In practice, an ordinary sheet can be folded only about 6 to 8 times. The catch is geometric: every fold doubles the thickness while halving the width, so the stack quickly becomes too thick and too small to fold again. Thinner paper (tissue, newsprint) can manage a few more folds.

The often-quoted “7 times” limit was beaten in 2002 by Britney Gallivan, who folded a very long, thin sheet 12 times. The rest of this chapter explores what would happen if we could keep folding forever.

AnswerUsually 6–8 folds for normal paper; the record is 12. The math below assumes ideal, unlimited folding.

The thickness doubles each fold, so after \(n\) folds it is

A few milestones from the completed table:

AnswerAfter 30 folds ≈ 10.7 km; after 45 ≈ 3.5 lakh km; after 46 the stack reaches beyond the Moon.

Each fold multiplies by 2, so over several folds the multipliers themselves multiply:

That is why the growth table shows the thickness multiplying by 1024 over every block of 10 folds.

Answer4 times · 8 times · 1024 times.

Ten folds multiply the start by 2 ten times, i.e. by \(2^{10}\). Starting from \(v\), the thickness is \(v \times 2^{10} = 2^{10}v\).

Answer(v) \(2^{10}v\).

Dividing repeatedly by primes: \(32400 = 2\times2\times2\times2 \times 3\times3\times3\times3 \times 5\times5\).

Answer\(2^4 \times 3^4 \times 5^2\).

A negative base raised to an odd power stays negative; to an even power it turns positive (the minus signs pair off and cancel).

And \((-2)^4 = (-2)(-2)(-2)(-2) = 16\) — yes.

Answer\((-1)^5 = -1\), \((-1)^{56} = +1\), and \((-2)^4 = 16\). ✓

Zero multiplied by itself any number of times is still zero.

AnswerAll are 0 (for \(n \ge 1\)). Note: \(0^0\) is left undefined.

Multiplying powers of the same base simply piles up the factors, so the exponents add:

The same logic gives \(p^4 \times p^6 = p^{4+6} = p^{10}\). In general, \(\;n^a \times n^b = n^{a+b}\).

AnswerBecause exponents add: \(3^2 \times 3^5 = 3^7\), and \(p^4 \times p^6 = p^{10}\).

Splitting to compute:

Power of a power uses \((n^a)^b = n^{a\times b}\), so an exponent can be split into factors:

And yes, \(2^{10} = (2^5)\times(2^5) = (2^5)^2\).

Answer\(2^9=512,\ 5^7=78125,\ 4^6=4096\); the power-of-power forms above; and \(2^{10}=(2^5)^2\). ✓

AnswerGroup equal factors and count them — see the table.

Answer648 = 2³×3⁴ · 405 = 3⁴×5 · 540 = 2²×3³×5 · 3600 = 2⁴×3²×5².

In (vi), \((-2)^5 = -32\) (odd power, negative) and \((-10)^6 = +10^6\) (even power, positive), so the product is negative: \(-32000000\).

Answer2000 · 392 · 768 · 225 · 90000 · −32000000.

At every stage the count is multiplied by 3, so the totals are powers of 3.

Diamonds need three more factors of 3 (tables, necklaces, diamonds): \(3^4 \times 3^3 = 3^7\). We can reuse work: \(3^7 = 3^4 \times 3^3 = 81 \times 27 = 2187\).

Answer81 rooms (= 3⁴) and 2187 diamonds (= 3⁷).

If the pond doubles each day and is full on day 30, then one day earlier it was exactly half as covered.

AnswerHalf full on day 29; fully covered = \(2^{30}\), half covered = \(2^{29}\).

After 4 days doubling: \(1\times2^4 = 2^4\). Then 4 days tripling multiplies by \(3^4\):

Order does not matter — multiplication can be regrouped: \(2^4\times3^4 = (3\times2)^4\). This is the law \(\;m^a \times n^a = (mn)^a\). Using it:

Answer\(6^4 = 1296\) either way · \(2^5\times5^5 = 10^5\) · \(10^4/5^4 = 16\).

For each independent choice we multiply the number of options.

AnswerEstu: 12 outfits · Roxie: 42 ways.

Each slot is an independent choice, so we multiply the options slot by slot. Building up: a 2-digit lock has \(10\times10 = 100\), a 3-digit lock \(100\times10 = 1000\), and so on.

For 6 slots each chosen from 26 letters:

Answer5-digit lock: \(10^5 = 1{,}00{,}000\) passwords · 6-letter lock: \(26^6\) passwords.

Dividing powers of the same base subtracts exponents (common factors cancel): \(n^a \div n^b = n^{a-b}\).

Answer\(2^{75}\).

Division by \(n\) is hidden inside the rule, and we can never divide by zero — so \(n=0\) is not allowed. To find \(n^0\), set \(a=b\):

(For \(n=0\) this would read \(0/0\), which is undefined — another reason to exclude it.)

AnswerBecause the rule divides by \(n\); and any non-zero number to the power 0 equals 1.

Yes — the laws keep working for all integers. A negative exponent simply means the reciprocal: \(n^{-a} = \dfrac{1}{n^{a}}\).

AnswerYes; each is the reciprocal of the positive power (see above).

1. \(2^{-4}\times2^{7} = 2^{-4+7} = 2^{3}\).
2. \(3^{2}\times3^{-5}\times3^{6} = 3^{2-5+6} = 3^{3}\).
3. \(p^{3}\times p^{-10} = p^{3-10} = p^{-7}\).
4. \(2^{4}\times(-4)^{-2} = 2^{4}\times\dfrac{1}{16} = 2^{4}\times 2^{-4} = 2^{0} = 1\).
5. \(8^{p}\times8^{q} = 8^{p+q}\).

Answer(i) 2³ · (ii) 3³ · (iii) p⁻⁷ · (iv) 1 · (v) 8^(p+q).

“How many times larger” means divide: \(4^{2} \div 4^{-2} = 4^{2-(-2)} = 4^{4} = 256\) times.

Reading values off the line and adding/subtracting exponents:

Answer\(4^2\) is \(4^4 = 256\) times larger than \(4^{-2}\); the powers of 7 are listed above.

1. Rewrite the base: \(4^{32} = (2^2)^{32} = 2^{64}\).
2. Divide: \(2^{224} \div 2^{64} = 2^{224-64} = 2^{160}\).
3. Units digits of \(2^1,2^2,2^3,2^4,\dots\) repeat in a cycle of 4: 2, 4, 8, 6. Since 160 is a multiple of 4, the units digit matches \(2^4\), which is 6.

AnswerThe units digit is 6.

One container (5 bottles) is added each day — this is linear, not exponential, growth.

Answer200 bottles (= 2×10²).

First find the prime form, then split the exponent however you like.

AnswerThree sample splittings for each are shown above (many others work too).

• Sometimes(i) Cube numbers are also square numbers. Only when the number is a 6th power, since it must be both a square and a cube. Example: \(64 = 2^6 = (2^3)^2 = (2^2)^3\).
• Always(ii) Fourth powers are also square numbers. \(n^4 = (n^2)^2\) — always a perfect square.
• Always(iii) The fifth power of a number is divisible by its cube. \(n^5 \div n^3 = n^2\), a whole number — always divisible.
• Always(iv) The product of two cube numbers is a cube number. \(a^3 \times b^3 = (ab)^3\) — always a cube.
• Never(v) \(q^{46}\) is both a 4th power and a 6th power (\(q\) prime). A 4th power needs the exponent divisible by 4, a 6th power by 6. But 46 is divisible by neither — never true.

Answer(i) Sometimes · (ii) Always · (iii) Always · (iv) Always · (v) Never.

1. \(10^{-2}\times10^{-5} = 10^{-7} = \dfrac{1}{10^{7}}\).
2. \(5^{7}\div5^{4} = 5^{3}\).
3. \(9^{-7}\div9^{4} = 9^{-7-4} = 9^{-11} = \dfrac{1}{9^{11}}\).
4. \((13^{-2})^{-3} = 13^{(-2)\times(-3)} = 13^{6}\).
5. \(m^{5}n^{12}(mn)^{9} = m^{5}n^{12}\cdot m^{9}n^{9} = m^{14}n^{21}\).

Answer(i) 10⁻⁷ · (ii) 5³ · (iii) 9⁻¹¹ · (iv) 13⁶ · (v) m¹⁴n²¹.

Each number is \(12\) shifted by powers of 10, and squaring shifts the decimal twice as far.

Answer1.44 · 0.0144 · 0.000144 · 14400.

Reduce each to powers of the primes 2 and 3.

The first, second and fourth all equal \(2^4\times3^6 = 11664\).

Answer2⁴×3⁶, 6⁴×3² and 18²×6² are the same (= 11664).

Answer(i) 3⁴ · (ii) 2⁸ · (iii) 2¹⁰⁰.

An \(n\)-digit code (digits 0–9) gives \(10^n\) possibilities. We need at least 8.5 billion = \(8.5\times10^9\):

So \(10^9\) is too few and \(10^{10}\) is enough — the code needs 10 digits.

Answer10 digits.

A number that is both a square and a cube must have every prime exponent divisible by both 2 and 3, hence by 6. So such numbers are exactly the 6th powers, \(n^6\).

There are infinitely many, and each \(n^6 = (n^3)^2 = (n^2)^3\).

AnswerYes — infinitely many; they are the 6th powers \(n^6\) (1, 64, 729, 4096, …).

Each slot can be any of 26 letters or 10 digits = 36 characters, and there are 5 independent slots.

Answer\(36^5\) codes (about 6 crore).

Adding equal powers is not the same as multiplying them — the exponents do not add. We simply add the counts:

So the correct expression is (vi) \(10^9 + 10^9\), which equals \(2\times10^9\) (option (v) is the same value).

Answer(vi) \(10^9 + 10^9 = 2\times10^9\).

(For (iv): assume ~70 years and ~1 hour = 3600 s of eating per day.)

Answer2.46×10¹¹ pieces · 5×10¹² bees · 3.116×10²³ bacteria · ≈ 9.198×10⁷ s.

Convert seconds to years: \(1\times10^9\) s \(\div\) (number of seconds in a year).

So count back about 31 years 8 months from today’s date to get the answer (the exact date depends on when you solve it).

AnswerAbout 31.7 years (≈ 11,574 days) before today.