+0

# pls help

0
205
8
+238

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

Oct 21, 2018

#1
+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.

Oct 21, 2018
#2
+27850
+2

99!→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.

Oct 21, 2018
#3
+1

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

Oct 21, 2018
#4
+27850
+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:

z→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
+128
+1

39 this is the correct answer

Oct 21, 2018
#6
+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
+238
0

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

Oct 21, 2018
#8
+22188
+8

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 $$3^j$$

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

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

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

Oct 22, 2018