
#1
Nov3004, 03:56 PM

P: 347

Find all integers n>1 such that n is a power of 3, and n1 is five times a power of 2.
Can anyone give me a push in the right direction? 



#2
Nov3004, 04:49 PM

Sci Advisor
HW Helper
P: 2,589

If n is a power of 3, then n is in the form [itex]3^k[/itex] for some natural k. Now, if n 1 is five times a power of two, then it is in the form [itex]5\cdot 2^m[/itex]. So, we have:
[tex]n = (n  1) + 1[/tex] [tex]3^k = 5 \cdot 2^m + 1[/tex] You know that [itex]5\cdot 2^m[/itex] is a multiple of 10, so the right side will have a "1" in it's units place. Of course, so must the left side. You can quickly convince yourself that for this to happen, k must be a multiple of 4, i.e. n must be a power of 81. So, we have: [tex]81^h = 5 \cdot 2^m + 1[/tex] Where [itex]h = k/4 \in \mathbb{N}[/itex]. Now, take the binomial expansion of the left side as follows: [tex]81^h = 5 \cdot 2^m + 1[/tex] [tex](80 + 1)^h = 5 \cdot 2^m + 1[/tex] [tex]\sum _{i = 0} ^h {h\choose i}80^i = 5 \cdot 2^m + 1[/tex] [tex]\sum _{i = 1} ^h {h\choose i}80^i = 5 \cdot 2^m[/tex] [tex]16 \sum _{i = 1} ^h {h\choose i}80^{i  1} = 2^m[/tex] [tex]\sum _{i = 1} ^h {h\choose i}80^{i  1} = 2^{m4}[/tex] Now, I did this quickly, so you have to consider the case when h = 0, or m < 4, or things like that, but I'll leave that to you. Now, it's really a simple matter of finding the values of h for which the sums in the form [itex]\sum _{i = 1} ^h {h\choose i}80^{i  1}[/itex] are multiples of 2. Now, that means you have to find the h for which the following is even: [tex]{h\choose 1} + {h\choose 2}80 + {h\choose 3}80^2 + \dots + {h\choose h}80^{h  1}[/tex] But you know that all the terms except the first are necessarily even. So, the whole thing is even iff [itex]{h\choose 1}[/itex] is even, which is obviously when h is even. So, all n that are even powers of 81 satisfy the conditions. 



#3
Nov3004, 04:55 PM

P: 347

Thanks. I'll try to see what I can do.




#4
Nov3004, 05:01 PM

P: 347

Modulo arithmetic (?) question
81^{2}=6561
3^{8}=6561, so it satisfies the first condition. But 6561 doesn't equal 5(2^{q})+1, since [itex]\frac{65611}{5}=1312[/itex] which isn't a power of 2. I believe the only solution is n=81. 



#5
Nov3004, 06:22 PM

Sci Advisor
HW Helper
P: 2,589

Oops, sorry. I went from talking about powers of 2 to multiples of 2. So it's not that simple. You have to find the values of h > 1 for which:
[tex]\sum _{i = 1} ^h {h\choose i}80^{i  1}[/tex] is a power of 2. We know that h=1 works (i.e. n = 81). So, dealing with h>1, if h is odd, this won't work. You can see if it won't work also if it's even (try using the fact that it doesn't work if its odd). 



#6
Dec104, 06:20 AM

P: 347

How did you jump from:
[tex](80 + 1)^h = 5 \cdot 2^m + 1[/tex] To: [tex]\sum _{i = 0} ^h {h\choose i}80^i = 5 \cdot 2^m + 1[/tex] To: [tex]\sum _{i = 1} ^h {h\choose i}80^i = 5 \cdot 2^m[/tex] To: [tex]16 \sum _{i = 1} ^h {h\choose i}80^{i  1} = 2^m[/tex] And to: [tex]\sum _{i = 1} ^h {h\choose i}80^{i  1} = 2^{m4}[/tex] ? 



#7
Dec104, 06:30 AM

P: 646

First transition :
Binomial expansion Second transition : Cancelling 1 from both sides Third Transition : Take 80 common on the LHS and divide both sides by 5 Fourth Transition : divide both sides by 16 (2^4 = 16)  AI 



#8
Dec104, 06:50 AM

Sci Advisor
HW Helper
P: 9,398

Some other observations. We have that 5.2^m+1 is zero mod 3 (assuming k>0)
mod 3 this tells us that m is even, so we're equivalently finding integers such that: 81^h=5.4^n+1 working mod 9 yields that n congruent to 2 mod 3, ie it is 3p+2 for some p 81^h=5.4^{3p+2}+1 = 80.64^p+1 working mod 81 shows 64^p=1... don't know where this is going now. Any ideas? 



#9
Dec1104, 12:58 AM

P: 12

Hi. Im trying to understand what is this problem 
Is it right for me to state the problem as follows? 3^x = 5*(2^y) + 1 , for some integers x and y. Thanks. 



#10
Dec1104, 03:06 AM

Sci Advisor
HW Helper
P: 2,589

okidream, yes.




#11
Dec1104, 07:03 AM

P: 12

If 3^x = 5*(2^y) + 1 , for some integers x and y is true, I had approached the same manner you did. But I got stuck at the 3rd transition. Can you explain how you pull 80 out? And please verify what I did here.... 1st Transition: I had obtained the Binomial Tree/expansion by splitting 3 into its integer sum with (i.e 2+1). 2nd Transition: 1 cancelled away 3rd Transition: Where is common in 80 from the binomial expansion? Except if I claim x must be greater or equal to 80....is this so? How you prove that the solutions only exist when the powers of 3 is greater than 80? Thanks for your sharing. 


Register to reply 
Related Discussions  
arithmetic modulo  Calculus & Beyond Homework  0  
arithmetic question  Calculus & Beyond Homework  1  
Question on Arithmetic Series  General Math  1  
modulo 1 arithmetic  General Math  5  
Quick question about modulo  Introductory Physics Homework  1 