A Square and a Cube

100 lockers, 100 people. Person $k$ toggles every $k$-th locker (opens it if closed, closes it if open). Which lockers stay open at the end?

• Times toggled = number of factors: locker $n$ is toggled by Person $k$ exactly when $k$ is a factor of $n$. Locker 6 is toggled by persons 1, 2, 3, 6 (four times, so it ends closed).
• Odd number of toggles ⇒ open; even ⇒ closed.
• Partner factors: factors come in pairs whose product is the number (for 6: $1 \times 6$ and $2 \times 3$), so the count is usually even.
• The count is odd only when one pair has both numbers equal (such as $6 \times 6$ for 36). Such numbers are squares.

• Why "square"? The area of a square of side $n$ is $n \times n = n^2$. Side 5 gives area 25; side 10 gives area 100.
• Not only whole numbers: sides can be fractions or decimals. $\left(\tfrac{3}{5}\right)^2 = \tfrac{9}{25}$ and $(2.5)^2 = 6.25$.
• The square of a negative number is positive: $(-8)^2 = 64$.

• Units digit of a square is always $0, 1, 4, 5, 6$ or $9$.
• Test for NON-squares: if a number ends in $2, 3, 7$ or $8$, it is definitely not a square.
• Warning: ending in $0, 1, 4, 5, 6, 9$ does not guarantee a square. $16$ and $36$ are squares but $26$ is not.
• Trailing zeros: a number ending in $k$ zeros has a square ending in $2k$ zeros (e.g. $700^2 = 490000$). So a square can only have an even number of zeros at the end.
• Parity: square of an even number is even; square of an odd number is odd.
• Quick check: $34^2, 46^2, 56^2, 74^2$ all end in 6 (units digit of the number is 4 or 6).

• Differences of consecutive squares: $4-1=3$, $9-4=5$, $16-9=7$, … and in general $(n+1)^2 - n^2 = 2n+1$.
• $n$th odd number $= 2n - 1$. So the 36th odd number is $71$.
• Using it: given $35^2 = 1225$, then $36^2 = 1225 + 71 = 1296$. Similarly $126^2 = 125^2 + 251$.
• Test for a perfect square: subtract $1, 3, 5, \dots$ in turn from the number. If you reach exactly $0$, it is a perfect square, and the number of subtractions is its root. $25 \to 24 \to 21 \to 16 \to 9 \to 0$ (5 steps), so $25 = 5^2$. For $38$ you go past $0$ (to $-11$), so it is not a square.
• Numbers between consecutive squares: between $n^2$ and $(n+1)^2$ there are $2n$ numbers (if $p, q$ are consecutive squares, $q - p - 1$). E.g. between $16^2$ and $17^2$ there are $32$; between $99^2$ and $100^2$ there are $198$.

Triangular numbers: $1, 3, 6, 10, 15, \dots$ (dots arranged in a triangle).

• Two roots: every perfect square has two integer square roots, one positive and one negative. $8^2 = 64$ and $(-8)^2 = 64$, so the roots of $64$ are $+8$ and $-8$.
• Symbol: in this chapter $\sqrt{\ }$ denotes the positive square root. $\sqrt{64} = 8$, $\sqrt{9} = 3$, $\sqrt{n^2} = n$ for positive $n$.

Units-digit rules can rule a number out (327 is not a square) but cannot confirm one (576?). Four methods:

• Make it a square (multiply): multiply by the unpaired primes. $9408 = 2^6 \times 3 \times 7^2$, the $3$ is unpaired, so multiply by $3$ to get $2^6 \times 3^2 \times 7^2$, with root $8 \times 3 \times 7 = 168$.
• Smallest square divisible by given numbers: take the LCM, then pair up leftover primes. For $4, 9, 10$: LCM $= 180 = 2^2 \times 3^2 \times 5$, multiply by $5$ to get $900$.

• Bracket it between consecutive perfect squares, then see which one is closer.
• $\sqrt{250}$: $15^2 = 225 < 250 < 256 = 16^2$, so $15 < \sqrt{250} < 16$. Since $256$ is much closer to $250$, $\sqrt{250}$ is a little less than 16 (approximately $16$).
• Real use: a square cloth of area $125$ cm² lies between $11^2 = 121$ and $12^2 = 144$, so the largest square handkerchief with an integer side is $11$ cm.

• Why "cube": a cube of edge $n$ is made of $n \times n \times n$ unit cubes. A $4$-unit cube has $4$ layers of $4 \times 4 = 16$, so $64$ unit cubes.
• Units digit: a cube can end in any digit. $\;1\to1,\ 2\to8,\ 3\to7,\ 4\to4,\ 5\to5,\ 6\to6,\ 7\to3,\ 8\to2,\ 9\to9,\ 0\to0$. Cubes ending in 8 exist ($2^3 = 8$).
• Trailing zeros: a number ending in $k$ zeros has a cube ending in $3k$ zeros, so a cube can never end with exactly two zeros.
• Parity: the cube of an odd number is odd; of an even number is even.
• Digits: there are $2$ one-digit cubes ($1, 8$), $2$ two-digit cubes ($27, 64$) and $5$ three-digit cubes ($125, 216, 343, 512, 729$).
• Other numbers: fractions, decimals and negatives can be cubed. $\left(\tfrac{4}{6}\right)^3 = \tfrac{64}{216}$, $(-6)^3 = -216$ (a negative number has a negative cube).
• Factors: cubes do not always have an odd number of factors ($8$ has $1, 2, 4, 8$). That property belongs to squares.

• $n^3$ is the sum of $n$ consecutive odd numbers, taken in order from the start of the odd numbers (1 term, then 2, then 3, …).
• The group of $10$ odd numbers $91 + 93 + \dots + 109$ is the $10$th group, so its sum is $10^3 = 1000$ (no addition needed).

• Make it a cube: multiply by what completes each incomplete triplet. $1323 = 3^3 \times 7^2$, so multiply by $7$ to get $3^3 \times 7^3$.
• Guessing a cube root (for a number you know is a cube):
  1. The units digit of the cube fixes the units digit of the root (use the table in Section 9).
  2. Bracket the cube between cubes of tens ($10^3 = 1000$, $20^3 = 8000$, $30^3 = 27000$, …) to get the tens digit.
  Example: $1331$ ends in $1$ and lies between $1000$ and $8000$, so the root is $11$. Similarly $\sqrt[3]{4913} = 17$, $\sqrt[3]{12167} = 23$, $\sqrt[3]{32768} = 32$.

• Squares need two levels of differences to become constant; cubes need three.

• Babylonians (about 1700 BCE) made the first known lists of squares and cubes on clay tablets, to find roots quickly in land measurement and building.
• Sanskrit terms: varga = square (figure, area and power); ghana = cube (solid and power); varga-varga = fourth power. In use at least from the 3rd century BCE. Aryabhata (499 CE) defines varga as a four-sided equal figure and its area.
• Why "root"? Sanskrit mula (root of a plant, origin) gave varga-mula (square root) and ghana-mula (cube root). Arabic jidhr and Latin radix followed. Brahmagupta (628 CE) also used pada for root.