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Prove or disprove that in any power of $20n+c$, where $c$ is an odd digit (i.e., $1,3,5,7,9$), the ten's digit is even.

This is probably a generalization of this.

I tried in the following way.

I observe that $(20m+c)(20n+d)$ is $20(20mn+m+n)+cd \equiv cd \pmod{20}.$

First of all, powers of $5$ end with $5$.

So, putting $c=d=5$, we get $25$, or directly $$(20a+5)(20b+5)=20(20ab+a+b+1)+5,$$ clearly the ten's digit is even.

Now the product of any two numbers in $S=\{1,3,7,9\}$ is of the form $20r+e$ where $r$ is a positive integer and $e \in S$.

If it obvious that the powers of any number ending with $e \in S$, will end with a member of $S$, we are done.

Any rectification or better solution is more than welcome.


I was trying to take the modulo 20 on the product of each member of $T=\{10a+b\}$ with each of the member of $U=\{10c+d\}$ where a,c are natural numbers, b,d are odd digits. I observed that the tens' digit is even when x,y are of same parity(for example, 11*11=121, 11*13=143, 11*17=187, 13*13=169, 13*17=221 etc.) except the cases of 3*5 and 5*7 where x,y need to be in the opposite parity to make the tens' digit even(for example, 13*15=195, but 13*25=325; 15*17=255, but 15*27=405 etc.). This is a further generalization from the power of a number to the product of different numbers and the numbers are of the form 10n+b instead of 20n+b.

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That means $(20n+c)^r$, for any $r$, right? –  SiliconCelery Aug 1 '12 at 17:23
    
@SiliconCelery,r should be positive integer. –  lab bhattacharjee Aug 1 '12 at 17:35
    
My simple answer in your link also works here. –  Bill Dubuque Aug 1 '12 at 17:40
    
I'm yet to know whether my approach is correct or not? –  lab bhattacharjee Aug 2 '12 at 4:44

2 Answers 2

up vote 3 down vote accepted

Suppose $(20n+c)^k = 20\alpha + c^k$ where $\alpha \in \mathbb{N} \cup \{ 0 \}$ and $c$ is an odd digit. (This certainly holds for $k=1$.)

Then

$$\begin{align}(20n+c)^{k+1} &= (20\alpha + c^k)(20n+c) \\ &= 20(20\alpha n + \alpha c + nc^k) + c^{k+1}\end{align}$$

So we've proved by induction that all powers of $20n+c$ are of the form $20\alpha + c^k$.

It suffices to show that the powers of $1,3,5,7,9$ less than $100$ have even tens digit. This is something you can verify directly. Indeed, they are $1, 3, 9, 27, 81, 5, 25, 7, 49, 9, 81$, and so we're fine.

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$$(20n+c)^m =\sum_{k=0}^m \binom{m}{k} (20n)^{k}c^{m-k} \equiv \binom{m}{k} 20nc^{m-1} + c^m \mod 20 $$

So the problem reduces to proving that the 10's digit of $c^m$ is even. This follows from a simple lemma.

Lemma Let $c$ be an odd digit, and let $k>1$ be so that $c^k \equiv c \mod 20$. If the tens digits of $c, c^2,..., c^{k}$ are even, then the tens digit of $c^m$ is even for all $m$.

This lemma is easy to prove (Prove it) and to complete the exercise you can observe that

$$1^2\equiv 1 , 3^5 \equiv 3, 5^2 \equiv 5, 7^5 \equiv 7, 9^3 \equiv 9 \mod 20 \,.$$

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$\phi(20)=8=>c^8≡1(mod\ 20)$ –  lab bhattacharjee Aug 1 '12 at 17:44
    
True, but that doesn't work for 5, and it is easier to check up to 3rd power for 9 than up to 5th. Since one has to check all powers anyhow, FLT doesn't make the solution shorter ;) –  N. S. Aug 1 '12 at 17:47
    
Sorry, I'd missed to identify $c^4≡1(mod\ 20)$ as 20=5*4 can not have any primitive root. For c=5, we can observe $5^2≡5(mod\ 20)=>5^n≡5\ where\ n>0$. According to order formulae, we only need to check the powers which divide 4(right?) i.e, 2 & 4. –  lab bhattacharjee Aug 2 '12 at 4:43
    
The point is that we need to check that $c^i$ has the tens digit even for ALL i's up to that power... –  N. S. Aug 2 '12 at 12:26
    
There's a comma missing in the last line (I tried to add it, but since it's a one-letter edit, SE doesn't let me). –  celtschk Aug 4 '12 at 7:55

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