Question

which decimal digits occur as the final digit of a fourth power of an integer? Please, help

Answer 1

well, the fifth happens to be the same as the original number's

Answer 2

as for this, I guess we can find any fourth power last decimals by simply doing those on 1-9

Answer 3

well, 0-9 but that's trivial

Answer 4

Hey, is there any other method than just list the digit out?

Answer 5

uhm, there is, but uh, number theory, uh I got around using the formulas by doing this last time, so uh, I don't remember :/

Answer 6

What does it mean by saying: the last digit of a^2 is a^2mod10?

Answer 7

whats the remainder when you divide 2315 by 10 ?

Answer 8

5

Answer 9

I think I got it.

Answer 10

Hey sorry there was some issue with electricity last night

Answer 11

I know and after sleeping, I figured it out :)

Answer 12

nice :) so what have you figrued out
```
What does it mean by saying: the last digit of a^2 is a^2mod10?
```

Answer 13

all of them :)
However, I have another question on this topic. Let me post it here.

Answer 14

If \(a^k \equiv b^k (mod m)\\a^{k+1}\equiv b^{k+1}(modm)\)
\(a, b, m, k \in \mathbb Z; m>0, k>0, (a,m)=1\)
Prove \(a\equiv b(modm)\)

Answer 15

My question: Why can't we go directly from the problem to get the answer? Since we have \(a^k \equiv b^k (mod m)\) and \(a^{k+1}= a^k *a\)
Hence \(a*a^k \equiv b*b^k \implies a\equiv b\)

Answer 16

are we given that \(a^k \equiv b^k \pmod{}\) is valid for all \(k\gt 0\) ?

Answer 17

Why do we have to expand \(a^{k+1}-b^{k+1}= (a-b)(a^{k} +a^{k-1}b +\cdots +b^{k})\) and then use the fact that (a, m) =1 to show that the LHS divided by m iff m | (a-b) to get \(a\equiv b (mod m)\)

Answer 18

Yes

Answer 19

then the question makes no sense, why can't we directly plugin \(k=1\) ?

Answer 20

since we are given that \(a^k \equiv b^k \pmod{m}\) is valid for all \(k\gt 0\)

plugging in \(k=1\) gives \(a^1 \equiv b^1 \pmod{m}\)

Answer 21

We cannot, since k is arbitrary,

Answer 22

you have said the given statemetn is true for all \(k\gt 0\) ?

Answer 23

yes

Answer 24

or do you mean \(k\) is some "fixed" positive integer ?

Answer 25

could you take a screenshot of the problem and post ?

Answer 26

For all k > 0

Answer 27

sure

Answer 28

Problem 26

Answer 29

Okay, we should interpret \(k\) as some specific "fixed" positive integer

Answer 30

ok, next?

Answer 31

My question: Why can't we go directly from the problem to get the answer? Since we have \(a^k \equiv b^k (mod m)\) and \(a^{k+1}= a^k *a\)
Hence \(a*a^k \equiv b*b^k \implies a\equiv b\)


How did you get that ?

Answer 32

oh, yeah, it is invalid!! we have \(a\equiv b \\c\equiv d \\ac \equiv bd\)
not backward.

Answer 33

it is perfectly fine, you need to add proper justifications, thats all

Answer 34

\(a^k\equiv b^k \pmod{m}\) and \(a^{k+1}\equiv b^{k+1}\pmod{m}\)

together imlies \(a*a^k \equiv b*b^k \equiv b*a^k \pmod{m}\)

Answer 35

so we have \(a*a^k \equiv b*a^k \pmod{m}\)

nothing fancy, fine so far, right ?

Answer 36

Yes

Answer 37

but what exactly does that mean

Answer 38

\(a\equiv b\)

Answer 39

\(a*a^k \equiv b*a^k \pmod{m} \iff m \mid (a*a^k - b*a^k)\)

Answer 40

factor out \(a^k\) and apply Euclid lemma

Answer 41

and factor a^k out and use the fact that m not | a to conclude that m | (a-b)

Answer 42

Looks perfect!

Answer 43

Thank you so much.

Answer 44

np