+0  
 
0
75
8
avatar+119 

What is the largest integer n such that 3^n is a factor of 1*3*5*...*97*99?

ANotSmartPerson  Oct 21, 2018
 #1
avatar
+2

Listfor(n, 1, 165, (99!! % 3^n) =(0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 508 3731656658, 508 3731656658, 5083 7316566580, 11946 7693931463, 53124 9958120761, 114892 .......etc.

I see that the first 26 powers of 3 are factors of 99!!. Therefore, the largest integer n = 26, or:

3^26 =2,541,865,828,329 which is a factor of 99!!.

Note: 99!! is the double factorial function, i.e., 99 x 97 x 95, 93 x.............x 5 x 3 x 1.

Guest Oct 21, 2018
 #2
avatar+27128 
+2

99!ā†’2^95*3^48*5^22*7^16*11^9*13^7*17^5*19^5*23^4*29^3*31^3*37^2*41^2*43^2*47^2*53*59*61*67*71*73*79*83*89*97

 

n is 48.

 

99! is 99*98*97*...*3*2*1    It is a single factorial, not a double factorial.

Alan  Oct 21, 2018
 #3
avatar
+1

Alan: Isn't the question "3^n is a factor of 1 x 3 x 5 x .........x 97 x 99"?

Guest Oct 21, 2018
 #4
avatar+27128 
+1

Oops! I read it as 1*2*3*...*97*98*99.   It isn't!  

 

With z, say, as the product of the odd integers from 1 to 99 we have:  

 

zā†’3^26*5^12*7^8*11^5*13^4*17^3*19^3*23^2*29^2*31^2*37*41*43*47*53*59*61*67*71*73*79*83*89*97

 

so n is 26

 

(and, yes, this is known as a double factorial.  Thanks for pointing out my error Guest.).

Alan  Oct 21, 2018
edited by Alan  Oct 21, 2018
edited by Alan  Oct 21, 2018
 #5
avatar+97 
+1

39 this is the correct answer

MATHEXPERTISE  Oct 21, 2018
 #6
avatar
+1

NO, IT IS NOT!!. 99!! does NOT divide 3^39 EVENLY. It gives the following fraction:

672512041624018765061367383616729238980006511176089107384077  +  502004/1594323

Guest Oct 21, 2018
 #7
avatar+119 
0

Thanks for all of y'alls help, the answer is 26, I think mathexpertise got confused somewhere.

ANotSmartPerson  Oct 21, 2018
 #8
avatar+20108 
+1

What is the largest integer n such that 3^n is a factor of 1*3*5*...*97*99?

 

\(\begin{array}{|rcll|} \hline 99!! &=& 1\cdot 3 \cdot 5\cdot 7\cdot \ldots \cdot 95\cdot 97\cdot 99 \\\\ &&\text{Formula:}~\boxed{(2n-1)!! = \frac{(2n)!}{2^n\cdot n!}} \\\\ && (2\cdot 50 - 1)!! = \dfrac{(2\cdot 50)!} {2^{50}\cdot 50! }\\\\ 99!! &=& \dfrac{100!} {2^{50}\cdot 50! } \\ \hline \end{array}\)

 

 

100!

The factor \(3^i\)

\(\begin{array}{|rcll|} \hline \frac{100}{3} &=& [33].\overline{3} \\ \frac{100}{3^2} &=& [11].\overline{1} \\ \frac{100}{3^3} &=& [3].\overline{703} \\ \frac{100}{3^4} &=& [1].\overline{234567901} \\ \frac{100}{3^5} &=& 0 \\ \hline 33+11+3+1 &=& 48 \\ i &=& 48 \\ \hline \end{array} \)

 

 

50!

The factor \(3^j\)

\(\begin{array}{|rcll|} \hline \frac{50}{3} &=& [16].\overline{6} \\ \frac{50}{3^2} &=& [5].\overline{5} \\ \frac{50}{3^3} &=& [1].\overline{851} \\ \frac{50}{3^4} &=& 0 \\ \hline 16+5+1 &=& 22 \\ j &=& 22 \\ \hline \end{array} \)

 

\(\begin{array}{|rcll|} \hline && \dfrac{3^{48}}{3^{22}} \\\\ &=& 3^{48-22} \\ &=& 3^{26} \\ \hline \end{array} \)

 

\(\text{The largest integer $n$ such that $3^n$ is a factor of $1*3*5*...*97*99$ is $\mathbf{26}$ }\)

 

laugh

heureka  Oct 22, 2018

36 Online Users

avatar
avatar
avatar
avatar
avatar
avatar

New Privacy Policy

We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive information about your use of our website.
For more information: our cookie policy and privacy policy.