2003 USAMO Problems/Problem 1


(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.


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 this 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.

Solution 3 (Construct digits of exemplar.)

$5^n \pmod {10} \equiv 5$.

$2^k 5^n \pmod {10^{k +1} } \equiv  10^k 5^{n-k} \equiv 5 \cdot 10^k$.

For any $n$, we can construct a $a = 5^n \cdot \sum_{0 \leq i < n}{2^{x_i}}$:

$x_0 = 1$

$a_i = 5^n \sum_{0 \leq j \leq i}{2^{x_j}}$

$x_{i} = \frac{a_{i-1}}{10^{i}} + 1 \pmod 2$. (This makes the $10^{i}$ place of $a_{i}$ odd, without changing the the smaller place digits).

$a = a_{n-1} = 5^n \cdot  \sum_{0 \leq j < n}{2^{x_j}}$ , with digits in place $10^i$ odd, for $0 \leq i<n$.

Since $\sum_{0 \leq j < n}{2^{x_j}}  < 2^n$, $a < 5^n 2^n = 10^n$.

By construction, the digits in all the places up through $10^{n-1}$ are odd, and since $a<10^n$, there are no other digits.

In fact, if $A_m = a$ that solves the case $n=m$, then $A_{m+1} \mod 10^ m \equiv A_m$.

Note that in some cases (like $n=5$) , both of $x_{n-1} \in \{0,1\}$ yield distinct numbers $a-5\cdot10^{n-1}$ and $a$ with all odd digits. $a-5\cdot10^{n-1}$ has $n-1<n$ digits, and so $x_{n-1}=1$ is needed for padding.

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

See also

2003 USAMO (ProblemsResources)
Preceded by
First question
Followed by
Problem 2
1 2 3 4 5 6
All USAMO Problems and Solutions

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