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 integers with exactly k digits 0 and 2, and integers of the form 20000***. So, the required number of integers is

Problem 2: Find the last two digits of the number:

Solution 2: This is a classic type of question: the product has 2, 5, and 10 as factors, therefore being divisible by 100. Hence, the last two digits of n! are zeros for any and it suffices to find the last two digits of . 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:

Find the sum of all possible values of the expression .

Solution 3: By Viete’s relations, the possible values of are included in the set of roots of the quadratic equation: .

*Work hard and play harder! 🙂*

Nalin Pithwa.

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