# Number puzzle...

• Apr 7th 2009, 08:59 AM
Vedicmaths
Number puzzle...
1) Which of the five numbers 2007, 2008, 2009, 2010 and 2011 has the largest number of factors and which one has the fewest number of factors?

2) And determine the total number of factors for the number:
2007 * 2008 * 2009 * 2010 * 2011

For example, 21 has 4 factors (1,3,7,21) and 20 has 6 factors (1,2,4,5,10,20)

I know this is a very stupid problem but I am tired of solving the factors and there are like millions of them...And I know there is a trick of doing it! can you please help me figuring it out?

Thanks a lot!
• Apr 7th 2009, 09:08 AM
ThePerfectHacker
Hint: If $n$ can be factorized as $p_1^{a_1}...p_k^{a_k}$ where $p_1,...,p_k$ are distinct prime factors and $a_1,...,a_k$ are their exponents then the number of factors of $n$ is given by $(a_1+1)(a_2+1)...(a_n+1)$.
• Apr 7th 2009, 09:33 AM
Vedicmaths
yeah but 2008 and 2010 can not be prime factorized...so how can we write this in that form anyway?
do we have to do them separately and them combine them together?
I tried both ways and end up getting a very big number..there should be some tricky way to combine them together...please help!

Thanks a lot!
• Apr 7th 2009, 01:37 PM
aidan
Quote:

Originally Posted by vedicmaths
yeah but 2008 and 2010 can not be prime factorized...

2008 = 2^3 * 251^1

2010 = 2^1 * 2^3 * 2^5 * 67^1
• Apr 7th 2009, 01:43 PM
aidan
Sorry for the above error.
Having problems with LaTex math/math tags & reading small fonts:

2010 = 2^1 * 3^1 * 5^1 * 67^1
• Apr 9th 2009, 04:48 PM
Media_Man
Unfortunately, no trick when it comes to factors. You only have to check for prime divisors up to sqrt(2011)~43 though:

2007=3^2*223
2008=2^3*251
2009=7^2*41
2010=2*3*5*67
2011=prime
• Apr 9th 2009, 07:30 PM
Vedicmaths
thanks for the help...but what do you mean by the largest number of factors? like you explained below that:

2007=3^2*223
2008=2^3*251
2009=7^2*41
2010=2*3*5*67
2011=prime...

so that means largest number of factors is 2010...right? since it has 2,3,5, 67 maximum numbers of factors...

thanks for the help!
• Apr 9th 2009, 08:41 PM
Soroban
Hello, Vedicmaths!

Here's a helpful theorem . . .

Given a positive integer $N$, its prime factorization is: . $N \;=\;p_1^a\cdot p_2^b\cdot p_3^c\:\hdots$
. . The number of factors of $N$ is: . $(a+1)(b+1)(c+1)\cdots$
. .
(Add 1 to each exponent, and multiply.)

This includes 1 and $N$ itself.

Quote:

1) Which of the five numbers 2007, 2008, 2009, 2010 and 2011
has the largest number of factors and which one has the fewest number of factors?

. . $\begin{array}{|c|c|c|}\hline
N & \text{factorization} & \text{no. of factors} \\ \hline
2007 & 3^2\cdot 223 & (3)(2) \:=\:6\\
2008 & 2^3\cdot 241 & (4)(2) \:=\:8\\
2009 & 7^2\cdot 41 & (3)(2) \:=\:6 \\
2010 & 2\cdot 3\cdot 5\cdot 67 & (2)(2)(2)(2) \:=\:{\color{red}16} \\
2011 & \text{prime} & {\color{blue}2} \\ \hline\end{array}$

Quote:

2) Determine the number of factors for the number: . $2007\cdot2008\cdot2009\cdot2010\cdot2011$

The product is: . $P \;=\;(3^2\cdot223)(2^3\cdot251)(7^2\cdot41)(2\cdot 3\cdot5\cdot67)(2011)$

. . . . . . . . . . . . . $= \;2^4\cdot3^3\cdot 5\cdot7^2\cdot41\cdot61\cdot223\cdot251\cdot2011$

The number of factors is: . $5\cdot4\cdot2\cdot3\cdot2\cdot2\cdot2\cdot2 \;=\;3840$

• Apr 9th 2009, 09:24 PM
Vedicmaths
thanks for the help Sir!

but do you think the last one, there is a calculation mistake ...when we multiply 5*4*2*3*2*2*2*2 = 1920...or should there be another 2 in between to make it 3840...but if I count I get the same number as you got...so I think total number should be 1920?...

thanks a lot for the help..that was very nice of you to explain that properly! :)