Some problems from Estonian Mathematical Olympiad: 2000

Problem 1: How many positive integers less than 20002001 and not containing digits other than 0 and 2 are there?

Solution 1: The set of integers under consideration consists of all integers with up to 7 digits containing only digits 0 and 2, all 8 digit integers of the form 20000***, and the integer 20002000. There are 2^{k-1} integers with exactly k digits 0 and 2, and 2^{3} integers of the form 20000***. So, the required number of integers is (2^{0}+2^{1}+2^{2}+ \ldots + 2^{6})+8+1=(2^{7}-1)+9=136

Problem 2: Find the last two digits of the number: 1! + 2! + 3! +4! + \ldots +2000!

Solution 2: This is a classic type of question: the product 1.2 \ldots 10 has 2, 5, and 10 as factors, therefore being divisible by 100. Hence, the last two digits of n! are zeros for any n \geq 10 and it suffices to find the last two digits of 1!+2! + 3! + \ldots + 9!. The last two digits of the summands are 01, 02, 06, 24, 20, 20, 10, 20 and 80, yielding 13 as the answer.

Problem 3: Real numbers x and y satisfy the system of equations:

x + y + \frac{x}{y}=10

\frac{x(x+y)}{y}=20

Find the sum of all possible values of the expression x+y.

Solution 3: By Viete’s relations, the possible values of x+y are included in the set of roots of the quadratic equation: a^{2}-10a+20=0.

Work hard and play harder! 🙂

Nalin Pithwa.

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