Question

when (67^(67) + 67) is divided by 68 the remainder is
1) 1
2) 63
3) 66
4) 67

Answer 1

calculator?

Answer 2

Calculator will not work.......... U need super computer or a super brain...lol

Answer 3

@lgbasallote

Answer 4

...or wolframalpha....

Answer 5

YEP

Answer 6

you can do it manually actually...

7^1 = 7
7^2 = 49
7^3 = xxx3
7^4 = xxxx1
7^5 = xxxx7

so in every four powers, there is a cycle

you have 7^67

67/4 = 16 R 3

so 7^67 has a last digit of 3

if you add that to 67...then the last digit is 0

now...when you divide that by 8...i have no idea what the remainder is

Answer 7

40 is actually exactly divisible by 8

Answer 8

@mukushla could help...... but is offline

Answer 9

I think @experimentX can also do this

Answer 10

any formulas solve this?

Answer 11

here is the answer :)

http://en.allexperts.com/q/Algebra-2061/2010/6/math-88.htm

Answer 12

67^5 mod 68 = 67
67^(5*13+2) mod 67 = 67^(1+2) mod 68
with this you get
(67^3+67) mod 68 = (67^2 + 1)*67 mod 68

67^2 mod 68 = 67*67 mod 68 = 68 - 67
n*67 mod 68 = 68 - n, if n<68

(67^2 + 1) mod 68 = 2

so we have
(67^2 + 1)*67 mod 68 = 2*67 mod 68 = 68 - 2 = 66

Answer 13

\[67^{67}+67\equiv(-1)^{67}+67\equiv-1+67\equiv66 \ \ \text{mod} \ 68\]

Answer 14

i can't beat you on algebra man!!

Answer 15

;)