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

By: Kevin

Kevin — Wed, 18 Nov 2009 15:44:12 +0000

I did this the same way Dave did it, by instinct.

10 mod 11 === –1 mod 11, so

10^100 mod 11 === (-1)^100 mod 11 === 1 mod 11

By: R V

R V — Thu, 15 Oct 2009 09:18:41 +0000

We can also prove this by the Method of Induction:

Problem: (100)^n / 11 gives remainder 1
To prove that any power of 100 gives Remainder 1.

(100)^1 / 11 gives remainder 1 (after dividing to 99) .… 1
(100)^2 / 11 gives remainder 1 (after dividing to 9999)

Thus (100)^n / 11 gives remainder 1 (our Hypothesis from the above statements)

Now, (100)^(n+1) = 100 * (100)^n
Dividing the above by 11 we see that (100)^n already gives remainder 1, and the other 100 also gives remainder 1 (as per statement .… 1).

Thus, all powers of 100 shall give remainder 1 when divided by 11. (Proved)

I also liked the solution given by: Anthuan

By: Deepthi

Deepthi — Fri, 27 Feb 2009 09:53:39 +0000

By: alexander

alexander — Mon, 10 Nov 2008 16:34:25 +0000

childs play got the answer immediately

By: jumbo

jumbo — Thu, 16 Oct 2008 15:46:54 +0000

answer is one
100 is of the form to 11k+1 .
now (11k+1)*(11k+1)=11p+1
hence you raise 100 to any power it will always give remainder 1 on being divided by 11

By: Luca

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

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

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

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

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