Archimedes of Syracusec. 287–212 BC · Mathematician & Engineer“Give me a place to stand, and I shall move the Earth.”

Three sides and no angle: this is Heron’s ground. Take the semiperimeter first.

s = (5 + 6 + 7)/2 = 9.

Area = √[s(s−a)(s−b)(s−c)] = √(9 · 4 · 3 · 2) = √216.

Simplify: 216 = 36 · 6, so the area is 6√6 ≈ 14.7.

A check by the law of cosines: cos of the angle between 5 and 6 is (25 + 36 − 49)/60 = 1/5, so its sine is √24/5, and the area is ½ · 5 · 6 · √24/5 = 3√24 = 6√6. ✓

Beware the right-triangle instinct: 5-6-7 is not right, since 25 + 36 ≠ 49. Treating it as such gives 15.

John Napier1550–1617 · Mathematician“Seeing there is nothing that is so troublesome as the multiplications and divisions.”

My logarithms turn multiplication into addition; read the equation backwards and the product reappears.

log₂[x(x − 2)] = 3, so x(x − 2) = 2³ = 8.

x² − 2x − 8 = 0, that is (x − 4)(x + 2) = 0, giving x = 4 or x = −2.

Now the step that decides the problem: check the domain. A logarithm demands a positive argument. At x = −2 both log₂(x) and log₂(x − 2) are undefined. At x = 4 we have log₂4 + log₂2 = 2 + 1 = 3. ✓

The answer is 4. Combining logarithms can create solutions the original equation never had; every candidate must be returned to the equation it came from.

Carl Friedrich Gauss1777–1855 · Mathematician“Mathematics is the queen of the sciences, and number theory is the queen of mathematics.”

The last two digits are the residue modulo 100. Find the period.

7² = 49,   7³ = 343 ≡ 43,   7⁴ = 7 · 43 = 301 ≡ 1 (mod 100).

So 7 has order 4 modulo 100 and the residues cycle 7, 49, 43, 1.

Since 2024 = 4 · 506, the exponent is a multiple of 4, and

7²⁰²⁴ = (7⁴)⁵⁰⁶ ≡ 1 (mod 100).

The last two digits are 01 — the leading zero matters, since we were asked for two digits.

Euler’s theorem promises only 7^φ(100) = 7⁴⁰ ≡ 1; the true order 4 divides 40, as it must, and finding it saves ten-fold work.

Pythagoras of Samosc. 570–495 BC · Mathematician & Philosopher“Number rules the universe.”

The law of cosines names the angle by the side that faces it. With b = 14 opposite the angle B, and the other two sides 13 and 15:

b² = a² + c² − 2ac cos B.

196 = 169 + 225 − 2(13)(15) cos B = 394 − 390 cos B.

So 390 cos B = 198, and

cos B = 198/390 = 33/65.

Sanity: 33/65 ≈ 0.508, an angle just under 60° — plausible for the middle side of a 13-14-15 triangle, whose angles are all acute. The famous area 84 confirms the shape: sin B = 2 · 84/(13 · 15) = 56/65, and indeed (33/65)² + (56/65)² = (1089 + 3136)/4225 = 1.

The distractor 56/65 is that sine — the right ratio, the wrong function.

Blaise Pascal1623–1662 · Mathematician & Philosopher“Chance is tamed the moment we learn to count the ways.”

Lay the ten balls in a row. Nine gaps separate them. Choosing where to cut the row into four nonempty runs means choosing 3 of those 9 gaps — and the boxes, being distinguishable, receive the runs in order.

Number of ways = C(9, 3) = (9 · 8 · 7)/(3 · 2 · 1) = 84.

Equivalently, substitute x_i = y_i + 1 to absorb the “nonempty” condition; then y₁ + … + y₄ = 6 with y_i ≥ 0, and the stars-and-bars count is C(6 + 3, 3) = C(9, 3) = 84.

Allowing empty boxes would give C(13, 3) = 286 — the distractor. The single word “nonempty” moves the answer by a factor of more than three.

Gottfried Wilhelm Leibniz1646–1716 · Mathematician & Philosopher“Nothing is more important than to see the sources of invention.”

See the source of the invention, not the invention. Each term hides a difference; rationalise it and the sum collapses.

Multiply the k-th term above and below by √(k+1) − √k:

1/(√k + √(k+1)) = (√(k+1) − √k) / [(√(k+1))² − (√k)²] = √(k+1) − √k,

because the denominator is simply (k+1) − k = 1.

The sum therefore telescopes:

(√2 − √1) + (√3 − √2) + … + (√100 − √99) = √100 − √1.

Every intermediate root cancels against its neighbour, leaving 10 − 1 = 9.

The last term is the one with √100, so the sum has 99 terms — not 100. Miscounting the final index is the only way to lose this problem.

René Descartes1596–1650 · Mathematician & Philosopher“Divide each difficulty into as many parts as is feasible and necessary to resolve it.”

Divide the difficulty. A real power equals 1 in exactly three ways, and I take them in turn.

Case 1 — base = 1. x² − 5x + 4 = 0, so (x−1)(x−4) = 0: x = 1, 4.

Case 2 — base = −1 with an even integer exponent. x² − 5x + 6 = 0 gives x = 2, 3. At x = 2 the exponent is 4 − 18 + 20 = 6, even ✓. At x = 3 it is 9 − 27 + 20 = 2, even ✓.

Case 3 — exponent = 0 with a nonzero base. x² − 9x + 20 = 0 gives x = 4, 5. The base at x = 4 is 1 ≠ 0, at x = 5 it is 5 ≠ 0 ✓.

Collecting without repetition: {1, 2, 3, 4, 5}. Their sum is 15.

x = 4 belongs to two cases at once; adding the cases instead of taking their union gives 19. And a base of 0 with exponent 0 would be undefined — worth the check, though it does not arise here.

Pythagoras of Samosc. 570–495 BC · Mathematician & Philosopher“Number rules the universe.”

Areas are the shortest road. The bisector cuts triangle ABC into ABD and ACD, and each of the three areas can be written with the sine formula.

Let t = AD. The bisector makes two angles of 30° at A.

[ABD] = ½ · 8 · t · sin 30°
[ACD] = ½ · 5 · t · sin 30°
[ABC] = ½ · 8 · 5 · sin 60°

The two pieces make the whole:

½t sin 30°(8 + 5) = ½ · 40 · sin 60°.

With sin 30° = 1/2 and sin 60° = √3/2:

t · (1/2) · 13 = 40 · (√3/2), so 13t = 40√3 and t = 40√3/13 ≈ 5.33.

The general formula follows the same way: AD = 2bc cos(A/2) / (b + c). Do not expect AD to be shorter than the shorter side: here it is 5.33, longer than AC = 5. Only an altitude is bounded by the sides it falls between; a bisector runs to the far side and may overshoot.

The distractor 40/13 drops the cosine; 80√3/13 forgets to halve.