Joseph-Louis Lagrange1736–1813 · Mathematician“When we have grasped the whole, the parts explain themselves.”

Do not compute 3¹⁰⁰. Ask instead how long it takes the powers of 3 to return to 1 modulo 100 — that period governs everything.

3² = 9, 3⁴ = 81, 3⁵ = 243 ≡ 43, 3¹⁰ ≡ 43² = 1849 ≡ 49, and 3²⁰ ≡ 49² = 2401 ≡ 1 (mod 100).

So the order of 3 modulo 100 is 20. Since 100 = 20 × 5,

3¹⁰⁰ = (3²⁰)⁵ ≡ 1⁵ = 1 (mod 100).

The last two digits of 3¹⁰⁰ are 01. Note the economy: once a power returns to 1, every multiple of that exponent returns to 1 as well. Squaring your way up (3⁵ → 3¹⁰ → 3²⁰) costs three multiplications, not ninety-nine.

Euclid of Alexandriac. 300 BC · Mathematician“There is no royal road to geometry.”

Place the right angle at the origin and let the legs lie along the axes; then the bisector of that right angle is the line making equal angles with both — the line y = x. Coordinates turn the whole question into one intersection.

Let B = (0, 0), A = (0, 9), C = (12, 0). The line AC has intercept form x/12 + y/9 = 1, that is 3x + 4y = 36.

D lies on that line and on y = x. Substituting: 3x + 4x = 36, so 7x = 36 and D = (36/7, 36/7).

Therefore BD = √[(36/7)² + (36/7)²] = (36/7)√2 = 36√2/7.

You may check the position of D against my bisector proposition: it must divide AC in the ratio of the adjacent sides, AD : DC = AB : BC = 9 : 12 = 3 : 4. The point (36/7, 36/7) does exactly that.

The distractor 36/7 is the answer to a different question — it is the coordinate of D, not the distance to it. The factor √2 is the diagonal you must not drop.

Nicolaus Copernicus1473–1543 · Astronomer & Mathematician“To know that we know what we know, and that we do not know what we do not know, is true knowledge.”

Average speed is total distance divided by total time. It is not the average of the speeds — that quantity answers a different question, and answers it about time rather than distance.

Let the whole journey be 2d kilometres, so each half is d.

Time for the first half: d/30 hours. Time for the second: d/60 hours.
Total time = d/30 + d/60 = 2d/60 + d/60 = 3d/60 = d/20.

Average speed = 2d ÷ (d/20) = 40 km/h.

The distance d cancels, as it must — the answer cannot depend on how long the road is.

The trap is 45, the arithmetic mean of 30 and 60. The car spends twice as long crawling at 30 as it does at 60, so the slow speed must weigh more. What we have computed is the harmonic mean, 2(30)(60)/(30+60) — and it is always less than the arithmetic mean.

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

Two remainder conditions at once — this is exactly the situation my congruences were built for. Write them as n ≡ 2 (mod 5) and n ≡ 3 (mod 7).

Start from the second condition, since it is the more restrictive of the two: the candidates are 3, 10, 17, 24, 31, … Now test each against n ≡ 2 (mod 5). The first to succeed is 17, because 17 = 3·5 + 2 and 17 = 2·7 + 3.

Once a single solution is found, every other solution differs from it by a multiple of the least common multiple of the two moduli. Since 5 and 7 are coprime, that period is 35. So the full solution set is n ≡ 17 (mod 35): 17, 52, 87, 122, …

Of these, only 17, 52 and 87 do not exceed 100. The count is 3.

Note how the answer depends on the period 35, not on the individual moduli. Had the moduli shared a factor, the system might have had no solution at all.

Ada Lovelace1815–1852 · Mathematician“That brain of mine is something more than merely mortal, as time will show.”

Parity is the only thing that matters, so replace each number by its parity and count the machine’s states. The set holds 5 even numbers and 5 odd ones.

A sum of three numbers is even precisely when the count of odd summands is even — that is, zero odds or two odds.

Zero odds: choose 3 of the 5 evens, C(5,3) = 10 ways.
Two odds: choose 2 of the 5 odds and 1 of the 5 evens, C(5,2) · 5 = 10 · 5 = 50 ways.

Favourable = 60. Total = C(10,3) = 120. The probability is 60/120 = 1/2.

The symmetry is not an accident, but neither is it obvious — with 4 evens and 6 odds the answer would not be 1/2. Do not assume balance; count it.

Gerolamo Cardano1501–1576 · Mathematician & Physician“The cube and its things are made equal to a number.”

Never chase a and b themselves. The two symmetric quantities a+b and ab generate every symmetric expression, so find ab first.

(a+b)² = a² + 2ab + b², so 100 = 68 + 2ab and ab = 16.

Now use the cubic identity

a³ + b³ = (a+b)³ − 3ab(a+b) = 1000 − 3(16)(10) = 1000 − 480 = 520.

You may confirm it: a and b are the roots of x² − 10x + 16, namely 2 and 8, and 8 + 512 = 520.

Answering 1000 means forgetting the correction term −3ab(a+b). A cube of a sum is never the sum of cubes.

Muḥammad ibn Mūsā al-Khwārizmīc. 780–850 · Mathematician & Astronomer“By restoring and by balancing, the unknown is brought to light.”

Let the two roots be p and q. Balancing the equation against its factored form (x − p)(x − q) tells us pq = 36 and p + q = −k.

So I need every pair of distinct integers whose product is 36. Both share a sign, since their product is positive.

Positive pairs: (1, 36), (2, 18), (3, 12), (4, 9). The pair (6, 6) is excluded — the roots must be distinct. That is 4 pairs, giving k = −37, −20, −15, −13.

Negative pairs: (−1, −36), (−2, −18), (−3, −12), (−4, −9), again excluding (−6, −6). That is 4 more, giving k = 37, 20, 15, 13.

All eight values of k are different, so the answer is 8.

Two traps: forgetting the negative factorisations halves the count to 4, and forgetting to exclude (6, 6) — whose discriminant is zero — adds a spurious ninth value.

Pierre-Simon Laplace1749–1827 · Mathematician & Astronomer“The theory of probabilities is at bottom nothing but common sense reduced to calculus.”

Count the bad integers and subtract — but count them so that nothing is counted twice. This is the principle of inclusion and exclusion.

Divisible by 2: 500. By 3: 333. By 5: 200. Adding these triple-counts the overlaps, so subtract them back: by 6: 166. By 10: 100. By 15: 66. That now removes the numbers divisible by all three once too often, so restore them: by 30: 33.

Bad = 500 + 333 + 200 − 166 − 100 − 66 + 33 = 734.

Good = 1000 − 734 = 266, and the probability is 266/1000 = 133/500.

Now the trap, and it is a beautiful one. The density of such integers is (1/2)(2/3)(4/5) = 4/15 = 0.2666…, and 4/15 sits among the choices. But 1000 is not a multiple of 30, so the density is only an approximation: 4/15 of 1000 is 266.67, and the true count is 266. The exact probability is 133/500 = 0.266, not 4/15. When the range does not tile the modulus, count — do not scale.

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

Every path is a word. Reaching B requires exactly 4 steps right and 3 steps up, in some order — no more, no fewer, since the grid is 4 wide and 3 tall.

So a path is nothing but an arrangement of the letters R R R R U U U, and two paths are different exactly when their words differ.

Choose which 3 of the 7 steps are the up-steps; the rest are forced:

C(7, 3) = (7 · 6 · 5)/(3 · 2 · 1) = 35.

Equivalently C(7, 4) = 35 — choosing the right-steps instead. The two counts must agree, and they do.

Counting ordered steps as distinct gives 7! / (4! 3!) — the same 35. But 7 · 6 · 5 = 210 without dividing by 3! treats the three identical up-steps as if they were labelled, and overcounts by a factor of six.