Prove 2^2013 + 13 is a multiple of 7

[@alphs4184]

Here's my take, no modular arithmatic, no binomial expansion, no number theory. Just pure brute force . Observe that 2^2013 + 13 is 2^2013 - 1 + 14. 14 is just 7x2 but 2^2013-1 can be expanded by using geometric series to 1 + 2 + 2^2 + 2^3 + ... + 2^2012, which is 111...111 in binary (2013 digits binary of all 1). Similarly, 7 is 2^3 -1, which can be expanded to 1 + 2 + 2^2 which is 111 in binary (3 digits of 1). Since 2013 is divisible by 3 (2+0+1+3 = 6), 111...111 must be divisible by 111. Why? Because the result is obviously 100100...1001 in binary (670 copies of100 followed by 1, 2011 digits total) which definitely is an integer. The cool thing when you brute force is that not only you've proven that 7 divides 2^2013+13 but you also got the quotient (although, it's in binary).

[@MyOneFiftiethOfADollar]

@@alphs4184 if you want to call appealing to the geometric series and binary “pure brute force”, then OK 😀 Geometric series is used frequently in number theory, e.g. the sum of the divisors of an integer formula. I used to code in assembly and have fond memories of binary and hexadecimal! Thx much for your point of view.

[@fgvcosmic6752]

(Before watching) My first thought is immediately Fermats Little Theorem, but I'm assuming we probably cant do that. Could be useful that 2³ = 1 mod 7.

[@MyOneFiftiethOfADollar]

Your way is better! 2^2013 + 13 = (2^3)^671 + 13 == 1 + 13 = 14 == 0 mod 7 I didn’t notice that since FLT gives 2^6 == 1 mod 7 Nice find! Thanks

[@whycantiremainanonymous8091]

Before watching: 8-1=7 If 8^n-1 is divisible by 7, 8^(n+1)-1=7×8^n+8^n-1 is also divisible by 7. The rest is basic arithmetic (8=2^3, 2013 is divisible by 3, so is a power of 8, 2^2013+13 is 2×7 more than 2^2013-1).

[@whycantiremainanonymous8091]

After watching: my way of doing it is actually easier 😀

[@ananyapatil7552]

Witty name for a channel 😊

[@MyOneFiftiethOfADollar]

Thx, nerd humor at its finest😀 2 cents = dollar/50