Elementary Math Olympiad - MOEMS Exams

Mathematical Olympiads for Elementary and Middle Schools (MOEMS) is a worldwide math competition, organized by a not-for-profit foundation with the same name. Two dozen other nations also participate in the competition. There are two divisions, Elementary and Middle School. Elementary level problems are for grades 4-6 and Middle School level problems are for grades 7-8, though 4-6 graders may participate in Middle School problems.

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Olympiad 1 Oct 1 to Oct 31st'2022
Olympiad 2 Nov 1 to Nov 31th'2022
Olympiad 3 Dec 1 to Dec 31st'2022
Olympiad 4 Jan 1 to Jan 31st'2023
Olympiad 5 Feb 1 to Feb 28th'2023

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Some sample of problem from this course with their details solution.

Problem: Evaluate: 51 – 33 + 42 – 33 + 24 – 33 + 15 – 33. ( Source: 2020 Contest 1 division E Q1)
Solution : Strategy: Find the difference between each pair of numbers and then find the sum.
Notice that each pair (51 – 33) + (42 – 33) + (24 – 33) + (15 – 33) = (+18) + (+9) + (– 9) + (–18).
The differences result in two pairs of additive inverses, (+28) + (–28) + (+9) + (–9), which sum to 0.

METHOD 2 Strategy: Use the Commutative and Associative Properties to find the sum of the positive numbers and the sum of the negative numbers and then combine them.
Looking at the problem, there are four groups of −33: 4 ✕ (−33) = −132. Also, there are two groups of +66 (15 + 51 and 24 + 42): 2 ✕ (+66) = +132. It follows that (−132) + (+132) = 0.
METHOD 3 Strategy: Do the arithmetic left to right.

Asking: A palindromic number is a number that reads the same of forwards as it does backwards (example: 282 or 51715). How many 5-digit palindromic numbers have digits that add up to 14? (Source: 2022 Contest 4 division E Q5)
Ans:- 165
Strategy: Consider digits that sum to 14.
Let the palindromic number be represented by abcda, where a, b, and c are digits. Then 2a+2b+c=14. Since 2a and ab are even, c must also be even so c must be 0,2,4, or 8. If c 0, a + b a 7. Since a can be any number from I to 7, there are 7 such palindromic numbers.
If c - 2, a +b a 6. Since a can be any number from I to 6, there are 6 such palindromic numbers.
If c is 4, a + b 5. Since a can be any number from I to 5, there are 5 such palindromic numbers.
If c in 6, a + b a 4. Since a can be any number from I to 4, there are 4 such palindromic numbers.
If c 8, a + b 3. Since a can be any number from I to 3, there are 3 such palindromic numbers.
All together there are 7 + 6 + 5 + 4 + 3 25 palindromic 5-digit numbers whose digits sum to 14.

Question: Find the sum of the following nine products: 1 × (10 – 1), 2 × (10 – 2), 3 × (10 – 3), …, 9 × (10 – 9). (Source: 2020 Contest 1 division E Q2)
Answer:165 Strategy: Look for patterns in the parentheses and the product terms.
Notice that the first term of each product increases consecutively from 1 to 9 and the second term for each difference in the parentheses increases by 1 so each resulting difference decreases consecutively from 9 to 1.
This pattern creates the products 1 ✕ 9 and 9 ✕ 1, 2 ✕ 8 and 8 ✕ 2, 3 ✕ 7 and 7 ✕ 3, 4 ✕ 6 and 6 ✕ 4, and 5 ✕ 5. The sum will be 9 + 9 + 16 + 16 + 21 + 21 + 24 + 24 + 25 = 18 + 32 + 42 + 48 + 25 = 165.

Hundreds of thousands of students participate annually in MOEMS events. We at our academy will help you to prepare for the exam with the help of our best educators. Not only that we provide free MOEMS mock tests, participation in the official contest, and more. Join Talent academy and get your MOEMS journey started.

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The competition is initializing from July 2022 to October 2022 at 11:59 PM ET.
TALENT ACADEMY is going to conduct an official MOEMS test. This test will give a glance at the MATHEMATICS Olympiad idea .
You can go through the details on the official website, the list provided above. Register on our website.

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