MOEMS Practice Problems with Solutions (Div. E and M)

MOEMS practice problems are easy to find online, but explanations are scarce. Most of what turns up is scanned contest pages with a bare answer key, which helps a child who already got the problem right and does little for the child who didn’t.

This page gives you two original practice olympiads, one for Division E and one for Division M, set up like the real thing: five problems, 30 minutes. Every solution names the strategy that cracks the problem, because MOEMS is a contest about strategy. None of these problems comes from a real MOEMS contest.

Quick refresher on the format: MOEMS runs five contests, November through March, and each one has five non-routine problems to solve in 30 minutes. A student’s score is the number of problems answered correctly, one point each. Calculators, rulers, graph paper, and other aids aren’t allowed, so practice with pencil and plain paper only. Our MOEMS preparation guide for parents covers the rest: awards, teams, and how to enter.

Division E or Division M?

MOEMS has two divisions, and the placement rules are simple:

Division E is for grades 4, 5, and 6. Fourth and fifth graders normally compete here.
Division M is for grades 6, 7, and 8. A team with any 7th or 8th grader must be in Division M.
Sixth graders can be on either, depending on how their school or program sets up its teams. Fourth and fifth graders can also be placed on a Division M team if that’s appropriate for them.

If your child is in 6th grade, try the Division E set first. If it goes smoothly, try the Division M set a few days later.

How to run a practice olympiad at home

Clear the table. Pencil and blank paper only, the same as contest day.
Set a 30-minute timer and hand over all five problems at once. That averages six minutes per problem (our arithmetic). Some problems will take two minutes and some ten, so teach your child to skip a stuck problem and come back to it.
Write a clear final answer for each problem, with the working on scratch paper. Every problem below has a single number as its answer.
Score it out of 5, then go over every problem together, including the ones your child got right. Ask “which strategy did you use?” before showing the solution.

For a first try, it’s fine to skip the timer entirely. The goal in October is learning the strategies; speed comes by the first contest in November.

Practice olympiad: Division E (grades 4-6)

E1. The ages of three siblings add up to 27 years. What will the sum of their ages be 4 years from now?

E2. Ana is standing in a line. She is 7th from the front and 12th from the back. How many people are in the line?

E3. How many whole numbers from 1 to 100 contain the digit 7 at least once?

E4. A 3 × 3 × 3 cube is painted red on all six outside faces. It is then cut into 27 small cubes, each 1 × 1 × 1. How many of the small cubes have red paint on exactly two of their faces?

E5. Use each of the digits 1, 2, 3, 4, 5, and 6 exactly once to write two 3-digit numbers. What is the smallest possible sum of the two numbers?

Practice olympiad: Division M (grades 6-8)

M1. The pages of a book are numbered 1, 2, 3, and so on. Printing all the page numbers uses exactly 151 digits. How many pages does the book have?

M2. The average of five numbers is 12. When one of the numbers is removed, the average of the remaining four numbers is 10. What number was removed?

M3. What is the value of 1 + 3 + 5 + 7 + … + 37 + 39, the sum of all the odd numbers from 1 to 39?

M4. What is the remainder when 2³⁰ (2 multiplied by itself 30 times) is divided by 7?

M5. A rectangle is divided into a grid of unit squares 3 rows high and 4 columns wide. How many rectangles of any size (including squares) can be traced along the lines of the grid?

Solutions: Division E

E1. Answer: 39 · Strategy: think about what changes

In 4 years, each sibling gets 4 years older. Three siblings × 4 years = 12 years added to the total. So the sum becomes 27 + 12 = 39.

The trap: 27 + 4 = 31, adding the 4 years once. A quick-looking problem still rewards reading carefully: the question is about three people, not one.

E2. Answer: 18 · Strategy: draw it

Being 7th from the front means 6 people are ahead of Ana. Being 12th from the back means 11 people are behind her. So the line has 6 + 1 + 11 = 18 people.

The trap: 7 + 12 = 19 counts Ana twice, once from each end. A quick sketch (a row of dots with Ana circled) makes this impossible to miss.

E3. Answer: 19 · Strategy: make an organized list

Count in two groups:

7 in the ones place: 7, 17, 27, 37, 47, 57, 67, 77, 87, 97. That’s 10 numbers.
7 in the tens place: 70, 71, 72, …, 79. That’s 10 numbers.

That’s 20, but 77 is in both lists, so it was counted twice. The answer is 20 − 1 = 19.

The habit: whenever you count two groups and add, ask “could anything be in both?” This overlap check comes up constantly in olympiad counting problems.

E4. Answer: 12 · Strategy: sort by position

Think about where each kind of small cube sits in the big cube:

Corner cubes show 3 painted faces. There are 8 corners.
Cubes in the middle of an edge (not at the corners) show 2 painted faces.
Cubes in the center of a face show 1 painted face.
The one cube in the very center shows none.

A cube has 12 edges, and on a 3 × 3 × 3 cube each edge has exactly one small cube between its two corners. So 12 small cubes have exactly two painted faces.

Check: 8 corners + 12 edges + 6 face centers + 1 hidden center = 27. ✓ Checking that the groups add back up to the whole is a good way to catch a miscount.

E5. Answer: 381 · Strategy: think about place value

In the sum of two 3-digit numbers, the hundreds digits are worth 100 each, the tens digits 10 each, and the ones digits only 1 each. To make the sum small, put the smallest digits where they’re worth the most:

Hundreds digits: 1 and 2
Tens digits: 3 and 4
Ones digits: 5 and 6

One way to write it: 135 + 246 = 381. (Any arrangement with 1 and 2 in the hundreds, 3 and 4 in the tens, and 5 and 6 in the ones gives the same sum, for example 136 + 245.)

The trap: 123 + 456 = 579 looks natural but wastes the small digits. The place-value idea is the whole problem: 100 × (1 + 2) + 10 × (3 + 4) + (5 + 6) = 300 + 70 + 11 = 381.

Solutions: Division M

M1. Answer: 80 · Strategy: count in blocks

Split the pages by how many digits their numbers have:

Pages 1 to 9 use 1 digit each: 9 digits.
That leaves 151 − 9 = 142 digits for the two-digit page numbers, at 2 digits each: 142 ÷ 2 = 71 pages.

The two-digit pages start at 10, so the 71 of them run from 10 to 80. The book has 80 pages.

Check: 9 + 2 × 71 = 151. ✓ (Pages 10 to 99 would need 180 digits, so the book never reaches three-digit page numbers.)

The trap: off-by-one errors. A student who knows there are 71 two-digit pages but forgets they start at page 10 might answer 71, or 81. Writing out “pages 10 to ___” and counting them avoids it.

M2. Answer: 20 · Strategy: turn averages into totals

An average is a total shared equally. Five numbers with an average of 12 add up to 5 × 12 = 60. Four numbers with an average of 10 add up to 4 × 10 = 40. The removed number is the difference: 60 − 40 = 20.

The habit: in almost every average problem, the first move is “average × how many = total.” Working with totals is much easier than working with averages.

M3. Answer: 400 · Strategy: pair from the ends

First count the terms. The odd numbers from 1 to 39 are the odd numbers in the first 40 whole numbers, and half of 40 is 20 terms.

Now pair the first term with the last, the second with the second-to-last, and so on:

1 + 39 = 40
3 + 37 = 40
5 + 35 = 40
… all the way to 19 + 21 = 40

Twenty terms make 10 pairs, and every pair is 40, so the sum is 10 × 40 = 400.

A bonus pattern: 1 = 1, 1 + 3 = 4, 1 + 3 + 5 = 9, 1 + 3 + 5 + 7 = 16. The sum of the first n odd numbers is always n × n, and 20 × 20 = 400 confirms the answer.

The trap: miscounting the terms. A student who thinks there are 39 terms (because the list ends at 39), or who makes 20 pairs out of 20 terms, gets an answer far too big. Counting the terms first, before any adding, prevents both mistakes.

M4. Answer: 1 · Strategy: look for a pattern

Nobody expects your child to multiply out 2³⁰. Instead, watch the remainders when powers of 2 are divided by 7:

Power	Value	Remainder ÷ 7
2¹	2	2
2²	4	4
2³	8	1
2⁴	16	2
2⁵	32	4
2⁶	64	1

The remainders repeat in a cycle of three: 2, 4, 1. Every power whose exponent is a multiple of 3 leaves remainder 1. Since 30 is a multiple of 3, 2³⁰ leaves a remainder of 1.

The habit: when a number is too big to compute, compute small cases until a pattern appears, then figure out where the big case lands in the pattern.

M5. Answer: 60 · Strategy: count choices, not shapes

Every rectangle in the grid is fixed by choosing two horizontal lines (its top and bottom) and two vertical lines (its left and right sides).

A grid 3 rows high has 4 horizontal lines. The number of ways to choose 2 of them is 6. (Label them 1 to 4: 12, 13, 14, 23, 24, 34.)
A grid 4 columns wide has 5 vertical lines. The number of ways to choose 2 of them is 10.

Every pair of horizontal lines combines with every pair of vertical lines, so there are 6 × 10 = 60 rectangles.

The trap: trying to draw and count the rectangles one by one. It can be done, but it’s slow and it’s almost impossible not to miss a few. Turning a counting problem about shapes into a counting problem about choices is one of the most useful ideas at the Division M level.

Scoring your practice olympiad

On the real contests, each correct answer is worth one point, so each practice set is scored out of 5. A few ways to read the result:

5 out of 5 without the timer: try the set again with the 30-minute clock, or move up (Division E students to the Division M set).
3 or 4: normal for a first practice olympiad. Look at why each miss happened. Most early misses are trap answers (31 in E1, 19 in E2), and those are fixed by checking, not by more practice.
0 to 2: fine in October. Go back through the solutions one strategy at a time, over several days, and try the set again in two weeks.

After each real contest, it also helps to ask your child’s PICO (the adult who runs the team) to go over the problems with the team, since they score the contest and have the answers.

Where to find more MOEMS practice problems

Free on moems.org: MOEMS posts sample contests for Division E and Division M, plus a Problem of the Week, for anyone to use.
With enrollment: a team’s PICO gets access to 100 more practice problems with solutions from MOEMS.
Our MOEMS Division E guide and MOEMS Division M guide: strategy-first chapters built around a problem-solving toolkit (drawing, organized lists, working backwards, patterns), with original practice problems, full practice olympiads, and step-by-step solutions. Both have a free sample.
Our free Problem of the Week: one original problem a week by email.

For this season’s contest windows and team fees, see the dates and deadlines in our MOEMS guide for parents.

This guide is not affiliated with or endorsed by Math Olympiads for Elementary and Middle Schools (MOEMS®). “MOEMS” is a trademark of its respective owners.