Question

Fool's problem of the day,

what is the remainder when we divide \( \large 9^{{100}^{{101}^{102}}}-5^{81^{82^{83}}} \) by \(13\).

Answer 1

Initial thoughts:

1. \(5^x\) will always end with the digits 25 for \(x\gt1\)

2. \(9^x\) ends in the following set of cyclicly recurring digits:\[\begin{align}
9^1&\to01\\
9^2&\to81\\
9^3&\to29\\
9^4&\to61\\
9^5&\to49\\
9^6&\to41\\
9^7&\to69\\
9^8&\to21\\
9^9&\to89\\
9^{10}&\to01\hspace{2cm}\text{(pattern starts repeating)}
\end{align}\]
Now in \(100^{101^{102}}\), will result in a number that is evenly divisible by 10, therefore \(9^{100^{101^{102}}}\) must end in the digits 01.

Not sure of the next step yet but I hope I am going along the right lines.

Answer 2

Another thought on this - using the binomial expansion:\[(a-b)^n=a^n-C_1^na^{n-1}b+...+(-1)^nb^n\]
then we can write:\[9=13-4\]and\[5=13-8\]so:\[9^n=(13-4)^n\]\[5^m=(13-8)^m\]therefore:\[9^n-5^m=(13-4)^n-(13-8)^m\]and if these are expanded using the binomial expansion, then all terms except the last one in \((13-4)^n\) will be divisible by 13 and, similarly, all terms except the last one in\((13-8)^m\) will be divisible by 13.
We can therefore just look at the last terms in order to work out the remainder when dividing by 13. Therefore:\[9^n-5^m\text{ mod }13=((-1)^n4^n-(-1)^m8^m)\text{ mod }13\]\[\qquad=((-1)^n(2^2)^n-(-1)^m(2^3)^m)\text{ mod }13\]In this case \(n=100^{101^{102}}\) which is even and \(m=81^{82^{83}}\) which is odd, therefore we get:\[9^{100^{101^{102}}}-5^{81^{82^{83}}}\text{ mod }13=((2^2)^{100^{101^{102}}}+(2^3)^{81^{82^{83}}})\text{ mod }13\]
Need to think if this is a fruitful avenue to pursue...

Answer 3

Keep trying, Asnaseer you will get it :)

Answer 4

Let\[9^{100^{101^{102}}} = b_{1} \mod 13\]\[5^{81^{82^{83}}} = b_{2} \mod 13\]Therefore,\[9^{100^{101^{102}}} - 5^{81^{82^{83}}} = b_{1} - b_{2} \mod 13\]With that established we could get the individual modulus then subtract them. Note that\[\forall a_{1}a_{2}b_{1}b_2 \in \mathbb{Z}(a_{1}a_{2} = b_{1}b_2 \mod n)\] and \[9^{100^{101^{102}}} = 9^{100 * 101 * 102} = 9^{1 030 200}= 9^{85 850 * 12} = 1 \mod 13\]\[5^{81^{82^{83}}} = 5^{81 * 82 * 83} = 5^{551 286} = 5^{49940 * 12}5^{6} = 5^{6} \mod 13\]by repeated use of Fermat's Little Theorem and the property above. Therefore,\[9^{100^{101^{102}}} - 5^{81^{82^{83}}} = 1 - 5^{6} \mod 13 = -15624 \mod 13 = 2 \mod 13\]
Therefore, the remainder is 2.