# Ask Uncle Colin: The Last Two Digits

Ask Uncle Colin is a chance to ask your burning, possibly embarrassing, maths questions -- and to show off your skills at coming up with clever acronyms. Send your questions to colin@flyingcoloursmaths.co.uk and Uncle Colin will do what he can.

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 in your name, who would have thought it? It's almost as if they're made up.

It's much less involved than it looks, once you start working modulo 100. This will always give you the last two digits of any calculation you care to compute, as long as you don't do anything silly like take roots or divide.

In particular, you can take powers in a very clever way. $19^2 = 361$, which is equivalent to 61 (modulo 100). Squaring that gives $19^4 \equiv 3721$, or $21 \pmod {100}$. But aha! $19 \times 21 = 399$, which means $19^5 \equiv 99 \pmod {100}$. Squaring that gives $19^{10} \equiv 99^2 = 9801 \equiv 1 \pmod {100}$.

So, if $19^{10} \equiv 1 \pmod {100}$, then $19^{100} = (19^{10})^{10} \equiv 1^{10} \equiv 1 \pmod {100}$.

That means, the last two digits of $19^{100}$ are 0 and 1, in that order.

Wolfram|Alpha says it's 75,051,624,198,251,984,443,456,989,853,061,891,539,043,939,434,909,537,798,332,873,934,101,480,896,578,056,472,849,915,762,891,214,746,171,016,655,874,432,115,640,378,001, by the way.

-- Uncle Colin ## Colin

Colin is a Weymouth maths tutor, author of several Maths For Dummies books and A-level maths guides. He started Flying Colours Maths in 2008. He lives with an espresso pot and nothing to prove.

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