Try to Solve Without Using a Calculator

Note that $5^n\equiv 5 \pmod{10}$ , so $13^{156}-5^{156}\equiv 3^{156}-5\equiv (3^4)^{39}-5\equiv 81^{39}-5\equiv 1^{39}-5\equiv \boxed{6} \pmod{10}$

An odd number minus another odd number will be even. 6 is the only even answer.

This problem uses very simple concept of noticing how number raised to the power of... ends. We know that $5$ to the power of anything ends with 5. But how about $13$ ? For that we need to take a piece of paper (or a calculator, if you don't mind cheating a little) and multiply:

We need $13$ raised to only even exponents, so:

$13^{2} = 46 \boxed{9}$

$13^{4} = 2856 \boxed{1}$

$13^{6} = 482680\boxed{9}$

$13^{8} = 81573072\boxed{1}$

$13^{10} = 13785849184\boxed{9}$

We could continue raising $13$ until we reach the number stated in the question, but we don't need to do that because we got a sequence! $9$ , $1$ , $9$ , $1$ , $9$ ... That means $13$ raised to any even power ends in either $9$ or $1$ .

Another thing that we could notice from above statements is: if the exponent of $13$ when divided by $2$ is even, then the number that we get from raising $13$ to the power of exponent will end in $1$ , and if the exponent of $13$ when divided by $2$ is odd, then the number that we get from raising $13$ to the power of exponent will end in $9$ .

$\frac{156}{2} = \boxed{78}$

$78$ is even, so $13^{156}$ will end in $1$ .

Now we will need to imagine two numbers: one that ends in $1$ and another one that ends in $5$ . The number that ends in $1$ is subtracted by the number that ends in $5$ . We need to remember, that both numbers are quite large, so it is possible to take $10$ and add it to $1$ , so that while subtracting we won't get negative answer.

$11 - 5 = \boxed{6}$

The number that we will get from $13^{156} - 5^{156}$ will end in digit $\boxed{6}$ .

The last digits when 13 is raised to any power follows a cyclic order- 3,9,7,1,3,9,7,1... That's why 13 raised to any power that is a multiple of 4 will end in 1. Secondly 5 raised to any power ends in 5. That means ....1-....5=....6

We know $4|156$ Using Euler's Totient Function, $\phi(10) = 4\\\Rightarrow 13^{\phi(10)}\equiv13^4\equiv 1~(\text{mod } 10)$ Also $5^{156}\equiv 5~(\text{mod } 10)$ Thus $13^{156}-5^{156}\equiv 1-5\equiv 6~(\text{mod } 10)$

13: Taking 3 for 3, 9, 7, 1, 3, 9, 7, 1, 3, 9, 7, 1...

5: Taking 5 for 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5...

156 = 4 (39) + 0 stands for 1 which is before 3;

Therefore, ?1 - 5: - 4 $\rightarrow 10 - 4 = 6$

Answer: $\boxed{6}$

Here's the way I did it: 5 raised to any power will always give 5 as the unit's digit. 13 raised to any power will always give an odd digit at unit's place. The different between to odd numbers can never be odd. Hence, 6 was the only possible answer! 🎉

the way I looked at it was that 5 and 13 are both odd numbers. The difference of two odd numbers is always even. The only given even answer was 6.

What is the solution?