0
$\begingroup$

I'd like to answer this question without using a calculator.

My first instinct is to find a pattern in the value of $2^n$:

$2^1 \equiv 2$, $2^2 \equiv 4$, $2^3 \equiv 8$, $2^4 \equiv 6$, $2^5 \equiv 2,...$

So units digit form a set of $4$ digits: $2,4,8,6,2,4,8,6...$

Now I'm stuck, how do I know which digit to pick from the set above

$\endgroup$
4
$\begingroup$

As you can see, it repeats with period $4$. The units digit of $2^k$ thus depends on $k \pmod 4$. Can you find $102 \pmod 4$?

$\endgroup$
  • 1
    $\begingroup$ By "(mod $4$)" Ross means the remainder of the integer division of the number by 4. $\endgroup$ – Dariel Rudt Oct 7 '17 at 3:00
  • $\begingroup$ Thanks for the help. So the remainder is 2 —> 2^2 = 4. The units digit is 4. How would this apply if the remainder is 0 meaning a whole number ? $\endgroup$ – Cyzanfar Oct 7 '17 at 4:26
  • $\begingroup$ It would be the same as remainder $4$, so $6$. If you follow your pattern all multiples of $4$ greater than $0$ end in $6$. There are often perturbations at the start like $2^0=1$. If you look at two digits you never again match $2^1=02$ You can see that all powers greater than $0$ need to be even, so cannot be $1$. $\endgroup$ – Ross Millikan Oct 7 '17 at 4:29
2
$\begingroup$

Because the period of the pattern is $4$, the units digit of $2^{102}$ is the same as that of $2^{102-4}=2^{98}$. Which is the same as that of $2^{98-4}=2^{102-2(4)}$. Use this idea and you shall see why we can say that because the remainder of 102 divided by 4 is 2, the answer is the same as the answer to the units digit of $2^2$.

$\endgroup$
1
$\begingroup$

You've made a good start; I'm sure you've noticed that the pattern $2,4,8,6,2,\ldots$ repeats every $4$ numbers. So then $2^5$ gives you the same answer as $2^1$ and $2^{5+4}$ or $2^{5+4+4}$. This must mean $2^{102}$ would give the same final digit as $2^{4+4+\ldots+4+?}$. So how many $4$'s can we fit into $102$? In other words, when does the sequence $4,8,12,16,\ldots$ surpass $102$? As Ross Millikan points out, this is the same as finding the remainder of $\frac{102}{4}$.

$\endgroup$
0
$\begingroup$

$2^{102}\pmod {10}\equiv 2^{2+10\times 10}\equiv 4\times (2^{10})^{10}\equiv 4\times (1024)^{10}\equiv 4\times 4^{10}\equiv 4\times (2^{10})^2\equiv 4\times (1024)^2\equiv 4\times 4^2\equiv 64\equiv 4$

$\endgroup$
0
$\begingroup$

If you want a general way of solving this problem you should look into Fermat's Little Theorem which states:

For a given integer $a$, $a^p \equiv a$ (mod $p$) for a prime number $p$. Or equivalently $a^{p-1} \equiv 1$ (mod $p$).

For your specific problem you could let $a=2$ and $p = 5$. Although you would ideally want to get equivalence modulo 10, in this case, you will be fine since $2^n$ is even for $n > 0$. Then you see that $2^4 \equiv 1$ (mod 5), so $2^{100} = {(2^4)}^{25} \equiv 1$ (mod 5). Therefore, $2^{102} = (2^{100})(2^2) \equiv 2^2 \equiv 4$ (mod 5). We now see that the units digits is $2^2 = 4$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.