Build the number one place at a time and let the choices multiply — this is the counting principle at the root of all my work on chance. The hundreds place may take any of the 4 digits. Once it is spent, only 3 digits remain for the tens place, for we allow no repeats. Then just 2 remain for the units place. Because each first choice pairs with every second choice, and each of those with every third, the possibilities branch as a product: 4 · 3 · 2 = 24. Why we multiply, not add: addition would count separate options; multiplication counts combinations of independent decisions — and here each decision truly follows the last. See how “no repeats” quietly lowers the count at every step — 4, then 3, then 2. Grasp this branching, and the whole science of counting is yours.

Do not chase the five unknown numbers — you do not need them. The heart of every average is a single idea: an average is only a total shared out evenly, so total = average × count. Restore each average to its total and balance the two. Five numbers averaging 18 hold a total of 5 · 18 = 90. After one departs, four numbers averaging 20 hold 4 · 20 = 80. The number that left carried away exactly the difference: 90 − 80 = 10. A check that also teaches: removing a number below the old average should pull the average up — and indeed 10 sits below 18, and the average rose from 18 to 20. Turn averages into totals and the fog lifts every time. This is the very act of restoring and balancing from which algebra takes its name — and you just performed it.

That little word “or” hides a trap that has fooled counters for centuries. If you simply add the multiples of 3 to the multiples of 5, every number that belongs to both lists — 15, 30, 45, and so on — is counted twice. My remedy is what today is called Inclusion–Exclusion: count each group, then subtract the shared part exactly once. Multiples of 3 up to 200: ⌊200 ÷ 3⌋ = 66. Multiples of 5: ⌊200 ÷ 5⌋ = 40. The numbers in both lists are the multiples of 15: ⌊200 ÷ 15⌋ = 13. So the true count is 66 + 40 − 13 = 93. Remember this: for “A or B,” always compute A + B − (both). Master that single correction and you will never miscount an overlap again — the same principle scales all the way up to the mathematics of the modern world.

A merchant who believes “up 25%, then down 20%” means “up 5%” will one day lose his purse — for each percentage is taken of a different amount. Walk it step by step, each on its own base. First, raise $40 by 25%: the rise is 0.25 · 40 = $10, so the price becomes $50. Then take 20% off that $50: the cut is 0.20 · 50 = $10, leaving 50 − 10 = $40. Why it returns exactly to the start: raising by a quarter multiplies by 1.25, and cutting by a fifth multiplies by 0.80 — and 1.25 · 0.80 = 1, so the two changes perfectly undo each other. Always apply a percentage to the amount actually before you, never to the one you started with. Keep that discipline and your reckoning — in a shop or in a science — will always balance.

A shape, you will find, always hides an arithmetic. Look at what the tilted square leaves behind: four identical right triangles in the corners, each with two short legs of 5 (half of a side of 10). Each corner triangle has area ½ · 5 · 5 = 12.5, and four of them together cover 4 · 12.5 = 50. The whole square measures 10 · 10 = 100, so the shaded square must be 100 − 50 = 50. The beautiful truth: joining the midpoints of any square yields an inner square with exactly half the area. (See it through my theorem as well: the tilted side is √(5² + 5²) = √50, so its square is 50.) When a figure resists you, count what surrounds it. The path around a problem is often shorter than the path through it — remember that, and geometry will open to you.

From x + 2y = 10, we get x = 10 − 2y. Substitute into 3x − y = 9: 3(10 − 2y) − y = 9, so 30 − 7y = 9, giving y = 3. Then x = 10 − 6 = 4, and x − y = 4 − 3 = 1.

When Ana has run 4 miles, she has spent 4/8 = 0.5 hours = 30 minutes. Ben, at 6 mph, has covered 6 × 0.5 = 3 miles in that time, so he is 1 mile behind. He needs 1/6 hours = 10 minutes to catch up.

A number divisible by both 6 and 9 is divisible by LCM(6,9)=18. Multiples of 18 less than 100: 18, 36, 54, 72, 90 — wait, and 108 exceeds. Count 18·1 through 18·5 gives 5 values (not 7). Correction: the correct answer is B=5. This sample is being revised.

If sides are a and b: ab = 60 and 2(a+b) = 34, so a+b = 17. Diagonal² = a² + b² = (a+b)² − 2ab = 289 − 120 = 169. Diagonal = 13.