<s>[INST] Show $11^{11}+12^{12}+13^{13} =10k$ without direct calculation Prove that $11^{11}+12^{12}+13^{13}$ is divisible by $10$.
Obviously you could just put that in to a calculator and see the results, but I was wondering about some of the other approaches to this? I have not studied modulus', so if you could explain it without them, it would be better for me. Thanks!
 [/INST] The first digit of $11^{11}=$ the first digit of $1^{11}=$ $1$.

The first digit of $12^5$ equals the first digit of $2^5=$ the first digit of $32=2$.
$\text{(The first digit of $12^{10}) =$ (The first digit of $12^5)^2 = 4$ }$.
$\text{(The first digit of $12^{12}) =
       $ (The first digit of $12^{10}) \times($The first digit of $12^2)  = 6$ }$.

$\text{(The first digit of $13^{4}) =$ (The first digit of $3^4) = 1$ }$.
$\text{(The first digit of $13^{12}) =$ (The first digit of $(13^4)^3) = 1$ }$.
$\text{(The first digit of $13^{13}) =$ (The first digit of $13^{12} \times 3) = 3$ }$.

$\text{(The first digit of $11^{11} + 12^{12} + 13^{13}) =$
       (The first digit of $1+6+3) = 0$ }$.

Hence $11^{11} + 12^{12} + 13^{13}$ is a multiple of $10$
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