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Difference between revisions of "2003 USAMO Problems/Problem 1"

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We proceed by induction.  For our base case, <math>n=1 </math>, we have the number 5.  Now, suppose that there exists some number <math>a \cdot 5^{n-1} </math> with <math>n-1 </math> digits, all of which are odd.  It is then sufficient to prove that there exists an odd digit <math>k </math> such that <math> k\cdot 10^{n-1} + a \cdot 5^{n-1} = 5^{n-1}(k \cdot 2^{n-1} + a) </math> is divisible by <math>5^n </math>.  This is equivalent to proving that there exists an odd digit <math>k </math> such that <math> k \cdot 2^{n-1} + a </math> is divisible by 5, which is true when <math> k \equiv -3^{n-1}a \pmod{5} </math>.  Since there is an odd digit in each of the residue classes mod 5, <math>k </math> exists and the induction is complete.
 
We proceed by induction.  For our base case, <math>n=1 </math>, we have the number 5.  Now, suppose that there exists some number <math>a \cdot 5^{n-1} </math> with <math>n-1 </math> digits, all of which are odd.  It is then sufficient to prove that there exists an odd digit <math>k </math> such that <math> k\cdot 10^{n-1} + a \cdot 5^{n-1} = 5^{n-1}(k \cdot 2^{n-1} + a) </math> is divisible by <math>5^n </math>.  This is equivalent to proving that there exists an odd digit <math>k </math> such that <math> k \cdot 2^{n-1} + a </math> is divisible by 5, which is true when <math> k \equiv -3^{n-1}a \pmod{5} </math>.  Since there is an odd digit in each of the residue classes mod 5, <math>k </math> exists and the induction is complete.
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=== Solution 2 ===
 
=== Solution 2 ===
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Note:  Not only does this prove the desired claim that there exists such a number, but it also proves that there is exactly one such number.   
 
Note:  Not only does this prove the desired claim that there exists such a number, but it also proves that there is exactly one such number.   
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{{alternate solutions}}
  
 
== Resources ==
 
== Resources ==

Revision as of 20:44, 6 April 2013

Problem

(Titu Andreescu) Prove that for every positive integer $n$ there exists an $n$-digit number divisible by $5^n$ all of whose digits are odd.

Solutions

Solution 1

We proceed by induction. For our base case, $n=1$, we have the number 5. Now, suppose that there exists some number $a \cdot 5^{n-1}$ with $n-1$ digits, all of which are odd. It is then sufficient to prove that there exists an odd digit $k$ such that $k\cdot 10^{n-1} + a \cdot 5^{n-1} = 5^{n-1}(k \cdot 2^{n-1} + a)$ is divisible by $5^n$. This is equivalent to proving that there exists an odd digit $k$ such that $k \cdot 2^{n-1} + a$ is divisible by 5, which is true when $k \equiv -3^{n-1}a \pmod{5}$. Since there is an odd digit in each of the residue classes mod 5, $k$ exists and the induction is complete.

Solution 2

First, we note that there are $5^n$ n digit numbers with only odd digits. Now, we will prove that none of these numbers have the same residue mod $5^n$, and therefore one of them must be 0 mod $5^n$.


Proof by contradiction: Assume we have two distinct numbers $A_1A_2A_3...A_n$ and $B_1B_2B_3...B_n$ with only odd digits that leave the same residue mod $5^n$. Then, subtracting the larger from the smaller would yield a new number that is a multiple of $5^n$ and has only even digits. We could then halve all of the digits in that number to get a second multiple of $5^n$ with at most n digits that only uses the digits 0 through 4.


Lemma: Every multiple of $5^n$ with n digits or less has a 5 as one of its digits.

All numbers of type can be written as $k5^n$. Then, let $k$ have $x$ factors of $2$ in it. ($x<n$, or else our number would have more than n digits). So, we have $k5^n=a2^x5^n=a*10^x*5^{n-x}$ for some odd a. Now $a*10^x*5^{n-x}$ is an odd multiple of 5 ($a5^{n-x}$) with x zeroes after it, and all multiples of 5 end in 5. Therefore, $a*10^x*5^{n-x}$ always contains a 5 as its $(x+1)^{st}$ digit, and we have proven our lemma.

By our lemma, our number with only the digits 0 through 4 cannot be a multiple of $5^n$, and so we have reached a contradiction. QED

Note: Not only does this prove the desired claim that there exists such a number, but it also proves that there is exactly one such number.

Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.

Resources

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