2021年5月21日 · Units digit of a product of numbers only depends on product of units digits. Let's say we have number 343. We can break it down into 340+3 and if we multiply it by a different number, for example $(340+3) (340+3)$ We'll have $340^2 + 2*340*3 + 3*3$ We know that $340^2$ and $2*340*3$ will not be units digits because $2*340*3 = 2*34*3*10$ and …

If you look at $2^1, 2^2, 2^3 ...$ you will quickly notice the unit digits follow the pattern $2,4,8,6,2,4,8,6...$ to prove that this pattern will continue to repeat note that when we multiply a number by $2$ it is the unit digit alone that determines what the unit digit of the answer will be.

How do you find the units digit in case of an expression like this. $$ 7^{8^7} $$ I know how to find the units digit when there is one integer and there is only one power. But how do I find it when the power has a power. Please provide a descriptive answer.

What is the units digit of $13^4\cdot17^2\cdot29^3$? 3. What will be the units digit of $7777^{8888}$? 0.

What is the units digit of the product $2^1·2^2· 2^3 · 2^4 · · · 2^{199}· 2^{200}$ where the numbers multiplied are the whole number powers from $2^1$ to $2^{200}$? exponentiation Share

2017年10月7日 · 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 $2^2$.

2016年10月17日 · How does this tie in with powers? Well, suppose you want to know the unit digit of $12^{12}$. The unit digit is the remainder mod 10, so mod 10 it is. $12^{12} \equiv 2^{12} \equiv (2^4)^3 \equiv 16^3 \equiv 6^3 \equiv 36 \times 6 \equiv 6 \times 6 \equiv 36 \equiv 6$ Presto. We got the unit digit without actually carrying on the full calculation.

$\begingroup$ When you say that $7$ is a really neat number in that the last digit of powers of $7$ follow a pattern. But, that has to do with primes, say $3$ has pattern: $3, 9, 7, 1$. Only, few primes like $5,11$ don't have such nice patterns. Say, $5$ has only one $5$, and $11$ has only one last digit $1$. $\endgroup$ –

We know if unit digits of two numbers are same, their difference is divisible by 10 and vice versa. Method $1a:$

2019年8月29日 · Is it possible to find the units digit of $2019^{2018^{2017^{.^{.^{.^{3^{2^{1}}}}}$? Where the expression contains all natural numbers $[1,2018]$ as powers and $2019$ as the main base.