Proving the Divisibility of 81^7 - 27^9 - 9^13 by 45

This article aims to demonstrate the divisibility of the mathematical expression 81^7 - 27^9 - 9^13 by 45 using a multi-step proof technique. The key idea is to leverage the properties of modulo arithmetic and factorization. By proving the divisibility by both 5 and 9, we can conclude that 81^7 - 27^9 - 9^13 is divisible by their product, 45.

Step 1: Checking Divisibility by 5

First, we need to check if the expression is divisible by 5. We will start by simplifying each term modulo 5:

81 1 mod 5 27 2 mod 5 9 4 mod 5

Now let's calculate each term modulo 5:

81^7 1^7 1 mod 5 27^9 can be simplified using the pattern of powers of 2 modulo 5. The cycle repeats every 4, thus:

2^1 2 2^2 4 2^3 3 2^4 1

Since 9 mod 4 1, we have: 2^9 2^1 2 mod 5

9^{13} 4^{13}. Using the pattern of powers of 4 modulo 5, the cycle repeats every 2, thus:

4^1 4 4^2 1

Since 13 mod 2 1, we have: 4^{13} 4^1 4 mod 5

Putting it all together, we get:

81^7 - 27^9 - 9^{13} 1 - 2 - 4 -5 0 mod 5

Step 2: Checking Divisibility by 9

Next, we need to check if the expression is divisible by 9. We will start by simplifying each term modulo 9:

81 0 mod 9 27 0 mod 9 9 0 mod 9

Now let's calculate each term modulo 9:

81^7 0^7 0 mod 9 27^9 0^9 0 mod 9 9^{13} 0^{13} 0 mod 9

Putting it all together, we get:

81^7 - 27^9 - 9^{13} 0 - 0 - 0 0 mod 9

Conclusion

Since 81^7 - 27^9 - 9^{13} is congruent to 0 mod 5 and 0 mod 9, it follows that 81^7 - 27^9 - 9^{13} is congruent to 0 mod 45.

Thus, we conclude that 81^7 - 27^9 - 9^{13} is divisible by 45. This can also be verified by the simplification:

81^7 - 27^9 - 9^{13} 3^{28} - 3^{27} - 3^{26} 3^{26}[3^2 - 3^1 - 3^0] 3^{26}[9 - 3 - 1] 3^{26}5 3^{24} 3^2 5 3^{24} 9 5 3^{24} 45

Hence, 45 is a factor, making the expression divisible by 45.