Problem (Number Theory):
Each positive integer less than or equal to 2019 is written on a blank sheet of paper, and each of the digits 0 and 5 is erased. Compute the remainder when the product of the remaining digits on the sheet of paper is divided by 1000 .

Solution:
Count the digits separately by position, noting that 1 is irrelevant to the product. There are a total of 20 instances of the digit 2 in the thousands place. The digit 0 only occurs in the hundreds place if the thousands digit is 2 , so look at the numbers 1 through 1999. Each non-zero digit contributes an equal number of times, so there are 200 each of $1,2,3,4,6,7,8,9$. The same applies to the tens digit, except there can be the stray digit of 1 among the numbers 2010 through 2019, but again, these do not affect the product. In the units place, there are 202 of each of the digits. Altogether, there are 602 each of $2,3,4,6,7,8$, 9, along with 20 extra instances of the digit 2 . Note that $9 \cdot 8 \cdot 7 \cdot 6 \cdot 4 \cdot 3 \cdot 2=3024 \cdot 24=72,576$ leaves a remainder of 576 when divided by 1000 . Also $2^{20}=1024^{2} \equiv 24^{2}(\bmod 1000)$, so $2^{20}$ contributes another factor of 576 . The answer is therefore the remainder when $576^{603}$ is divided by 1000 . This computation can be simplified by using the Chinese Remainder Theorem with moduli 8 and 125 , whose product is 1000 . Note $576^{603} \equiv 0(\bmod 8)$ because 576 is divisible by 8 . Also $576 \equiv 76(\bmod 125)$. By Euler's totient theorem, $576^{100} \equiv 1(\bmod 125)$, so $576^{603} \equiv 76^{3}(\bmod 125)$. This can quickly be computed by noting that $76^{3}=(75+1)^{3}=75^{3}+3 \cdot 75^{2}+3 \cdot 75+1 \equiv 3 \cdot 75+1 \equiv-24(\bmod 125)$. Observing that $-24 \equiv 0(\bmod 8)$, it follows that $576^{603} \equiv-24(\bmod 1000)$, hence the desired remainder is 976 .

Final answer: 976