1 raised to any power is always one, and 2 raised to any power divisible by 4 has a 6 in the ones place. The same applies to 4. Three raised to any power divisible by 4 ends in a one. Any number divisible by 5 has either a 0 or a 5 in the ones place. 1+6+1+6 is not divisible by 5. Therefore, if n is divisible by 4, 1^n +2^n +3^n +4^n is not divisible by 5.
But I think this is only part of what you want. I'm not sure how to continue from here.
This is a good start. You've got half of the "if and only if" claim here. Now what you need to do is show that if n is not divisible by 4, then 1^n + 2^n + 3^n + 4^n is divisible by 5.
Try putting everything in terms of remainders when divided by 5. What you've observed is that when n is divisible by 4, each of 1^n, 2^n, 3^n, and 4^n has a remainder of 1 when divided by 5. Now try for each case writing out the sequence of remainders. So for 2:
so you raise the numbers one through four to the same power. You divide each of them by 5. If the remainders added together are divisible by 5, then when you add up your numbers, the result is divisible by 5. There is a pattern when you figure out the remainders for different powers of each of these numbers divided by 5. For one, it is always 1. For two, the pattern is 1,2,4,3,1,2,4,3,etc. For three, the pattern is 1,3,4,2,1,3,4,2, etc. For four, it is simply 1,4,1,4,1,4,etc. the first remainder in each list being the remainder of the zero power, which happens to be divisible by 4.
When 1, 2, 3, and 4 are raised to the n power, so long as n is not divisible by 4, the remainders add up in one of 3 ways:
1+2+3+4=10 1+4+4+1=10 1+3+2+4=10
When n is divisible by four, the remainders add up in this way:
1+1+1+1=4
Not divisible by 5.
ergo, unless n is divisible by 4, 1^n + 2^n + 3^n + 4^n is always divisible by 5.
1 raised to any power is always one, and 2 raised to any power divisible by 4 has a 6 in the ones place. The same applies to 4. Three raised to any power divisible by 4 ends in a one. Any number divisible by 5 has either a 0 or a 5 in the ones place. 1+6+1+6 is not divisible by 5. Therefore, if n is divisible by 4, 1^n +2^n +3^n +4^n is not divisible by 5.
ReplyDeleteBut I think this is only part of what you want. I'm not sure how to continue from here.
This is a good start. You've got half of the "if and only if" claim here. Now what you need to do is show that if n is not divisible by 4, then 1^n + 2^n + 3^n + 4^n is divisible by 5.
ReplyDeleteTry putting everything in terms of remainders when divided by 5. What you've observed is that when n is divisible by 4, each of 1^n, 2^n, 3^n, and 4^n has a remainder of 1 when divided by 5. Now try for each case writing out the sequence of remainders. So for 2:
2^1: remainder 2
2^2: remainder 4
2^3: remainder 3
2^4: remainder 1
2^5: remainder 2
2^6: remainder 4
etc.
okaaay...
ReplyDeleteso you raise the numbers one through four to the same power. You divide each of them by 5. If the remainders added together are divisible by 5, then when you add up your numbers, the result is divisible by 5.
There is a pattern when you figure out the remainders for different powers of each of these numbers divided by 5.
For one, it is always 1.
For two, the pattern is 1,2,4,3,1,2,4,3,etc.
For three, the pattern is 1,3,4,2,1,3,4,2, etc.
For four, it is simply 1,4,1,4,1,4,etc.
the first remainder in each list being the remainder of the zero power, which happens to be divisible by 4.
When 1, 2, 3, and 4 are raised to the n power, so long as n is not divisible by 4, the remainders add up in one of 3 ways:
1+2+3+4=10
1+4+4+1=10
1+3+2+4=10
When n is divisible by four, the remainders add up in this way:
1+1+1+1=4
Not divisible by 5.
ergo, unless n is divisible by 4, 1^n + 2^n + 3^n + 4^n is always divisible by 5.
Okay, so am I going in the right direction?