QuanticaPrealgebraOpen in the courseSquaring a number means multiplying it by itself, written \(b^2\) and read as “b squared.” A product like \(3\times5\) counts a grid with 3 rows and 5 columns, so \(5^2\) counts a grid with 5 rows and 5 columns, an actual square. That shape is worth holding onto, since most of the shortcuts in this lesson come from picturing it.
Problem
Here are the first six perfect squares. $$1^2=1,\quad 2^2=4,\quad 3^2=9,\quad 4^2=16,\quad 5^2=25,\quad 6^2=36$$ Look at how much each square grows over the one before it, then use that growth pattern to predict \(8^2\) without multiplying \(8\times8\) directly.
Show a hint
- The growth from one square to the next goes 3, 5, 7, 9, 11, all odd numbers increasing by 2. What should the next two terms in that list be?
Show the full solution
The growth from each square to the next is \(3, 5, 7, 9, 11, \ldots\), the odd numbers, each one 2 more than the last. The next two terms are 13 and 15. So \(7^2 = 36+13=49\), and $$8^2 = 49+15=\boxed{64}.$$ Every step from one perfect square to the next adds the next odd number.
A square is what you get from two copies of a number multiplied together. Use three copies and you get a cube. Where \(b^2\) counted the cells of a flat square, \(b^3\) counts the little cubes packed inside a solid one.
Problem
Stack unit cubes into a solid block. One cube along each edge is just \(1\) cube, two along each edge make \(8\), three along each edge make \(27\). How many unit cubes are in a block that is \(4\) along each edge?
Show a hint
- Each flat layer is a \(4\times4=16\) square of cubes, and the block is \(4\) layers tall.
Show the full solution
Each layer is a \(4\times4=16\) square of unit cubes, and there are \(4\) such layers stacked up. $$4\times4\times4=\boxed{64}$$ That count is \(4^3\), and the list \(1,8,27,64\) is the start of the perfect cubes.
Problem
Compute \(2\times3^2\). Is it \(2\times(3^2)\) or \((2\times3)^2\)? Those give different values, 18 and 36, so the placement of the exponent matters.
Show a hint
- An exponent attaches only to the symbol written directly beneath it, not to anything sitting in front of it.
Show the full solution
The exponent sits on the \(3\), so evaluate \(3^2\) first. $$2\times3^2 = 2\times9 = \boxed{18}$$ An exponent applies only to the symbol written directly beneath it, never to anything in front of it. Squaring the whole \(2\times3\) needs parentheses, and \((2\times3)^2=36\) is a different expression.
Problem
Two students evaluate \(-5^2\). One says the answer is \(-25\), the other says \(25\). Using the rule that an exponent applies only to its own base, who is right, and what does \((-5)^2\) equal instead?
Show a hint
- In \(-5^2\), is the minus sign part of the base, or is it applied to the result afterward? Compare \(-5^2\) to \(-1\times5^2\).
Show the full solution
The exponent in \(-5^2\) applies to the \(5\) alone, so the minus sign is applied last and \(-5^2=-(5^2)=-25\). The first student is right. Parentheses change the base itself. $$(-5)^2=(-5)\times(-5)=\boxed{25}$$ Same digits, opposite signs, because one expression puts the negative inside the base and the other does not.
Problem
Compute \((-2)^3\). Then compute \(-2^3\). With squares these two came out opposite, since \((-2)^2=4\) but \(-2^2=-4\). What happens this time?
Show a hint
- \((-2)^3=(-2)(-2)(-2)\). Multiply the first two, then bring in the third.
- An exponent still binds only to its own base, so \(-2^3\) means \(-(2^3)\).
Show the full solution
Multiply the three factors of \((-2)^3\) one at a time. \((-2)(-2)=4\), then \(4\times(-2)=-8\). And \(-2^3=-(2^3)=-8\) as well, since the exponent binds only to the \(2\). Both come out to $$\boxed{-8}.$$ This time the two expressions are equal, not opposite. Three negative factors pair off with one left over, so the cube of a negative stays negative.
Problem
Evaluate \(2x^2-x\) when \(x=-3\).
Show a hint
- Substitute \(-3\) for \(x\) and put parentheses around it everywhere it appears, so the expression becomes \(2(-3)^2-(-3)\).
Show the full solution
Substitute \(x=-3\), keeping it inside parentheses. $$2(-3)^2-(-3)$$ The exponent applies to the whole parenthesized \(-3\), so \((-3)^2=9\). $$2(9)-(-3)=18+3=\boxed{21}$$ Dropping the parentheses and writing \(2\times-3^2\) would turn the first piece into \(-2\times9=-18\), a different and wrong expression.
Squares and cubes also split across multiplication and division. That split turns awkward products into easy ones, and the next few problems build the rule.
Problem
Compute \(6^2\) directly. Then compute it a second way: write \(6=2\times3\), expand \((2\times3)\times(2\times3)\) into four factors, and use the reordering freedom from Chapter 1 to group the two 2's together and the two 3's together. Do the two methods agree, and what does the regrouped form become?
Show a hint
- After regrouping, the four factors split into \(2\times2\) and \(3\times3\), which are \(2^2\) and \(3^2\).
Show the full solution
Directly, \(6^2=36\). Expanding the second way, $$(2\times3)\times(2\times3)=2\times3\times2\times3$$ Reordering and regrouping these four factors gives $$(2\times2)\times(3\times3)=2^2\times3^2=4\times9=\boxed{36}$$ Both routes agree, and along the way \((2\times3)^2\) turned into \(2^2\times3^2\). Squaring a product is the same as squaring each factor and multiplying the results.
Problem
Compute \(2^3\times5^3\) without cubing either number on its own first.
Show a hint
- A cube distributes over a product, so \(2^3\times5^3=(2\times5)^3\).
Show the full solution
Cubing distributes over multiplication, so gather the bases before cubing. $$2^3\times5^3=(2\times5)^3=10^3=\boxed{1000}$$ The same pairing that turned an awkward product into a round number for squares works again for cubes.
Problem
Compute \(25^2\times4^2\) without ever multiplying two multi-digit numbers together.
Show a hint
- The pairing trick from Chapter 1 is back. What is \(25\times4\)?
Show the full solution
Since \(a^2b^2=(ab)^2\), $$25^2\times4^2=(25\times4)^2=100^2=\boxed{10000}$$ The same pairing that turned \(25\times4\) into a clean \(100\) now turns two awkward squares into one easy one.
Back at the start of this lesson, the jump from one perfect square to the next was always an odd number, \(2n+1\) more than the square before it. That pattern is not just a curiosity. It is a mental-math shortcut.
Problem
Use \((n+1)^2=n^2+2n+1\) to compute \(41^2\) without multiplying two 2-digit numbers.
Show a hint
- Let \(n=40\), a very friendly number. What are \(n^2\), \(2n\), and the final \(+1\)?
Show the full solution
Take \(n=40\). Then \(n^2=1600\) and \(2n=80\). $$41^2=1600+80+1=\boxed{1681}$$ Every piece is a number you can add in your head.
Problem
The identity also runs in reverse. If \(n=m+1\), then \(n^2=m^2+2m+1\), which rearranges to \((n-1)^2=n^2-2n+1\) (writing \(m=n-1\)). Use this reversed identity to compute \(99^2\) from the friendly number \(100\).
Show a hint
- Let \(n=100\), so \(99^2=(n-1)^2=n^2-2n+1\).
Show the full solution
Take \(n=100\). Then \(n^2=10000\) and \(2n=200\). $$99^2=10000-200+1=\boxed{9801}$$ One identity, run forward for a number just above a round number and run backward for a number just below one.
Problem
Compute \(12^3\div3^3\) without ever finding \(12^3\).
Show a hint
- A cube distributes over division, so \(12^3\div3^3=(12\div3)^3\).
Show the full solution
Since \(\left(\frac{a}{b}\right)^3=\frac{a^3}{b^3}\), read the division with the cube on the outside. $$12^3\div3^3=(12\div3)^3=4^3=\boxed{64}$$ One small cube, \(4^3\), in place of a four-digit division.
Practice these ideas
Practice
A square garden is \(40\) tiles along each side. The owner widens it to \(41\) tiles along each side. Find how many tiles get added, without ever computing \(40^2\) or \(41^2\).
Show the solution
Going from a \(40\)-by-\(40\) square out to \(41\)-by-\(41\) wraps an L-shaped gnomon around the old garden. The strip is two sides of length \(40\) plus the single corner where they meet, so it holds $$2\times40+1=\boxed{81}$$ tiles. That is the jump \(41^2-40^2\), and it came out odd, the way every step between consecutive squares does.
Practice
Two perfect squares sit side by side on the staircase of squares, and the larger is exactly \(159\) more than the smaller. The two numbers being squared are whole numbers one apart. Find the larger of those two numbers.
Show the solution
The step from one perfect square to the next is the gnomon \(2n+1\), where \(n\) is the smaller number being squared. Here that step is \(159\), so \(2n+1=159\). Peeling off the \(1\) leaves \(2n=158\), and half of \(158\) is \(n=79\). The two squares are \(79^2\) and \(80^2\), so the larger number being squared is \(\boxed{80}\). As a check, \(80^2-79^2=2\times79+1=159\).
Practice
You handled both of those without squaring anything. The step between one perfect square and the next is always the odd number \(2n+1\), the gnomon strip wrapped around the smaller square. Read it forward to get the jump. Read it backward to get the side.
Practice
Add the odd numbers from \(11\) up to \(39\). $$11+13+15+\cdots+37+39$$ The first \(n\) odd numbers add up to \(n^2\). Use that twice instead of adding all fifteen terms.
Show the solution
The odd numbers from \(1\) to \(39\) are the first \(20\) odds, since \(39\) is the twentieth odd number, so they sum to \(20^2=400\). The run you want skips the small odds \(1,3,5,7,9\), which are the first \(5\) odds and sum to \(5^2=25\). Trim that front block off. $$11+13+\cdots+39=20^2-5^2=400-25=\boxed{375}$$ A run of consecutive odds is one square with a smaller square sliced off its start.
Practice
A run of odd numbers starting at \(1\) folds into a single square, because the first \(n\) odds build an \(n\)-by-\(n\) square one gnomon at a time. A run that starts higher up is just that same idea, one square with a smaller square trimmed off the front.
Practice
A cubical crate is packed snug with small cubical boxes, no gaps. The boxes sit \(5\) across the floor, \(5\) deep, and the stack is \(5\) boxes tall. How many small boxes fill one full crate?
Show the solution
Each layer resting on the floor is a \(5\times5=25\) square of boxes, and \(5\) such layers stack from bottom to top. $$5\times5\times5=5^3=\boxed{125}$$ A cubical crate packed with cubical boxes always holds a perfect cube of them.
Practice
A square photo has side \(4\) cm, so its area is \(4^2=16\) square cm. A cube-shaped box also has side \(4\) cm, so its volume is \(4^3=64\) cubic cm. Now double every side, sending each to \(8\) cm. The box's volume grows by more cubic cm than the photo's area grows by square cm. How many more?
Show the solution
Because a square distributes over a product, \((2b)^2=2^2b^2=4b^2\), so doubling the side multiplies the area by \(4\). The photo goes from \(16\) to \(8^2=64\), a growth of \(48\) square cm. A cube distributes the same way, \((2b)^3=2^3b^3=8b^3\), so doubling the side multiplies the volume by \(8\). The box goes from \(64\) to \(8^3=512\), a growth of \(448\) cubic cm. The volume outgrew the area by $$448-48=\boxed{400}.$$
Practice
A square counts a flat area and a cube counts a solid volume, and that one extra dimension is the whole story. Stretch the side and a cube grows more than a square. Double the side and the area climbs by \(2^2\), while the volume climbs by \(2^3\).
Practice
Two expressions go head to head. $$A=-4^2+(-3)^3 \qquad B=(-4)^2+(-3)^2$$ A classmate says they must be equal since they use the same numbers. Evaluate each one, watching every minus sign, and enter the value of the larger.
Show the solution
Take \(A\) one piece at a time. In \(-4^2\) the exponent attaches only to the \(4\), so \(-4^2=-(4^2)=-16\). And \((-3)^3\) cubes a negative, and an odd power keeps the sign, so \((-3)^3=-27\). Thus \(A=-16+(-27)=-43\). Now \(B\). Here \((-4)^2\) squares the whole \(-4\), and an even power of a negative is positive, so \((-4)^2=16\), and likewise \((-3)^2=9\). Thus \(B=16+9=25\). The two are nowhere near equal. The parentheses flipped two of the four signs, and the larger value is \(\boxed{25}\).
Practice
A detective is handed four numbers and told that exactly one is a perfect square. $$31427 \qquad 64008 \qquad 57121 \qquad 90003$$ She names it in seconds by glancing only at the last digit of each, never computing a square. Which number is the perfect square?
Show the solution
The last digit of \(b^2\) depends only on the last digit of \(b\). Running through \(0^2,1^2,\ldots,9^2\), the endings that ever appear are \(0,1,4,9,6,5,6,9,4,1\), so a perfect square can only finish in \(0,1,4,5,6,\) or \(9\), never in \(2,3,7,\) or \(8\). Now scan. \(31427\) ends in \(7\), \(64008\) ends in \(8\), and \(90003\) ends in \(3\), all impossible. Only \(57121\) ends in a legal digit, so it is the square. Indeed \(57121=239^2\). The answer is \(\boxed{57121}\).
Practice
A secret two-digit whole number \(N\) is cubed, and only the cube is shown. $$N^3=50653$$ Two clues pin \(N\) with no guessing. The size of the cube traps \(N\) in one narrow band, and the last digit of \(50653\) reveals the last digit of \(N\) outright. Find \(N\).
Show the solution
First trap \(N\) by size. Since \(30^3=27000\) and \(40^3=64000\), and \(50653\) sits between them, \(N\) is somewhere in the thirties. That still leaves ten numbers, so bring in the last digit. When you cube \(0\) through \(9\), the last digits come out \(0,1,8,7,4,5,6,3,2,9\), every digit exactly once. A cube ending in \(3\) can only come from a base ending in \(7\). The one number in the thirties ending in \(7\) is \(37\), and \(37^3=50653\) checks out. So \(N=\boxed{37}\).
Practice
The last digit of a power leans only on the last digit of the base. For squares that digit blurs several bases together, since \(2\) and \(8\) both square to an ending of \(4\), so it can rule a number out but not pin it down. For cubes every digit appears exactly once, so a cube's last digit fingerprints its base with no doubt at all.
Practice
A machine takes a whole number, squares it, adds twice the number, then adds \(1\). You feed it \(79\), and it prints a single perfect square. Find that square, but do it without multiplying two two-digit numbers. Name the perfect square first, then split it across a friendly product.
Show the solution
The machine builds \(79^2+2\times79+1\), which is exactly \(n^2+2n+1\) with \(n=79\). That pattern folds up into \((n+1)^2\), so the output is \((79+1)^2=80^2\). Now dodge squaring \(80\) head on. A square distributes over a product, so split the base. $$80^2=(8\times10)^2=8^2\times10^2=64\times100=\boxed{6400}$$ The expansion collapsed into one square, and that square broke into two easy pieces.
Practice
The signs alternate all the way down this chain of squares. $$20^2-19^2+18^2-17^2+\cdots+2^2-1^2$$ Do not square twenty numbers. Group the chain into pairs from the left, turn each pair into something simpler, then add.
Show the solution
Pair the chain from the left, \((20^2-19^2)+(18^2-17^2)+\cdots+(2^2-1^2)\). Each pair is a step between two consecutive squares, and that step is the odd number \(2n-1\). So \(20^2-19^2=39\), then \(18^2-17^2=35\), on down to \(2^2-1^2=3\). The ten pairs become the ten odd numbers \(39,35,31,27,23,19,15,11,7,3\). Add them by matching the ends inward, \(39+3=42\), \(35+7=42\), and so on, which gives five pairs of \(42\). $$5\times42=\boxed{210}$$
Practice
The holding-hands fact says \(1^3+2^3+\cdots+n^3=(1+2+\cdots+n)^2\). Use it on a slice to find \(9^3+10^3\) without cubing \(9\) or \(10\) directly. Write the two-cube sum as one stack of cubes minus a shorter stack, then finish.
Show the solution
Take the whole stack of cubes up to \(10\) and remove the stack up to \(8\). What survives is exactly \(9^3+10^3\). By the holding-hands fact each stack is the square of a plain running sum, and those sums are \(1+\cdots+10=55\) and \(1+\cdots+8=36\). So $$9^3+10^3=55^2-36^2=3025-1296=\boxed{1729}.$$ A pair of cubes became a difference of two squares. \(1729\) is famous as the smallest number that is a sum of two cubes in two different ways.
Practice
Stack up the odd numbers starting from 1, and look at the totals. \(1\), then \(1+3=4\), then \(1+3+5=9\), then \(1+3+5+7=16\). Every one of those totals is a perfect square, and it never stops. The first \(n\) odd numbers always add up to \(n^2\). The ancient Greek Pythagoreans drew exactly this picture, building each new square by wrapping an L-shaped strip of dots, called a gnomon, around the previous square, the same L-shape you just saw in this lesson's figure. It is the same pattern you found at the start of this lesson, just run all the way back to the beginning. 🚀
Practice
Add up the first few cubes and look at the totals. \(1^3=1\), then \(1^3+2^3=9\), then \(1^3+2^3+3^3=36\), then \(1^3+2^3+3^3+4^3=100\). Every total is a perfect square, and not a random one. \(9=3^2\), \(36=6^2\), \(100=10^2\), and \(3,6,10\) are exactly the running sums \(1+2\), \(1+2+3\), \(1+2+3+4\). So a stack of cubes always equals the square of a plain sum, \(1^3+2^3+\cdots+n^3=(1+2+\cdots+n)^2\). The two halves of this lesson, squares and cubes, turn out to be holding hands. 🚀