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.

François Viète1540–1603 · Mathematician“There is no problem that cannot be solved.”

The roots need never be found. My relations give their sum and product at a glance:

r + s = 3,   rs = 1.

The sum of cubes is built from exactly these two:

r³ + s³ = (r+s)³ − 3rs(r+s) = 27 − 3(1)(3) = 18.

The roots themselves are (3 ± √5)/2 — irrational, and utterly unnecessary. That is the point: a symmetric function of the roots is a polynomial in the coefficients.

Answering 27 forgets the term 3rs(r+s).

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.

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

My Elements state it as a proposition about the square on a side; today it is written as the law of cosines. For an angle A between sides b and c:

a² = b² + c² − 2bc cos A.

Here b = 8, c = 7, and cos 60° = 1/2:

BC² = 64 + 49 − 2(8)(7)(1/2) = 113 − 56 = 57.

So BC = √57 ≈ 7.55 — sensibly between 7 and 8, as the picture demands.

Dropping the cosine term entirely gives √113, which is the answer for a right angle at A. A 60° angle is sharper, so the opposite side must be shorter than the right-angle case — and √57 < √113. Use the estimate to catch sign errors.

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.

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

The binomial theorem tells us the general term of (a + b)ⁿ is C(n, k) a^n−kb^k.

Here a = 1, b = 2x, n = 6, and we want k = 3:

C(6, 3) · 1³ · (2x)³ = 20 · 8x³ = 160x³.

The coefficient is 160.

The whole difficulty is that the 2 travels with the x and is therefore cubed as well. Answering 20 reports C(6,3) alone; answering 40 cubes nothing but doubles once. The coefficient is C(6,3)·2³, never C(6,3)·2.

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

A sum of logarithms is the logarithm of a product. Undo the logarithm once and a quadratic remains.

log₃[x(x + 6)] = 3, so x(x + 6) = 3³ = 27.

x² + 6x − 27 = 0, that is (x + 9)(x − 3) = 0, giving x = −9 or x = 3.

Now the domain decides. A logarithm demands a positive argument, and x = −9 makes both log₃x and log₃(x+6) undefined. Only x = 3 survives.

Check: log₃3 + log₃9 = 1 + 2 = 3. ✓

Combining logarithms can manufacture solutions the original equation never had. Every candidate must be returned to the equation it came from.

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

A sum of logarithms is the logarithm of a product. Write the whole sum as one logarithm and watch the product telescope.

The sum equals log₂[(2/1) · (3/2) · (4/3) · … · (64/63)].

Inside the bracket, every numerator cancels the next denominator, leaving only the first denominator and the last numerator:

product = 64/1 = 64.

So the sum is log₂64 = 6, since 2⁶ = 64.

Equivalently, each term is log₂(k+1) − log₂k, and the sum telescopes to log₂64 − log₂1 = 6 − 0.

There are 63 terms, not 64 — but the count of terms is not the answer. Answering 63 confuses the two.

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

Find the period, and the exponent 2024 loses its terror.

3¹ ≡ 3,   3² ≡ 9,   3³ = 27 ≡ 1 (mod 13).

So 3 has order 3 modulo 13, and the powers cycle 3, 9, 1 forever. Only 2024 modulo 3 matters.

2024 = 3 · 674 + 2, so 2024 ≡ 2 (mod 3), and

3²⁰²⁴ = (3³)⁶⁷⁴ · 3² ≡ 1 · 9 = 9 (mod 13).

Fermat guarantees only that 3¹² ≡ 1; the true order 3 divides 12, as it must. Working with 12 instead of 3 also gives 9 — but with four times the labour.