Practice your exponents

I even did it Caleb Townsend's way, but was interested to post another type of solution not requiring Modular Arithmetic. Observation:- Last digit of $57^{4} = 1$ .So, $57^{89} = ((57^{4})^{22}) \times 57 \implies$ last digit of $57^{89} = 1\times 7 = 7$ .

The last digit of $57$ is $7,$ so a simpler problem to answer is "what is the last digit of $7^{89}?$ "

Consider $x = 7^n,\ n\in\mathbb{Z}^+.$ The first few values of $x \text{ mod } 10$ are $7,\ 9,\ 3,\ 1,\ 7,\ 9,\ ...$ If $n\equiv 0 \text{ (mod } 4),$ then the last digit is $1.$ If $n\equiv 1 \text{ (mod } 4),$ then the last digit is $7.$ If $n\equiv 2 \text{ (mod } 4),$ then the last digit is $9.$ Lastly, if $n\equiv 3 \text{ (mod } 4),$ then the last digit is $3.$

In this case, $n = 89 \equiv 1 \text{ (mod } 4),$ so the last digit is $7.$

Since we want the last digit of the number, it suffices to consider it modulo $10$ . It is easy to see that $57\equiv 7\pmod {10}$ So, we have $57^{89}\equiv 7^{89}\pmod {10}$ Since $7^4\equiv 1\pmod {10}$ it is seen that $7^{89}=7\times (7^4)^{22}\equiv 7\pmod {10}.$ Hence our required answer is $\boxed{7}$ .

57^1= 7; 57^2= 9; 57^3= 3; 57^4= 1; 57^5= * 7; And it continues like this.It will repeat.So at 23th step the last digit will be 7.

I noticed that last digit is of 57^1 is 7 The last digit of 57^2 will be the result of multiplying 7 of 57^1 by the 7 of the new 57^1. It will be the last digit of 7x7 which is 9 Fillowing this procedure we cab see 57^3 ends with the last digit of 9x7, which is 3 and 57^4 ends with 1 We can notice a pattern of 7,9,3,1 repeats with increasing the exponent. it is a 4 units series, so for 89th position we get 7

When we multiply last digit of 57 that is 7 with 1,2,3,4 5 times ,we get last digits as 7,9,3,1 and again repitition of the sequence,so 57^89 will haveninetimes aslast digit as 7 ans

It's all abot the rule of Cyclicity.

The order if cycle is 4a and it's 44timeso both side multiply 7we get 7

divide 89 by 4 remainder left is 1 ans is 7^1.

pattern :

• 7 ^ 0 = 1

• 7 ^ 1 = 7

• 7 ^ 2 = 49

• 7 ^ 3 = 343

• 7 ^ 4 = 2401 ...

thus the exponent cycle of 7 is mod 4

end_digit(57^{89}) = (7^{89 mod 4}) = 7^1 = 7

I just find the last digit of 7^9

If unit's digit is 7 , then units digit for 7^4 is 1...here we take 57^89 as (7^4)^22 7^1...7^4 is 1 therefore 1 7=7 answer

For every 4 multiplications of 57, last digit will be 1. So, 88 times of multiplication of 57 will result 1 at last and with multiply with last 57 will give last figure as 7

Last digit will be 7 as unit digit of any number repeating itself after every fifth power. Hence after 88the power the 89th term will be having 7 as last digit.

Last digit is 7

since after 7^4,again 7 comes in the first place,so 9=4*2+1.thats why 7 will come again

Get the answer of 7 to the power 9 you will get the last digit

Power cycle 7,9,3,1

last no from 57^(89) is evidently last no from 7^(9) 7 7 7 7 7 7 7 7 7=40353607 take the last no from it that is 7

7^5 the last digit will be 7 again so in 7^85 it will be 7 so just do the sequence 7^86 will be 9, 7^87 will be 3 7^88 will be 1 then the last one 7^89 will be 7 again

My method was to use the last digit by power: 57 to the first power would have a last digit of 7, with the second power having a last digit of 9. You keep doing this until you have reached the number 7 again, which you will leave out. Then, count how many of the different various last digits there are and record. Divide the power of 89 by the number of possible last digits to receive the minimum power. Now look back at the list you made and write down the minimum power where you will follow up with the last digits per power again (notice there is a pattern). So, if you were given 13.5 as minimum power, then you will go through the list until you have ended up at the power of 13 which the pattern should reveal what the last digit is.

(57)^89 = (57)(3249)^44 = (57)(................... 1) = ................... 7

So , the answer is 7

Another solution :

(57)^89 = (5 X 10 + 7)^(22 X 4 + 1) = (7)^(1) (mod 10) = 7 (mod 10)

The question is asking the remainder of the number $(mod 10)$ . Thus it makes sense to look for a pattern in the powers of 7 since 7 is congruent to ${57} mod { 10}$ . The powers of 7 have the following classes $(mod 10)$ : 1, 7, 9, 3. Since the pattern repeats after every 4th power, we must look at ${89} mod {4}$ to see which class $57^{89}$ falls into. We find that 89 is congruent to 1 mod 4, so $57^{89}$ is in the class of 7. Therefore, $57^{89}$ is congruent to ${7} mod {10}$ .

Last digit of any number whose unit digit is 7 are 7,9,3,1 with power 1,2,3,4 then repeating , that with power 5 last digit is 1 and so on. Using this trick we find with power 88 last digit is 1 and with power 89 the last digit is 7

The fifth power of every integer has the same final digit as the original integer. Furthermore, the final digits repeat every four successive powers. I suppose this fact is related to the reduced residue system modulo 10 $(\{1, 3, 7, 9\}, \times_{10})$ happening to be the cyclic group of order 4.

This reduces the whole calculation from $57^{89}$ to $7^1$ .