Find the last digit of $7^{802} -3^{683}$

Question

Find the last digit of $$7^{802} -3^{683}$$

$$7^{802} -3^{683}=(50 -1)^{401} -3(10 -1)^{341}$$ $$=50k -1 -3(10m -1) \space \text{where k,m are fixed integer } $$ $$=50k -30m +2 $$ So,the last digit will be $2$

Someone just added this question, but deleted it. I solved this question and got this thing. I just want to know if my answer correct or not.

Yes, your answer and your steps are correct. Why doubt them? — N.S.JOHN, Aug 24 '16 at 14:56
@N.S.JOHN No doubt just verification because I don't know what is the actual answer of this question ? — Aakash Kumar, Aug 24 '16 at 14:58
If you are into experimenting (a good idea IMHO, if you are not in a position to study congruences and/or modular arithmetic yet), you can also simply observe that the last digit of powers of $7$ repeat periodically $7^0$ ends in $1$, $7^1$ in a $7$, $7^2$ in a $9$, $7^3$ in a $3$, and then the period of length four repeats. So $7^{802}$ will share its last digit with $7^2$. Powers of $3$ also repeat their last digits in a period of length four. Meaning that $3^{683}$ has the same last digit as $3^3$. Given this the rest is easy. Your solution is, of course, correct. — Jyrki Lahtonen, Aug 24 '16 at 15:53
@JyrkiLahtonen Really a great method , Instead of $3^3$ it will be $3^1$ — Aakash Kumar, Aug 24 '16 at 16:06

Bill Dubuque · Answer 1 · 2016-08-24T15:22:09.540

4

Yes, your proof using the Binomial Theorem is correct. It is simpler using congruences

$$\begin{align} {\rm mod}\ 10\!:\,\ \color{#0a0}{7^{\large 2}}\equiv \color{#c00}{-1}\equiv \color{#0a0}{3^{\large 2}}\ \Rightarrow\, &\ \ (\color{#0a0}{7^{\large 2}})^{\large 401}\!-3(\color{#0a0}{3^{\large 2}})^{\large 341}\\ \equiv\ &(\color{#c00}{-1})^{\large 401}-3(\color{#c00}{-1})^{\large 341}\\ \equiv\ &\ \, {-}1\ \ \ + \ \ \ 3\\ \equiv\ &\ \ \ \ 2\end{align}\qquad\qquad $$

This is precisely the same as your proof, except we eliminated the Binomial Theorem in favor of congruences. Notice how the BT calculation becomes obvious in congruence language, reducing to the trivial result that $\,{\rm mod}\ 10\!:\ (10j\!-\!1)^{2n+1}\equiv (-1)^{2n+1}\equiv {-}1.\ $ Similarly congruence arithmetic serves to trivialize many other arithmetical computations.

edited Aug 24 '16 at 15:22

answered Aug 24 '16 at 15:01

Bill Dubuque

272,048

I don't know modular arithmetic but I want to learn about this .Can you provide any online good source to study this ? – Aakash Kumar Aug 24 '16 at 15:05
@AakashKumar See the introduction to congruences (aka modular aorithmetic) in any textbook on elementary number theory. – Bill Dubuque Aug 24 '16 at 15:09
@BillDubuque is there a special reason to take mod 10 ? – Aug 24 '16 at 16:35
@A---B We seek the last decimal digit, which is the remainder mod $,10\ $ – Bill Dubuque Aug 24 '16 at 16:40
@BillDubuque As i thought. so if we want last binary digit then we take mod 2 right ? – Aug 24 '16 at 16:41
@A---B Yes, in radix $,b,$ we have $,d_0!+d_1 b + \cdots + d_k b^k\equiv d_0 + b(\cdots)\equiv d_0\pmod b, $ or, in polynomial notation $\ f(b)\equiv f(0)\pmod{b}\ $ by the Polynomial Congruence Law. – Bill Dubuque Aug 24 '16 at 16:43
@BillDubuque Thanks. $\ddot \smile$ – Aug 24 '16 at 16:46

fleablood · Answer 2 · 2016-08-24T16:32:13.000

Okay, after reading the comments I realize my answer is completely out of the experience level of the OP. But read ahead for a taste of exciting things to come.

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Very nice!

I want to introduce you to the idea of Fermat's Little Thereom however.

If $\gcd(a, m) = 1$. And $\phi(m) = $ the number of natural numbers less than $m$ that are relatively prime to $m$ then:

$a^{\phi (m)} \equiv 1 \mod m$

(or in other words $a^{\phi(m)} = km + 1$ for some $k$.)

So... as $1,3,7$ and $9$ are relatively prime to $10$ so $\phi(10) = 4$. And $\gcd(7,10) = \gcd(3,10) = 1$.

So $7^4 = 1 \mod 10$ ($7^4 = 2401$) and $3^4 \equiv 1 \mod 10$ ($3^4 = 81$).

So $7^{802} - 3^{683} = 7^{2 + 4k} - 3^{3 + 4j}$

$= 7^2*(7^4)^k - 3^3*(3^4)^j \equiv 7^2*1^k - 3^3*1^j \mod 10$

$\equiv 49 - 27 \mod 10 \equiv 9 -7 \mod 10 = 2 \mod 10$

So $2$ is the last digit.

.........

Okay, I admit, now that I've typed the whole thing out that seems a LOT more complicated than your elegant fast solution.

But it's useful to know.

If I asked "what is the remainder when $8^{386} - 6^{216}$ is divided by $13$" it might be easier to use Fermat's Little Theorem, rather than "$8^2 = 64 = 5*13 - 1$ and $6^6 = 46656 = 3589*13 - 1$ so $(-1)^{193} - (-1)^36 = -1 -1 =-2$ so $11$ is the remainder"

[$13$ is prime so all numbers less than $13$ are relatively prime so $\phi(13) = 2$. So $8^{386} \equiv 1 \mod 13$ and $6^216 \equiv 1 \mod 13$.

[So $8^{386=2 + 12k} - 6^{216=12j} \equiv 8^2 - 1 \mod 13$

[$\equiv 63 \mod 13 \equiv 11 \mod 13$]

score 1 · Answer 3 · answered Sep 04 '16 at 18:05

1

The last digit of 7^802 is the same as 7^2, which is 9. The last digit of 3^683 is the same as 3^3, which is 7. Then 9-7 = 2.

answered Sep 04 '16 at 18:05

John BEAUNICE

11
1

Find the last digit of $7^{802} -3^{683}$

3 Answers3