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Divisibility of a 180-Digit Number by 99

Date: 05/10/2007 at 11:36:04
From: Nathan
Subject: divisibility of 180 digit number by 99

Hello, I was observing a math contest recently that had the 
following question:

Consider the 180 digit number formed by putting the numbers 10 to 
99 together (1011121314...99).  If you divide this number by 99, 
what is the remainder?

I don't know how to begin.  Thanks for your help.




Date: 05/10/2007 at 21:01:11
From: Doctor Greenie
Subject: Re: divisibility of 180 digit number by 99

Hi, Nathan --

The answer is zero; the 180-digit number is evenly divisible by 99.  
We can prove that result as described below.

The number is divisible by 99 if it is divisible by both 9 and 11.

For a number to be divisible by 9, the sum of the digits has to be 
divisible by 9.  So in this problem, the sum of the digits used to 
write the numbers from 10 to 99 must be divisible by 9.

For a number to be divisible by 11, the difference between the sum 
of the odd-numbered digits (1st, 3rd, 5th, etc.) and the sum of the
even-numbered digits (2nd, 4th, 6th, etc.) must be divisible by 11.  

In this number, made up by concatenating all the 2-digit numbers from
10 to 99, the sum of the even-numbered digits is the sum of the digits
in the ones column of the numbers from 10 to 99, and the sum of the
odd-numbered digits is the sum of the digits in the tens column of
those numbers.

The digits in both the ones column and the tens column for the 
numbers 10 to 99 consist of 9 each of the digits 0 through 9.  That 
sum is

  9(0 + 1 + 2 + ... + 8 + 9) = 9(45) = 405

Thus the sum of all 180 digits in the number is 405 + 405 = 810.  This
sum is clearly divisible by 9; therefore the 180-digit number is
divisible by 9.

And the sums of the digits in the tens column and ones column are 
the same, so the difference between those sums is zero, which is a 
multiple of 11.  So the 180-digit number is also divisible by 11.

And since the 180-digit number is divisible by both 9 and 11, it is 
divisible by 99.  Therefore, the remainder when this number is 
divided by 99 is zero.

Cool problem!  I work with high school math teams and often write 
questions for contests; this is a good one for me to use.

- Doctor Greenie, The Math Forum
  http://mathforum.org/dr.math/ 
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