What is the last digit of (1997)^(1997) - (1994)^(1994)?

[maverick280857]

The problem:

Find the last digit of

$$(1997)^{1997} - (1994)^{1994}$$

The Answer: 1 (one)

Will post a solution soon...I would like to know if there is a different way to do it.

Cheers
Vivek

 

[maverick280857]

Solution 1: Find the last digit of $$m^{n}$$

Divide the exponent n by 4. Let n = 4p + q. Let the last digit of m be t. Then the last digit of $$m^{n}$$ is simply the last digit of $$t^{p}$$.

Do this twice to get 7 and 6 for $$(1997)^{1997}$$ and $$(1994)^{1994}$$ respectively. Subtract to get 1.

Solution 2: Find the last digit of $$m^{n}$$

Let a denote the last digit of m. Enumerate the last digits of $$a, a^{2}, a^{3}...$$ and denote this series by $$\lambda$$. The series clearly repeats after T terms where T is the period of the sequence. Note that

Let r be the remainder when n is divided by T. The remainder has the values (1, 2, 3, ..., (T-1), 0) [note that zero has been placed after the (T-1)th term so that $$\lambda_{T} = 0$$]. The answer then is the r-th term of the $$\lambda$$ sequence, that is $$\lambda_{r}$$.

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What I want to know is: is there any other way to do this apart from the two ways mentioned above?

Cheers
Vivek

 

[faust9]

4 to an even power will always have a last digit of 6.
if 7 is raised to a power of some number n where (n-1)%4=0 then the resulting last digit will be a 7.

1994 is even thus 6
(1997-1)%4=0 thus 7
7-6=1

basically the last digit will follow some pattern in this case, the powers of 4 flip flop between 4 and 6 while the powers of 7 rotate through 7,9,3,1.

 

[maverick280857]

Yes, instead of evolving general rules I found it quite enlightening to figure out the last digit without using the methods outlined in my second post :-D

Cheers
Vivek

 

[arildno]

Why didn't you just multiply out by hand and subtract?

 

[maverick280857]

Arildno, do you really think that would be possible?! Thats $$1997^{1997}$$ and $$1994^{1994}$$ we're talking about! It would take me days (if at all I managed to do it).

Cheers
Vivek

 

[arildno]

Oh, I hadn't thought of that! (Or, possibly I had..)

 

[Galileo]

Wer're only interested in the last digit of $1997^{1997}$ and $1994^{1994}$, so consider these numbers modulo 10.
$$1997^{1997} \equiv 7^{1997} \mod 10$$
$$1994^{1994} \equiv 4^{1994} \mod 10$$
Since $\left(\mathbb{Z}\slash_{10}\mathbb{Z}\right)^*$ forms a group under multiplication, raising an element of the group to the power equal to the order of the group equals 1. The order of $\left(\mathbb{Z}\slash_{10}\mathbb{Z}\right)^*$
is $\varphi(10)=\varphi(2)\varphi(5)=4$.
Since
$$1997 \equiv 1 \mod 4$$ and
$$1994 \equiv 2 \mod 4$$, we have
$$7^{1997} \equiv 7^1 \mod 10$$ and
$$4^{1994} \equiv 4^2 \mod 10$$

$$7-4^2 \equiv 1 \mod 10$$
So the last digit is 1.

This is the method I learned from my algebra class, but it's principally the same as Fausts' solution.

 

[maverick280857]

Yes it is, except that it is more formal.

Thanks and cheers,
Vivek

 

[NateTG]

It's not that hard to exponentiate out, especially if you realize that you only care about the last digit:

$$1994=1024+512+256+128+8+2$$
and
$$1997=1024+512+256+128+8+4+1$$

Recognizing that we can throw away everything except the last digit:
$$7^1 \equiv 7$$
square to get
$$7^2 \equiv 9$$
square to get
$$7^4 \equiv 1$$
So $$7^{4n} \equiv 1$$
so we have
$$1997^{1997} \equiv$$
$$7 ^{1024+512+256+128+8+4+1} \equiv$$
$$7^{1024} \times 7^{512} \times 7^{256} \times 7^{128} \times 7^{8} \times 7^4 \times 7^1\equiv$$
$$1 \times 1 \times 1 \times 1 \times 1 \times 1 \times 7 \equiv$$
$$7$$

Similarly:
$$4^1 \equiv 4$$
$$4^2 \equiv 6$$
$$4^4 \equiv 6$$
So $$4^{2^n} \equiv 6$$ for $$n>1$$
now
$$1994^{1994} \equiv$$
$$4^{1024+512+256+128+8+2} \equiv$$
$$4^{1024} \times 4^{512} \times 4^{256} \times 4^{128} \times 4^{8} \times 4^2 \equiv$$
$$6 \times 6 \times 6 \times 6 \times 6 \times 6 \equiv$$
$$6^6 \equiv$$
$$6$$

Even without that $$x^y(mod z)$$ is $$O(\log_2(y) {\log_2(z)}^2)$$ or so. Certainly within paper and pencil calculation range for the numbers you gave

 

[maverick280857]

Thanks faust9, Galileo, arildno and NateTG

I learned a lot from the approaches you folks suggested...thanks.


Cheers
Vivek