Posted in number theory, proof, there's more than one way to do it.

This puzzle was in February's MathsJam Shout, contributed by the Antwerp MathsJam. Visit mathsjam.com to find your nearest event! Consider the set ${1, 11, 111, ...}$ with 2017 elements. Show that at least one of the elements is a multiple of 2017. The Shout describes this one as tough; you

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Posted in number theory.

Pick a number, any number1 Let's say 24,876,028, one of my favourites in the 20-millions. I can tell by looking at it that it's one more than a multiple of 9. It's not that clever a trick: I just added up the digits (getting 37), and repeated the trick with

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Posted in ask uncle colin, number theory.

Dear Uncle Colin, I've been asked to find the last two digits of $19^{100}$. For what reason, I cannot tell. However, my calculator bums out before I get to $19^{10}$! -- Many Other Digits, Unfindable Last Ones Hi, there, MODULO! What do you know, the clue to your problem is

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Posted in ask uncle colin, number theory.

Dear Uncle Colin, I wanted to work out $3^{41}\mod 13$: Wolfram|Alpha says it's 9, but MATLAB says it's 8. They can't both be right! What gives? MATLAB Obviously Doesn't Understand Logical Operations Hi, MODULO! First up, when computers disagree, the best thing to do is check by hand. Luckily, you

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Posted in number theory, surds.

Funny thing. Type $(2 + \sqrt{3})^{20}$ into Wolfram Alpha. (Or, if you're really lazy, click this link.). It's 274,758,382,273.999999999996 or so. The higher the power you pick, the closer $(2 + \sqrt{3})^n$ gets to an integer value -- although it never quite gets there, because $\sqrt{3}$ is irrational. So, how

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Posted in number theory.

This post is inspired by a question asked by Dan - thank you, Dan! So here's the gist of Dan's question: Take a random sum, e.g. 496866 + 446221 = 943087. Add up all the digits in each number (39, 19 and 31). Keep adding up the digits in each

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