</math>.
  
−
===Solution 2===
+
== Solution 2 ==
  
−
Clearly, <math>P</math> can not have a prime factor other than <math>2</math>, <math>3</math> and <math>5</math>.
+
Clearly, <math>P</math> cannot have a prime factor other than <math>2</math>, <math>3</math> and <math>5</math>.
  
 
We can not guarantee that the product will be divisible by <math>5</math>, as the number <math>5</math> can end on the bottom.
 
We can not guarantee that the product will be divisible by <math>5</math>, as the number <math>5</math> can end on the bottom.
Line 20: Line 20:
 
Finally, there are three even numbers, hence two of them are always visible and thus the product is divisible by <math>2^2</math>. This is the most we can guarantee, as when the <math>4</math> is on the bottom side, the two visible even numbers are <math>2</math> and <math>6</math>, and their product is not divisible by <math>2^3</math>.
 
Finally, there are three even numbers, hence two of them are always visible and thus the product is divisible by <math>2^2</math>. This is the most we can guarantee, as when the <math>4</math> is on the bottom side, the two visible even numbers are <math>2</math> and <math>6</math>, and their product is not divisible by <math>2^3</math>.
  
−
Hence <math>P=3\cdot 2^2 = \boxed{12}</math>.
+
== Solution 3 ==
 +
 
 +
The product P can be one of the following six numbers excluding the number that is hidden under, so we have:
 +
\begin{align*}
 +
2 \cdot 3 \cdot 4 \cdot 5 \cdot 6 = 2^4 \cdot 3^2 \cdot 5 \\
 +
1 \cdot 3 \cdot 4 \cdot 5 \cdot 6 = 2^3 \cdot 3^2 \cdot 5 \\
 +
1 \cdot 2 \cdot 4 \cdot 5 \cdot 6 = 2^4 \cdot 3 \cdot 5 \\
 +
1 \cdot 2 \cdot 3 \cdot 5 \cdot 6 = 2^2 \cdot 3^2 \cdot 5 \\
 +
1 \cdot 2 \cdot 3 \cdot 4 \cdot 6 = 2^4 \cdot 3^2 \\
 +
1 \cdot 2 \cdot 3 \cdot 4 \cdot 5 = 2^3 \cdot 3 \cdot 5
 +
\end{align*}
 +
 
 +
The largest number that is certain to divide product P is basically GCD of all the above 6 products which is <math>2^2 \cdot 3</math>.
 +
 
 +
Hence <math>P=3\cdot2^2=\boxed{\mathrm{(B)}\ 12}</math>.
  
 
== See also ==
 
== See also ==

Latest revision as of 22:42, 16 January 2024

Problem

A standard six-sided die is rolled, and $P$ is the product of the five numbers that are visible. What is the largest number that is certain to divide $P$?

$\mathrm{(A) \ } 6 \qquad \mathrm{(B) \ } 12 \qquad \mathrm{(C) \ } 24 \qquad \mathrm{(D) \ } 144\qquad \mathrm{(E) \ } 720$

Solution 1

The product of all six numbers is $6!=720$. The products of numbers that can be visible are $720/1$, $720/2$, ..., $720/6$. The answer to this problem is their greatest common divisor -- which is $720/L$, where $L$ is the least common multiple of $\{1,2,3,4,5,6\}$. Clearly $L=60$ and the answer is $720/60=\boxed{\mathrm{(B)}\ 12}$.

Solution 2

Clearly, $P$ cannot have a prime factor other than $2$, $3$ and $5$.

We can not guarantee that the product will be divisible by $5$, as the number $5$ can end on the bottom.

We can guarantee that the product will be divisible by $3$ (one of $3$ and $6$ will always be visible), but not by $3^2$.

Finally, there are three even numbers, hence two of them are always visible and thus the product is divisible by $2^2$. This is the most we can guarantee, as when the $4$ is on the bottom side, the two visible even numbers are $2$ and $6$, and their product is not divisible by $2^3$.

Solution 3

The product P can be one of the following six numbers excluding the number that is hidden under, so we have: \begin{align*} 2 \cdot 3 \cdot 4 \cdot 5 \cdot 6 = 2^4 \cdot 3^2 \cdot 5 \\ 1 \cdot 3 \cdot 4 \cdot 5 \cdot 6 = 2^3 \cdot 3^2 \cdot 5 \\ 1 \cdot 2 \cdot 4 \cdot 5 \cdot 6 = 2^4 \cdot 3 \cdot 5 \\ 1 \cdot 2 \cdot 3 \cdot 5 \cdot 6 = 2^2 \cdot 3^2 \cdot 5 \\ 1 \cdot 2 \cdot 3 \cdot 4 \cdot 6 = 2^4 \cdot 3^2 \\ 1 \cdot 2 \cdot 3 \cdot 4 \cdot 5 = 2^3 \cdot 3 \cdot 5 \end{align*}

The largest number that is certain to divide product P is basically GCD of all the above 6 products which is $2^2 \cdot 3$.

Hence $P=3\cdot2^2=\boxed{\mathrm{(B)}\ 12}$.

See also

2004 AMC 10B (Problems • Answer Key • Resources)
Preceded by
Problem 3 		Followed by
Problem 5
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 10 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2004_AMC_10B_Problems/Problem_4&oldid=210983"