Canadian Innovation Math Contest · 2026
The 2026 theme: Mathematics of Compassion
Every problem in the 2026 contest grew from an act of care — packing care kits for newcomer families, walking to a rescue centre, arranging flowers for seniors. The mathematics is real contest mathematics. The situations are stories of people helping people. Each walkthrough below pauses for you to try the problem, then names the trick that solves it.
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Add every whole number from 1 to 100. In 1786, nine-year-old Carl Friedrich Gauss solved this in under a minute. This trick — Gauss pairing — comes back in Levels 2, 3, and 5.
Six problems that climb from spotting a pattern (Level 0) to combining two named techniques (Level 5).
Maya makes snack bags every day at a welcome centre. On Monday she makes 14 bags, Tuesday 19, Wednesday 24, and Thursday 29. If the pattern keeps going, how many bags will she make on Friday?
Each day grows by the same amount. Spot the daily step, then extend the pattern one more day.
A team writes welcome cards for new families, but only every other day. On Day 1 they write 20 cards, Day 3 they write 25, Day 5 they write 30. If the pattern continues, how many cards will they write on Day 17?
The step is +5 every two days. From Day 1 to Day 17 is fewer steps than you might think. Counting steps carefully — not days — is the whole game.
A community book corner has 15 shelves. The first shelf holds 7 books, the second holds 8, the third holds 9, and so on — each shelf has one more book than the one before. What is the total number of books across all 15 shelves?
Write the sum forwards and backwards, then add the two rows column by column. Every column lands on the same total. There are 15 columns. The rest is mental math.
At a weekend meal program, the number of lunch boxes packed grows by 6 each day. On Day 1 she packs 12 boxes. After several days the total number of boxes packed is 210. How many days did the packing continue?
Every term is a multiple of the same number. Divide the pattern and the total by that factor — the big numbers become small ones you can add in your head. Then Gauss pair what is left.
A support line starts a phone tree to recruit volunteers. In Round 1, five new volunteers join. In each new round after that, the number of new volunteers triples. The tree stops before any round would add more than 1000 new volunteers. How many new volunteers join altogether?
For a geometric series, write the sum once, then multiply the whole sum by the ratio, then subtract. Every middle term cancels — same idea as Gauss pairing, different operation.
A meal-delivery van visits apartments numbered 1 to 199. It stops at every apartment that is a multiple of 11 and every apartment that is a multiple of 13. If an apartment is a multiple of both, it still counts once. What is the sum of all apartment numbers where the van stops?
Add the two sets, then subtract the overlap (multiples of both 11 and 13). Each piece is an arithmetic series you can sum using Gauss pairing.
Six disguises of one big idea — the bridge between two plans that share the same total. Leftover, shortage, swing, difference, adjustment, schedule.
A volunteer is packing care kits for new families. If she packs 2 kits, 6 juice boxes are left over. If she packs 5 kits, the juice boxes fit exactly. How many juice boxes go in each kit?
When two plans share the same total, the leftover from the smaller plan has to fill the extra groups in the larger plan. Divide the leftover by the gap and you have the per-group amount.
A volunteer is packing welcome boxes for new families. If she packs 8 boxes, she still needs 10 more soap bars. If she packs 6 boxes, the soap bars fit exactly. How many soap bars go in each box?
Same family as the Leftover Trick, just flipped. The shortage equals the gap times the per-group amount. Whether the difference is a leftover or a shortage, the bridge is the same.
A volunteer at an animal shelter is packing pet-food bags for rescued animals. If she fills 5 bags, 7 food packs are left over. If she fills 7 bags, she needs 3 more food packs. How many food packs are there altogether?
When one plan leaves a leftover and the other comes up short, the total swing is leftover plus shortage. The swing equals the gap times the per-group amount, so divide to find the per-group, then count the supply.
A volunteer at a community centre is preparing book bundles for newcomer families. If she makes 5 bundles, 16 books are left over. If she makes 7 bundles, 2 books are left over. How many books are there altogether?
When both plans leave a leftover, subtract the smaller pile from the bigger one. The difference is what filled the extra groups in the larger plan. Same bridge — subtraction instead of addition.
A volunteer is arranging flowers into vases for seniors at a care home. If she uses 3 vases, 18 flowers are left over. If she uses 5 vases with 2 more flowers in each vase than before, they fit exactly. How many flowers are there altogether?
When the per-group sizes differ between the two plans, match them first. Add the offset to the smaller plan, then the usual bridge works — difference divided by gap gives the per-group amount.
Mia walks to an animal rescue centre to help with evening care. If she walks at 40 metres per minute, she arrives 7 minutes late. If she walks at 50 metres per minute, she arrives 2 minutes late. How far is the rescue centre from her home?
The strongest disguise. Speed and time both change but the distance stays the same. Time Plan B saves at the slower speed equals the extra distance her faster speed gives over her full walking time.
One transferable technique — the Pretend Trick — wears new disguises across animals, vehicles, and shapes. Pretend everything is one kind, count what is missing, then swap to fix.
Nori is feeding the chickens and rabbits at a rescue farm. There are 5 animals altogether. Counting their legs, there are 14. How many rabbits live at the farm?
When two types of things share a known total of some unit (legs, wheels, sides), pretend everything is one kind first. Count what is missing. Divide by the per-swap gap. That is how many to swap.
Devon is at a community repair day, fixing bikes and trikes to donate. There are 12 vehicles altogether. They have 28 wheels in total. How many tricycles did Devon fix?
Same Pretend Trick, new surface. The unit pair shifts from legs to wheels, so the gap per swap changes. The lesson: the technique transfers. A math tool you know can travel to a new problem and still work.
Asha is at a Saturday craft workshop. The kids are gluing sticks into shapes. Triangles have 3 sticks. Hexagons have 6 sticks. They built 15 shapes altogether. Counting all the sticks, there are 75. How many hexagons did the kids build?
Third repetition of the same trick — bigger numbers, new gap of 3 sticks per swap. Practice the same recipe on bigger numbers, and the bigger numbers stop feeling big.
Mira is weaving friendship bracelets for a children's hospital. Small bracelets use 2 beads. Large bracelets use 7 beads. There are 6 more large bracelets than small. Counting all the beads, there are 78. How many small bracelets did Mira weave?
A new tool for a new clue. When you know one type has K more than the other, set those K extras aside and subtract their cost. What remains forms matched pairs the old Pretend Trick can finish.
Riya is packing gift baskets for elderly residents at a retirement home. Regular baskets hold 3 items each. Big baskets hold 5 items each. Riya packs twice as many big baskets as regular ones. Counting all the items, there are 78. How many regular baskets did Riya pack?
When the clue is a ratio, build one block that matches the ratio (1 regular + 2 big). Count items per block, then divide the total. The block count is the answer for whichever type appears once per block.
Sora is taking a school math quiz. Each correct answer earns 5 points. Each wrong answer LOSES 2 points. Sora answered all 30 questions and scored 80 points. How many did Sora get right?
The hardest disguise. Same Pretend Trick, but one per-unit value is negative. Pretend all are the negative type first; subtracting a negative becomes adding. The differential per swap is the gap between positive and negative.
These walkthroughs cover real CIMC problem-solving questions. Register for the next contest and put these techniques to work.
