Comments on: Brain Teaser: The Really, Really, Really Big Number

by: Luca

Fri, 13 Jun 2008 11:29:22 +0000

I used an invariant property of multiples of 11:

n=k*11 in n the sum of odd position digits is equal to the sum of even position digits.

Thus, 100^n=10^2n has 2n+1 digits, and the previous multiple of 11 must have 2n digits.

so it can only be a number like 999...999, and the remainder must be 1

by: T-Mak1978

Sat, 24 May 2008 01:24:52 +0000

Raising 100 to the 100 power is really not relevant. When you divide 100 by 11 the remainder is 1. In this case, no matter how many zeros will be in the "big number", you will still come out with a remainder of 1.

by: Anthuan

Tue, 13 May 2008 12:03:03 +0000

My solution was a bit different.

10^0 mod 11 = 1
10^1 mod 11 = 10
10^2 mod 11 = 1
10^3 mod 11 = 10

Thus

10^n mod 11 = 1, if n is even, and
10^n mod 11 = 10, if n is odd!

by: Aaron

Mon, 24 Mar 2008 06:00:51 +0000

I think it's one, I don't have any crazy solution it's just my gut instinct =P

by: RC

Thu, 06 Mar 2008 20:33:09 +0000

The number 100^100 may be written as,

(99 + 1)^(99 + 1). By the exponent law, a^m * a^n = a^(m + n),

we can rewrite to this,

(99 + 1)(99 + 1)^99

and futher,

(99 + 1)(99 + 1)(99 + 1)^98

and so on, down the line. Let a = (9)(11). We get,

(a^2 + 2a + 1)(a + 1)^98 =

(a^3 + 3a^2 + 3a + 1)(a + 1)^97 =

(a^4 + 4a^3 + 6a^2 + 4a + 1)(a + 1)^96 = .. = a^99 + [98 terms] + 1

We find that every term is divisible by 11 (a is divisible) except the very last one, 1.

The remainder is 1.

Here's another quick one, which is learned from the above.

Using only 9s, express 100^100.

Well, it's no trick at all. 9/9 = 1.

Ans. (99 + 9/9)^(99 + 9/9)

or (99 + 9/9)(99 + 9/9)^99

by: Alvaro

Mon, 24 Dec 2007 14:10:19 +0000

Hello Jonathan, thanks for sharing that experience! We are very glad your students enjoyed the teaser, very impressive

Happy holidays

by: Jonathan

Sat, 22 Dec 2007 18:49:06 +0000

So, I finally used the puzzle in class, though perhaps not exactly as you intended. It was with 2 9th grade, pre-vacation algebra classes, and the results are here.

by: Teaching: The Really Big Number (applied) « JD2718

Sat, 22 Dec 2007 18:45:22 +0000

[...] What is the remainder when is divided by 11? [...]

by: » Brain Teasers and Games with a neuroscience angle: our Top 50 « Brain Fitness Revolution at SharpBrains

Sat, 03 Nov 2007 13:03:33 +0000

[...] 45. Find the the Really, Really, Really Big Number. [...]

by: Rami Lehti

Thu, 18 Oct 2007 08:10:51 +0000

Ok. The remainder is 1.
What is the solution?

The generic solution for all positive exponents is:
9*100^0+9*100^1+9*100^2+...+9*100^(n-1)

For this particular problem just substitute n=100.