Find the last two digits of $3^{3^{2014}}$.   Attempt:   First I try to work with $3^{2014}$. So we can work on $\text{mod 10}$.  Then, $$\begin{align}&amp;3^1 \equiv 3\pmod{10}\\ &amp;3^2 \equiv 9 \pmod{10}\\ &amp;3^3 \equiv 27 \equiv 7\pmod{10}\\ &amp;3^4 \equiv 81 \equiv 1\pmod{10}\end{align}$$  So  $3^{2014} = 3^{(4\cdot503) + 2} \equiv 3^0 \cdot 3^2 \equiv 9 \pmod{10}$.  So the last digit is 9. However I am ask to find the last two digit. Can anyone please help me? I would really appreciate all the help.
122227
Can anyone please help me?  1) Find the last digit of $7^{12345}$   2) Find the last 2 digits of $3^{3^{2014}}$.  Attempt: 1)  By just setting the powers of $7$ we have $7^1 = 7$, $7^2=49$, $7^3=343$, $7^4 = 2401$, $7^5 = 16807$, $7^6 = 117649$, $\dots$  After the power of $4$, the last digits will repeat. Then by noticing the pattern the digits will end in $7,9,3$ and $1$.  Then we can divide the exponent $(12345)$ by $4$ since this is the cycle that makes it repeat. Then $12345 : 4$ has remainder $1$. So $7^1 = 7$ is the unit digit to $7^{12345}$.   So the last digit is $7$.   I know how to do it like this, the problem does not state how to find the last digit, but I know it has something do do with Euler's theorem.  for part $2$) I don't know how to start. Can anyone please help me?   Thank you for the help.
87190
Does math have an incorrect notation / syntax? I don't mean writing misaligned notation (google), but when you take something like a number to powers to powers to powers, $${{2^2}^2}^3$$ (I was told this is incorrect notation by a teacher). Is it really incorrect, or does it just need to be simplified with parentheses? Do people write maths like this?  a radical expression with the root being a radical expression? $$\sqrt[\sqrt{2^3}]{2}$$
66574
Show that $13$ divides $2^{70} + 3^{70}$.   My main problem here is trying to figure out how to separate the two elements in the sum, and then use Fermat's little theorem. So how can I separate the two?  Thanks!
29747
To find last two digits  of $2^{100}$  I have just learned about modular arithmetic and I wanted to solve this problem. I only know about equivalence classes and about  $a=b \pmod n$. I have also learned about multiplication and addition of classes.  Can someone explain to me step by step on how to apply modular arithmetic to this problem?  Thanks
64420
What should the 'efficient' way of finding the last two digits of $2^{2016}$ be? The way I found them was by multiplying the powers of $2$ because $2016=1024+512+256+128+64+32$. I heard that one way would be with the Chinese Remainder Lemma but I don't really know how I should start?
1672
I have had a go at this for a while now and can't seem to get anywhere.  $$2^{(2^{517})} \pmod {23}$$  I see that Fermat's Little Theorem must come into play but I can't see where to pull $22$ from. Any help would be appreciated.  Here is my attempt:  $$2^{(2^{517} \!\!\mod {22})} \equiv  2^{(2^{517})} \pmod {23}$$ Then proceed to do Chinese remainder theorem to find the exponent?
65418
I have to find the last two decimal digits of the number $9^{{9}^{9}}$.  That's what I did:  $$m=100 , \phi(m)=40, a=9$$ $$(100,9)=1, \text{ so from Euler's theorem,we have :} 9^{40} \equiv 1 \pmod{100}$$  $$9^{{9}^{9}} \equiv 9^{9^{2+2+2+2+1}} \equiv 9^{9^2 \cdot 9^2 \cdot 9^2 \cdot 9^2 \cdot 9} \equiv 9^{81 \cdot 81 \cdot 81 \cdot 81 \cdot 9} \equiv 9^{(40+40+1) \cdot (40+40+1) \cdot (40+40+1) \cdot (40+40+1) \cdot 9} \equiv (9^{(40+40+1)})^{(40+40+1) \cdot (40+40+1) \cdot (40+40+1) \cdot 9} \equiv 9^9 \equiv 9^4 \cdot 9^4 \cdot 9 \equiv 6561 \cdot 6561 \cdot 9 \equiv 3721 \cdot 9 \\ \equiv 21 \cdot 9 \equiv 89 \pmod{100}$$  So,the last two digits are $8 \text{ and } 9$. $$$$But,is there also an other way two calculate the last two digits of $9^{{9}^{9}}$ or is the above the only one?
105719
I wish to explain to a grade 5 kid, the answer to the question - 'When can we know that the observing the last digit of a number will not be sufficient to determine its divisibility by a given N?'  For example - It doesn't work for 3, because we get all sorts of digits, i.e. from 0 to 9, at the units place. The same reason works for 7 and 9.   But what would be a convincing argument for 4 and 8? Even though the last digits form a pattern - 0, 2, 4, 6 and 8 - but we're still not sure about its divisibility.
41272
I am trying to find a general formula for triangular square numbers. I have calculated some terms of the triangular-square sequence ($TS_n$):  $TS_n=$1, 36, 1225, 41616, 1413721, 48024900, 1631432881, 55420693056 1882672131025, 63955431761796, 2172602007770041, 73804512832419600  I have managed to find that:  $m^2 = \frac{n(n+1)}{2}$  where $m$ is the $m^{th}$ term of the square number sequence, and $n$ is the $n^{th}$ term of the triangular numbers sequence.  Would anyone be able to point out anything that could help me derive a formula for the square-triangular sequence?
26520
I am not looking for an answer on this. Just need to clarify why my approach is failing -  $N_1 + N_2 + N_3$, i.e. single digit, double digit, 3 digit  single $= 2, 4, 6, 8$, i.e 4   double = X non-zero $= 8 \cdot 4 = 32$ X zero $= 9 \cdot 1 = 9$  Now the confusing part three digit, breaking into 4 cases   X zero zero $= 9$   X nz nz $= 7 \cdot 8 \cdot 4 = 224$   X z nz $= 8 \cdot 1 \cdot 4 = 32$   X nz z $= 8 \cdot 9 \cdot 1 = 72$  Three digit total comes to $= 9 + 224 + 32 + 72 = 337$. This answer is wrong and it should be $328$. What am I missing in the logic? Please suggest.
63034
If you have a fraction such as $$\frac{7}{26}=0.269230\overline{769230}$$ where there are a number of digits prior to the repeating section, how can you tell how many digits there will be given just the fraction?  I believe I could run through the standard long division algorithm until I come across the same remainder for the second time and then use the location of the first instance of this remainder to calculate the number of digits before the repeating section, but this feels very cumbersome.  After lots of reading online, I came across what looks like a formula for it from :     When a rational number $\frac{m}{n}$ with $(m,n)=1$ is expanded, the period begins after ${s}$ terms and has length ${t}$, where ${s}$ and ${t}$ are the smallest numbers satisfying $10^s\equiv10^{s+t}\pmod{n}$.   I know how to calculate the length of the period of a fraction, and so in the case of my original fraction we have $10^s\equiv10^{s+6}\pmod{26}$, but I don't know how to solve for ${s}$ in this equation!  So there are really two questions here - the one in the title, and a sneaky one about how to take logs in a modulo arithmetic equation.
13240
Is there a direct way to determine how many digits a power of 2 will contain without actually performing the multiplication?  An estimation would help as well if there is no absolute solution.  EDIT: In both decimal and binary bases.
59304
I am finding a 4-digit number $abcd$ (in base-10 representation) satisfying the following property.  $$abcd=a^b c^d$$  I have been running my mind around this problem since a long time, but got no success.  I wrote a python program for a number with this property, and got the answer to be $2592$ since   $$2592=2^5\times 9^2$$  But I am looking for a purely mathematical way to solve this problem.   Thanks!
31508
Let $S(n)$ be the sum of digits of $n$.  Numerically, I've found that for all $n$ where $S(n) = S(n^2)$, $n \equiv 0,1 \bmod 9$.  Why is that? Is there a proof to show that this is always the case?
92549
Problem: For what pair of positive integers $(a,b)$ is $3^a + 7^b$ a perfect square.  First obviously $(1,0)$ works since $4$ is a perfect square, $(0,0)$ does not work, and $(0,1)$ does not work, so we can exclude cases where $a$ or $b$ are zero for the remainder of this post.  I have a few observations made but not much room for a full solution.   First, since powers of an odd number are odd, and the sum of two odd numbers is even, so the base of the square must be an even number.  Secondly, the last digit of the powers of $3$ are $\{1,3,7,9 \}$ , whereas the last digits of the powers of $7$ are $\{7,9,3,1 \}$.  From here I am not sure how to proceed and any hints much appreciated. I'm not sure if there a finite amount of pairs or not either.
80157
Question:     show that: the beautiful ${\tt sqrt}$-identity:      $$ \left({2 \over \sqrt{\vphantom{\Large A}\, 4\ -\ 3\,\sqrt[4]{\,5\,}\ +\ 2\,\sqrt[4]{\,25\,}\ - \,\sqrt[4]{\,125\,}\,}\,}\ -\ 1\right)^{4} =5 $$   Can you someone have methods to prove this by hand? (Maybe this problem have many methods?because this result is integer. It's a surprise to me.) Thank you   Because I found this $$4\ -\ 3\sqrt[4]{\,5\,}\ +\ 2\sqrt[4]{\,25\,}\ -\ \sqrt[4]{\,125\,}$$ is not square numbers.
121952
My question is this : how many distinct two digit numbers can be produced from numbers $4, 3, 3, 1$? When applying the formula $$\frac{4!}{(4-2)!2!}$$ you come up with $6$, yet when doing the problem manually, I come up with $7$ numbers, namely $41, 43, 14, 13, 33, 31, 34$. Can anybody explain this discrepancy?  Thanks.