Daily Math Visuals

Every number has a hidden recipe — and it's always the same.
Take 60. Break it apart.
• 60 = 6 × 10 = 2·3 · 2·5 → primes {2, 2, 3, 5}
• 60 = 4 × 15 = 2·2 · 3·5 → primes {2, 2, 3, 5}
• 60 = 5 × 12 = 5 · 3·(2·2) → primes {2, 2, 3, 5}
Three different starting points. Same four primes. Always.
That's the Fundamental Theorem of Arithmetic — every whole number has exactly ONE prime recipe.
And from those primes alone, you get every single divisor of 60. All twelve of them — hidden in 2, 2, 3, 5.
The primes ARE the number's DNA.
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How do you find a loop in a chain of arrows?
Picture a chain of stones, each with an arrow pointing to the next one. Sometimes the last arrow points back to an earlier stone — the chain loops forever.
The easy way: write down every stone you visit. Step on one you've already written down → loop. Works, but costs memory.
Floyd's trick costs nothing.
Send two friends. The tortoise hops one stone at a time. The hare hops two. If the chain has an end, the hare reaches it. If it loops, the hare goes around — and one day lands on the same stone as the tortoise.
Why? Once both are inside the loop, the hare gains exactly 1 step on the tortoise per round. After at most (loop size) rounds, the hare catches up. Always.
Two pointers. Zero memory.
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Two Sum — the classic pattern that takes a quadratic problem to linear.
Problem: given an array of numbers and a target, find two numbers that sum to the target. Return their indices.
Brute force checks every pair. For n items that's n(n−1)/2 pairs — quadratic time. With a million elements, that's 500 billion comparisons.
The reframe: don't think "find a pair". For each number x, what you actually want is the complement — target minus x. That's a LOOKUP problem. Hash maps do lookups in O(1).
Walk the array once. For each x:
• if (target − x) is already in the seen map → return both indices
• else → store x, continue
O(n) time. One pass. Half a million times faster on a million elements.
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All 92 solutions of the 8 queens puzzle, overlaid on one board.
The puzzle: place 8 queens on a chessboard so no two attack each other. There are exactly 92 ways.
Most videos show one solution. We're showing every single one — on top of each other.
For each square, count how many of the 92 solutions place a queen there. Then color it by that count.
The result is a heatmap with hidden structure:
• Corners — visited by only 4 solutions each.
• 8 peak squares — visited by 18 each.
• Perfect 4-fold symmetry: flip left-right, top-bottom, or rotate 180° → same heatmap.
No single solution shows this. You only see it when you stack all 92.
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A viewer asked for the Seed Corn Problem — a classic arithmetic puzzle.
A farmer sold 1/4 of his corn to one merchant, 1/3 of what was left to another, and 1/2 of what was left to a third. He had 6 bushels remaining — his seed corn for next year. How much did he start with?
The beautiful trick: each of those three sales is exactly 1/4 of the original. Look:
• 1/4 of x = x/4
• 1/3 of 3x/4 = x/4
• 1/2 of x/2 = x/4
Three different-looking fractions, same quarter every time. Plus the leftover quarter = seed corn. Four equal pieces of 6 = 24.
Moral: don't eat your seed corn. Save what you need to plant again.
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Around 500 BCE, Pythagoras found two numbers that secretly feed each other.
220's proper divisors: 1, 2, 4, 5, 10, 11, 20, 22, 44, 55, 110. Sum: 284.
284's proper divisors: 1, 2, 4, 71, 142. Sum: 220.
Each is the sum of the other's pieces. The Pythagoreans called them friendship numbers.
For 800 years afterward, nobody found a second pair. The next one wasn't discovered until around 850 AD by the mathematician Thabit ibn Qurra.
Only 13 amicable pairs exist below 10,000. They're genuinely rare.
Numbers really do have stories.
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Sliding window — one of the most reusable DSA patterns.
Problem: given an array and k, find the max sum of any k consecutive elements.
Brute force re-sums every window from scratch → O(n·k). For each of n positions, you add k elements. Wasteful.
The insight: when the window slides one step right, only ONE new element enters and only ONE leaves. The middle k−2 elements are reused. So:
new_sum = old_sum − leaving + entering
Constant work per slide → O(n) total.
On arr = [2, 6, 3, 5, 8, 1, 4] with k = 3: the window slides through five positions, max sum is 16 at [3, 5, 8].
The one-line trick — cur += arr[i] − arr[i−k] — is the whole pattern.
Same technique applies to longest substring without repeating characters, anagrams, max-of-subarrays, and dozens more.
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