--- title: "Powers" book: "Primary & Middle School Mathematics" subject: math language: en chapter: 56 exercises: 11 source: https://one-course.com/books/math/1/en/chapter/56-powers --- # Chapter 56 — Powers Fold a sheet of paper in [half](https://one-course.com/books/math/1/en/chapter/17-sharing-and-division#def-g3-division-half) $10$ times (if you can!): its thickness doubles each time, and $10$ doublings multiply it by $2^{10} = 1024$. Powers are the shorthand of repeated multiplication; this chapter sets up the notation and its rules, with a special role for the powers of $10$. ## 56.1 Definition **Definition 56.1 (Power).** For a number $a$ and a whole number $n \geq 1$: $$ a^n = \underbrace{a \times a \times \dots \times a}_{n \text{ factors}}, $$ read “$a$ to the (power) $n$”; $a$ is the *base*, $n$ the *exponent*. Special names: $a^2$ is “$a$ squared”, $a^3$ “$a$ cubed”. By convention: $$ a^1 = a, \qquad a^0 = 1 \ (a \neq 0), \qquad a^{-n} = \frac{1}{a^n}. $$ **Example 56.2.** $3^4 = 3 \times 3 \times 3 \times 3 = 81$; $10^3 = 1000$; $5^{-2} = \frac{1}{25} = 0.04$; $(-2)^3 = -8$ and $(-2)^4 = +16$ ([Example 54.5](https://one-course.com/books/math/1/en/chapter/54-multiplying-negative-numbers#ex-g8-negprod-powers)). Watch out: $a^n$ is *not* $a \times n$: $2^5 = 32$, not $10$. ![Powers of 2: each bar is twice the previous one. Growth by repeated multiplication runs away much faster than growth by repeated addition.](https://one-course.com/images/onecourse/chapters/math-1/g8-powers/fig-73c0637d8084.svg) *Powers of $2$: each bar is twice the previous one. Growth by repeated multiplication runs away much faster than growth by repeated addition.* ## 56.2 The rules of exponents **Theorem 56.3 (Rules of exponents).** For a nonzero base $a$ and whole-number [exponents](#def-g8-powers-def) $m$, $n$: $$ a^m \times a^n = a^{m+n}, \qquad \frac{a^m}{a^n} = a^{m-n}, \qquad \left(a^m\right)^n = a^{m \times n}, \qquad (ab)^n = a^n b^n . $$ **Proof by counting factors.** $a^m \times a^n$ lines up $m$ factors $a$ followed by $n$ more: $m + n$ in total. $\left(a^m\right)^n$ repeats a block of $m$ factors $n$ times: $mn$ factors. $(ab)^n$ contains $n$ letters $a$ and $n$ letters $b$, which can be regrouped. The [quotient](https://one-course.com/books/math/1/en/chapter/17-sharing-and-division#def-g3-division-remainder) rule comes from canceling $n$ of the $m$ factors; with the conventions $a^0 = 1$ and $a^{-n} = \frac{1}{a^n}$, it remains true even when $n \geq m$. ∎ **Example 56.4.** $$ 2^3 \times 2^5 = 2^8 = 256, \qquad \frac{7^6}{7^4} = 7^2 = 49, \qquad \left(5^2\right)^3 = 5^6, \qquad 2^4 \times 5^4 = 10^4 . $$ A trap: the rules apply to a *common base* (or a common [exponent](#def-g8-powers-def), for the last one). No rule simplifies $2^3 \times 5^2$ — just compute: $8 \times 25 = 200$. ## 56.3 Powers of ten **Proposition 56.5 (Powers of ten).** For $n \geq 1$: $10^n = 1\underbrace{0\dots0}_{n}$, and $10^{-n} = 0.\underbrace{0\dots0}_{n-1}1$. The rules of [exponents](#def-g8-powers-def) read: multiplying powers of ten adds the [exponents](#def-g8-powers-def). **Example 56.6.** $10^4 \times 10^3 = 10^7$; $\dfrac{10^2}{10^5} = 10^{-3} = 0.001$. Large and small quantities become readable: $$ \text{one billion} = 10^9, \qquad \text{one millionth} = 10^{-6}. $$ Combined with decimals: $3.2 \times 10^5 = 320\,000$ and $4.7 \times 10^{-3} = 0.0047$ (shift the [decimal point](https://one-course.com/books/math/1/en/chapter/25-tenths-and-hundredths#def-g4-decimals-point), [Proposition 38.7](https://one-course.com/books/math/1/en/chapter/38-decimal-numbers#prop-g6-decimals-shift)). The systematic use of this writing — *scientific notation* — is developed in [Chapter 63](https://one-course.com/books/math/1/en/chapter/63-fractions-and-powers#ch-g9-fractions). **Example 56.7 (Orders of magnitude).** Light travels about $3 \times 10^8$ m/s; a year has about $3.2 \times 10^7$ seconds. A light-year is therefore about $$ 3 \times 3.2 \times 10^{8+7} \approx 10 \times 10^{15} = 10^{16} \text{ m} $$ — ten million billion meters. Powers of ten make astronomical computations fit on one line. **Method 56.8 (Simplifying an expression with powers).** 1. Group the factors base by base; 2. apply the [exponent](#def-g8-powers-def) rules within each base; 3. compute the remaining small powers, or leave the answer as a power if it is large. **Example 56.9.** $$ \frac{3^5 \times 3^{-2} \times 4^2}{3^2} = \frac{3^{5 + (-2)}}{3^2} \times 16 = 3^{3 - 2} \times 16 = 3 \times 16 = 48 . $$ ## 56.4 Exercises **Exercise 56.1 ★.** Compute: $$ 2^6, \qquad 3^3, \qquad 10^5, \qquad 1^{100}, \qquad 0.1^2, \qquad 6^0 . $$ **Solution of Exercise 56.1.** $2^6 = 64$; $3^3 = 27$; $10^5 = 100\,000$; $1^{100} = 1$; $0.1^2 = 0.01$; $6^0 = 1$. **Exercise 56.2 ★.** Compute: $$ (-3)^2, \qquad -3^2, \qquad (-1)^{15}, \qquad (-5)^3, \qquad \left(\tfrac{2}{3}\right)^2 . $$ **Solution of Exercise 56.2.** $(-3)^2 = 9$; $-3^2 = -9$; $(-1)^{15} = -1$; $(-5)^3 = -125$; $\left(\frac23\right)^2 = \frac49$. **Exercise 56.3 ★.** Write as a single power: $$ 7^4 \times 7^5, \qquad \frac{2^9}{2^3}, \qquad \left(10^3\right)^4, \qquad 5^6 \times 5^{-2}, \qquad 3^4 \times 7^4 . $$ **Solution of Exercise 56.3.** $7^4 \times 7^5 = 7^9$; $\dfrac{2^9}{2^3} = 2^6$; $\left(10^3\right)^4 = 10^{12}$; $5^6 \times 5^{-2} = 5^4$; $3^4 \times 7^4 = 21^4$ (same [exponent](#def-g8-powers-def): multiply the bases). **Exercise 56.4 ★.** Write as a [decimal number](https://one-course.com/books/math/1/en/chapter/38-decimal-numbers#def-g6-decimals-places): $10^{-2}$; $4 \times 10^3$; $2.5 \times 10^{-4}$; $10^0$. **Solution of Exercise 56.4.** $10^{-2} = 0.01$; $4 \times 10^3 = 4000$; $2.5 \times 10^{-4} = 0.00025$; $10^0 = 1$. **Exercise 56.5 ★.** Write with a power of ten: one hundred thousand; one tenth; ten billion; $0.000\,001$. **Solution of Exercise 56.5.** $10^5$; $10^{-1}$; $10^{10}$; $10^{-6}$. **Exercise 56.6 ★.** Simplify, then compute: $$ \frac{10^7 \times 10^{-3}}{10^2}, \qquad \frac{2^5 \times 2^4}{2^6}, \qquad \left(2^2\right)^3 \times 2^{-4} . $$ **Solution of Exercise 56.6.** $\dfrac{10^7 \times 10^{-3}}{10^2} = \dfrac{10^4}{10^2} = 10^2 = 100$. $\dfrac{2^5 \times 2^4}{2^6} = \dfrac{2^9}{2^6} = 2^3 = 8$. $\left(2^2\right)^3 \times 2^{-4} = 2^6 \times 2^{-4} = 2^2 = 4$. **Exercise 56.7 ★.** True or false? Correct the false ones. $$ 2^3 \times 2^4 = 2^{12}; \qquad 5^2 + 5^3 = 5^5; \qquad (3^2)^4 = 3^8; \qquad 10^3 \times 10^3 = 100^3 . $$ **Solution of Exercise 56.7.** $2^3 \times 2^4 = 2^{12}$: *false* — [exponents](#def-g8-powers-def) add: $2^7$. $5^2 + 5^3 = 5^5$: *false* — no rule for [sums](https://one-course.com/books/math/1/en/chapter/2-addition-first-steps#def-g1-addition-def): $25 + 125 = 150$, while $5^5 = 3125$. $(3^2)^4 = 3^8$: *true*. $10^3 \times 10^3 = 100^3$: *true* — both equal $10^6$ (left: $10^{3+3}$; right: $(10^2)^3$). **Exercise 56.8 ★★.** A rumor spreads: on day 1, three people know it; each day, every person who knows tells three new people. Write with a power the number of *new* people informed on day $4$, and compute how many people know the rumor at the end of day 4 (including the original three). **Solution of Exercise 56.8.** New people on day $4$: each of the $3^3 = 27$ people informed on day 3 tells three others: $3^4 = 81$. Knowing at the end of day 4: $3 + 9 + 27 + 81 = 120$ people. **Exercise 56.9 ★★.** A sheet of paper is $0.1$ mm thick, i.e. $10^{-4}$ m. Folding it doubles its thickness each time. 1. Express the thickness after $10$ folds as a [product](https://one-course.com/books/math/1/en/chapter/10-multiplication-first-steps#def-g2-mult-def) , and compute it in centimeters ( $2^{10} = 1024$ ). 2. After $42$ folds the thickness would be $2^{42} \times 10^{-4}$ m, with $2^{42} \approx 4.4 \times 10^{12}$ . Show that this exceeds the Earth–Moon distance, about $3.8 \times 10^8$ m. **Solution of Exercise 56.9.** *1.* Thickness: $2^{10} \times 10^{-4}$ m $= 1024 \times 10^{-4}$ m $\approx 0.1$ m $= 10$ cm. *2.* $2^{42} \times 10^{-4} \approx 4.4 \times 10^{12} \times 10^{-4} = 4.4 \times 10^8$ m, larger than $3.8 \times 10^8$ m: after $42$ (theoretical!) folds, the wad of paper would pass the Moon. **Exercise 56.10 ★★.** Order from smallest to largest, without a calculator: $$ 2^{10}, \qquad 10^3, \qquad 3^6, \qquad 5^4 . $$ (Compute each one; $2^{10}$ and $10^3$ are famous neighbors.) **Solution of Exercise 56.10.** $2^{10} = 1024$; $10^3 = 1000$; $3^6 = 729$; $5^4 = 625$. Order: $$ 5^4 < 3^6 < 10^3 < 2^{10} . $$ **Exercise 56.11 ★★★.** Which is larger, $2^{100}$ or $10^{30}$? Use $2^{10} = 1024 > 10^3$ to compare $2^{100} = \left(2^{10}\right)^{10}$ with $\left(10^3\right)^{10}$. **Solution of Exercise 56.11.** $2^{100} = \left(2^{10}\right)^{10} = 1024^{10}$ and $10^{30} = \left(10^3\right)^{10} = 1000^{10}$. Since $1024 > 1000$, multiplying ten copies of each keeps the inequality: $2^{100} > 10^{30}$. ## 56.5 Problem: The chessboard and the powers of two **Problem 56.1.** Weekend problem — the geometric [sum](https://one-course.com/books/math/1/en/chapter/2-addition-first-steps#def-g1-addition-def) $1 + 2 + 4 + \dots + 2^{n-1} = 2^n - 1$, from a famous legend to binary numbers The legend: as a reward for inventing chess, the sage Sissa asked his king for one grain of wheat on the first square of the board, two on the second, four on the third — doubling from square to square, up to the sixty-fourth. The king laughed at such modesty. This problem computes what the king promised, using the [exponent](#def-g8-powers-def) rules of [Theorem 56.3](#thm-g8-powers-rules), and ends where the story secretly leads: the binary numbers inside every computer. **Part I — The doubling trick.** For $n \geq 1$, let $S_n$ be the total number of grains on the first $n$ squares. 1. Express the number of grains on square $k$ as a power of $2$ . Which power sits on square $64$ ? 2. Compute $S_1$ , $S_2$ , $S_3$ , $S_4$ and $S_5$ , and compare each with a nearby power of $2$ . Conjecture a formula for $S_n$ . 3. The *doubling trick*: write the [sums](https://one-course.com/books/math/1/en/chapter/2-addition-first-steps#def-g1-addition-def) $S_n$ and $2 \times S_n$ one under the other, subtract, and prove your conjecture: $$S_n = 1 + 2 + 4 + \dots + 2^{n-1} = 2^n - 1 .$$ 4. How many grains did the king promise in total? Express the answer with a power of $2$ , and complete the classic remark: “the whole board holds one grain fewer than a single sixty-fifth square would.” 5. Show that the second [half](https://one-course.com/books/math/1/en/chapter/17-sharing-and-division#def-g3-division-half) of the board (squares $33$ to $64$ ) holds *exactly* $2^{32}$ times as many grains as the first [half](https://one-course.com/books/math/1/en/chapter/17-sharing-and-division#def-g3-division-half) . **Part II — How big is $2^{64}$?** The comparison $2^{10} = 1024 > 10^3$ of [Exercise 56.11](#exo-g8-powers-11) is the key to all the estimates below. 6. Show that $2^{64} = 2^4 \times \left(2^{10}\right)^6 > 1.6 \times 10^{19}$ . 7. A grain of wheat weighs about $0.05$ g, i.e. $5 \times 10^{-2}$ g. Show that the promised wheat weighs more than $8 \times 10^{17}$ g, and convert this into tonnes ( $1$ tonne $= 10^6$ g). 8. The whole world currently harvests about $8 \times 10^8$ tonnes of wheat per year. At least how many years of world harvest did the king promise? 9. Find the smallest whole number $n$ such that $2^n > 10^6$ — that is, how many doublings it takes to pass one million. (Compute $2^{19}$ and $2^{20}$ exactly, using $2^{10} = 1024$ .) 10. A trickster offers you a month’s salary: $1$ cent on day $1$ , then double the previous day’s pay each day. On which day does the *daily* pay alone first exceed one million euros ( $10^8$ cents)? (Compute $2^{26}$ and $2^{27}$ exactly.) **Part III — Binary weights.** A merchant owns five weights: $1$, $2$, $4$, $8$ and $16$ grams, one of each. She places some of them on one pan of a balance to weigh goods on the other pan. 11. Which weights does she place to weigh $21$ g? To weigh $27$ g? 12. Explain why any target of $16$ g or more *must* use the $16$ g weight, and why any target of $15$ g or less must *not* use it. (Question 3 tells you what the weights $1, 2, 4, 8$ can reach at most.) Explain why the same reasoning repeats with the next-largest weight, at every stage. 13. Deduce that every whole target from $1$ to $31$ g can be weighed, and *in exactly one way* : each number between $1$ and $31$ is a [sum](https://one-course.com/books/math/1/en/chapter/2-addition-first-steps#def-g1-addition-def) of distinct powers of $2$ in a single manner. 14. The merchant buys a sixth weight, of $32$ g. Up to what target can she now weigh? Write $45$ g as a [sum](https://one-course.com/books/math/1/en/chapter/2-addition-first-steps#def-g1-addition-def) of distinct powers of $2$ . 15. In a computer, a “64-bit” number is stored on $64$ squares, each holding a $0$ or a $1$ — square $k$ contributing $2^{k-1}$ when it holds a $1$ , like the grains of the legend. Using Part I, explain why the whole numbers such a machine can store run exactly from $0$ to $2^{64} - 1$ . **Solution of Problem 56.1.** **1.** The grains double from square to square starting at $1 = 2^0$: square $k$ holds $2^{k-1}$ grains ([Definition 56.1](#def-g8-powers-def)). Square $64$ holds $2^{63}$. **2.** $S_1 = 1$, $S_2 = 1 + 2 = 3$, $S_3 = 3 + 4 = 7$, $S_4 = 7 + 8 = 15$, $S_5 = 15 + 16 = 31$: always one less than the next power of $2$ ($2$, $4$, $8$, $16$, $32$). Conjecture: $S_n = 2^n - 1$. **3.** Doubling every term of $S_n$ shifts each power up by one ($2 \times 2^{k} = 2^{k+1}$): $$ \begin{align*} S_n &= 1 + 2 + 4 + \dots + 2^{n-1}, \\ 2 \times S_n &= \phantom{1 + {}} 2 + 4 + \dots + 2^{n-1} + 2^n . \end{align*} $$ Subtracting the first line from the second, every term from $2$ to $2^{n-1}$ appears in both and cancels: $$ 2 \times S_n - S_n = 2^n - 1, \qquad\text{that is}\qquad S_n = 2^n - 1 . $$ **4.** The total is $S_{64} = 2^{64} - 1$ grains. A sixty-fifth square would hold $2^{64}$ grains: the whole board carries exactly one grain fewer than that single square would. **5.** The first [half](https://one-course.com/books/math/1/en/chapter/17-sharing-and-division#def-g3-division-half) holds $S_{32} = 2^{32} - 1$ grains. The whole board holds $2^{64} - 1$, so the second [half](https://one-course.com/books/math/1/en/chapter/17-sharing-and-division#def-g3-division-half) holds $$ \left(2^{64} - 1\right) - \left(2^{32} - 1\right) = 2^{64} - 2^{32} = 2^{32} \times \left(2^{32} - 1\right) $$ (factor out $2^{32}$, using $2^{32} \times 2^{32} = 2^{64}$, [Theorem 56.3](#thm-g8-powers-rules)): exactly $2^{32}$ times the first [half](https://one-course.com/books/math/1/en/chapter/17-sharing-and-division#def-g3-division-half). **6.** By the [exponent](#def-g8-powers-def) rules, $2^{64} = 2^{4 + 60} = 2^4 \times \left(2^{10}\right)^6$. Since $2^{10} = 1024 > 10^3$, $$ 2^{64} > 16 \times \left(10^3\right)^6 = 16 \times 10^{18} = 1.6 \times 10^{19} . $$ **7.** More than $1.6 \times 10^{19}$ grains at $5 \times 10^{-2}$ g each: $$ 1.6 \times 10^{19} \times 5 \times 10^{-2} = 8 \times 10^{17} \text{ g} . $$ Dividing by $10^6$ g per tonne: more than $8 \times 10^{11}$ tonnes — eight hundred billion tonnes. **8.** $\dfrac{8 \times 10^{11}}{8 \times 10^{8}} = 10^3$: the king promised at least a *thousand years* of today’s entire world harvest. (The legend says his advisers told him as much.) **9.** $2^{19} = 2^9 \times 2^{10} = 512 \times 1024 = 524\,288 < 10^6$, while $2^{20} = \left(2^{10}\right)^2 = 1024^2 = 1\,048\,576 > 10^6$. So the smallest [exponent](#def-g8-powers-def) is $n = 20$: twenty doublings pass the million. **10.** The pay on day $n$ is $2^{n-1}$ cents (day 1: $2^0 = 1$). Now $$ 2^{26} = 2^6 \times \left(2^{10}\right)^2 = 64 \times 1\,048\,576 = 67\,108\,864 < 10^8, $$ $$ 2^{27} = 2 \times 2^{26} = 134\,217\,728 > 10^8 . $$ So the daily pay first exceeds $10^8$ cents when $n - 1 = 27$: on day $28$. **11.** $21 = 16 + 4 + 1$: weights $16$, $4$ and $1$ g. $27 = 16 + 8 + 2 + 1$: weights $16$, $8$, $2$ and $1$ g. **12.** By question 3, the weights $1, 2, 4, 8$ together weigh $S_4 = 2^4 - 1 = 15$ g. So without the $16$ g weight the merchant cannot pass $15$ g: any target of $16$ g or more must use it. And a target of $15$ g or less must not use it, since the $16$ g weight alone already exceeds the target. The choice of the largest weight is therefore *forced*. What remains is a target of at most $15$ g to be formed with $1, 2, 4, 8$ — and the same argument repeats: $8$ is forced (used if the remaining target is $\geq 8$, unused otherwise, because $1 + 2 + 4 = 7$), then $4$ (because $1 + 2 = 3$), then $2$, then $1$. **13.** Following the forced choices, the remaining target after each stage is at most the total of the remaining weights, so the process ends with [remainder](https://one-course.com/books/math/1/en/chapter/17-sharing-and-division#def-g3-division-remainder) $0$: every target from $1$ to $31$ is reached. And since every choice along the way was forced, no other selection of weights can reach the same target: the writing of each number from $1$ to $31$ as a [sum](https://one-course.com/books/math/1/en/chapter/2-addition-first-steps#def-g1-addition-def) of distinct powers of $2$ exists and is *unique*. **14.** The six weights total $S_6 = 2^6 - 1 = 63$ g, and the same forced-choice argument covers every target from $1$ to $63$ g. For $45$: the target is $\geq 32$, so use $32$; remains $13 < 16$, so skip $16$; use $8$ (remains $5$), use $4$ (remains $1$), skip $2$, use $1$: $$ 45 = 32 + 8 + 4 + 1 = 2^5 + 2^3 + 2^2 + 2^0 . $$ **15.** Choosing a $0$ or a $1$ on each of the $64$ squares amounts to choosing which powers $2^0, 2^1, \dots, 2^{63}$ to include in a [sum](https://one-course.com/books/math/1/en/chapter/2-addition-first-steps#def-g1-addition-def) — exactly the merchant’s weighing with $64$ weights. The smallest storable number is $0$ (all squares at $0$); the largest is the [sum](https://one-course.com/books/math/1/en/chapter/2-addition-first-steps#def-g1-addition-def) of *all* the powers, which is the king’s total: $S_{64} = 2^{64} - 1$ (Part I). By the forced-choice argument, every whole number in between is reached exactly once: a $64$-bit machine stores precisely the whole numbers from $0$ to $2^{64} - 1$.