#1**+10 **

Let's attack 4^34 first:

4^1 = 4

4^2 = 16

4^3 = 64

4^4 = 256 ---> This indicates that if the exponent is odd, the last digit will be a 4; if the exponent is even, the last digit will be a 6.

Now 7^35:

7^1 = 7 7^5 = 16807

7^2 = 49 7^6 = 117649

7^3 = 343 7^7 = 823543

7^4 = 2401 7^8 = 5764801

This indicates that it cycles in groups of 4, so you can keep doing this until you get to 7^35. (You really don't have to keep doing the multiplications.)

Another way is to divide the exponent by 4:

if the remainder is 1, the answer ends in a 7;

if the remainder is 2, the answer ends in a 9;

if the remainder is 3, the answer ends in a 3; and

if the remainder is 0, the answer ends in a 1.

Dividing 35 by 4, the remainder is 3; therefore the last digit is a 3.

Now, the problem is reduced to adding 1 to a number whose last digit is 6 to another number whose last digit is 3. So what is the last digit of 1 + 6 + 3?

geno3141 Sep 29, 2014

#1**+10 **

Best Answer

Let's attack 4^34 first:

4^1 = 4

4^2 = 16

4^3 = 64

4^4 = 256 ---> This indicates that if the exponent is odd, the last digit will be a 4; if the exponent is even, the last digit will be a 6.

Now 7^35:

7^1 = 7 7^5 = 16807

7^2 = 49 7^6 = 117649

7^3 = 343 7^7 = 823543

7^4 = 2401 7^8 = 5764801

This indicates that it cycles in groups of 4, so you can keep doing this until you get to 7^35. (You really don't have to keep doing the multiplications.)

Another way is to divide the exponent by 4:

if the remainder is 1, the answer ends in a 7;

if the remainder is 2, the answer ends in a 9;

if the remainder is 3, the answer ends in a 3; and

if the remainder is 0, the answer ends in a 1.

Dividing 35 by 4, the remainder is 3; therefore the last digit is a 3.

Now, the problem is reduced to adding 1 to a number whose last digit is 6 to another number whose last digit is 3. So what is the last digit of 1 + 6 + 3?

geno3141 Sep 29, 2014

#2**+5 **

$$\\1+4^{34}+7^{35}\\

4^n=4,16,64,256,.... $ when n is even it ends in 6$\\

7^n=7,49,343,2401, 16807, $ So the last digits are 7,9,3,1,7,...$\\

35=4*8+3$ so the last digit of $ 7^{35}$ is 3 (3 is the third number in the sequence)$\\

$Edited: Geno is right, these three digits need to be added$\\\\

$ The last digit will be hte last digit of $ 6+3+1 $ Which is zero $$$

Melody Sep 29, 2014