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For how many positive integers n less than 100 is 5^n + 8^(n + 1) + 13^(n + 2) + 14^(n + 3) a multiple of 6?

 Apr 23, 2021
 #1
avatar+438 
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6=2*3, so we need to check if the number is divisible by both 2 and 3.

The first term will be an odd number, the second term will be even, the third will be odd, and the fourth will be even, and odd+even+odd+even = even, so we don't need to check if it's divisible by 2, because it must be for all positive integers.

That means we only need to check if it's divisible by 3. Notice that the middle 2 terms will both always leave a remainder of -1 when divided by 3 because 8 is congruent to -1 mod 3 and so is 13. 

Because of that, we only need to check if the sum of the first term and the last term will leave a remainder of 2 when divided by 3, since when that remainder is added to the sum of the remainder of the middle two terms, which is -1 + -1 = -2, the remainder becomes 0 since -2+2 leaves a remainder of 0 when divided by 3.

Since 5 is congruent to (-1) mod 3 and so is 14, n must be even so that the sum of the remainders of the first and last terms is equal to 1+1 = 2, which is congruent to 2 mod 3 (as mentioned before that's what we want). If n is odd, then the sum of the remainder of the first and last terms will be equal to (-1)^(2n+1) + (-1)^(2n+1) = 1 mod 3, which is not what we want.

There are exactly 49 even numbers under 100, which means that the answer is \(\boxed{49}\)

 Apr 23, 2021
 #2
avatar+438 
+1

Sorry, the previous solution was full of mistakes. here's a better one:

6=2*3, so we need to check if the number is divisible by both 2 and 3.

The first term will be an odd number, the second term will be even, the third will be odd, and the fourth will be even, and odd+even+odd+even = even, so we don't need to check if it's divisible by 2, because it must be for all positive integers.

 

The sum of the first and second terms mod 3 will always be 0 when n is a positive integer because when n is odd the first term will leave a remainder of 2 when divided by 3 and the second term will leave a remainder of 1 divided by 3, and when n is odd the first term will leave a remainder of 1 when divided by 3 and the second term will leave a remainder of 2 divided by 3. (that's because they are both congruent to 2 mod 3, but the powers of the two are always different in terms of parity). Either way, 2+1=3=0 mod 3, so the sum of the first 2 terms will always divide 3.

 

The third term will always leave a remainder of 1 when divided by 3, so we need the last term to leave a remainder of 2 when divided by 3. That only happens when n is an even number because n+3 will be odd, and when 14 is taken to a power that is odd, it will always leave a remainder of 2. (The reason behind that is that 14 is congruent to -1 mod 3, which means the remainder alternates between -1 = 3-1 = 2 and 1 mod 3).

 

That means that only even numbers work, so 49 is still our answer.

textot  Apr 24, 2021

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