QuanticaPrealgebraOpen in the courseMultiplication is how you count without counting one at a time. Nobody finds the number of seats in a stadium by tapping each one. You spot the structure, multiply, and jump straight to the total. Multiplication has the same two freedoms as addition, any order and any grouping, and it also has a rule of its own that ties it back to addition. That rule will do more work for you than anything else in this course. As before, we earn the rules by using them.
Problem
A chocolatier packs a gift box with $5$ rows of $8$ truffles. The box sits sideways in the shop window, so a passer-by sees $8$ rows of $5$. The chocolatier computes \(5 \times 8\), and the passer-by computes \(8 \times 5\). How many truffles do they each count?
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5 rows of 8 is \(8 + 8 + 8 + 8 + 8 = 40\), and 8 rows of 5 is also 40.$$5 \times 8 = 8 \times 5 = \boxed{40}.$$It is one box, and tilting it cannot change the count. Every product is a grid, and a grid counts the same whichever way you sweep it.
Problem
Multiply in the easy order: $$5 \times 9 \times 2.$$
Show a hint
- Pair the 5 and the 2 to make 10 first!
Show the full solution
Reorder to $(5\times2)\times9=10\times9=90$. Find a pair that makes 10 before multiplying anything.
Problem
Partner up: $$4 \times 7 \times 25.$$
Show a hint
- 4 and 25 are partners, and together they make 100.
Show the full solution
Group the partners: $(4\times25)\times7=100\times7=700$.
Problem
Find the friendly pair first: $$2 \times 13 \times 5.$$
Show a hint
- 2 and 5 make 10.
Show the full solution
Reorder: $(2\times5)\times13=10\times13=130$.
Problem
An arena has $2$ seating decks. Each deck is divided into $26$ sections, and each section holds $50$ seats. Multiply the three numbers in whatever order is most painless: how many seats does the arena have?
Show a hint
- Nothing forces you to multiply left to right. Which two of the three factors make a round number together?
Show the full solution
The 2 and the 50 make 100, so pair those first.$$(2 \times 50) \times 26 = 100 \times 26 = \boxed{2600}.$$Order and grouping are free, so look for a pair that builds a power of ten before multiplying anything.
Problem
Go partner-hunting in \(8 \times 73 \times 125\), entirely in your head.
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8 and 125 are partners.$$8 \times 125 = 1000,$$so \(73 \times 1000 = \boxed{73000}\). Keep the three pairs in mind, \(2 \times 5 = 10\), \(4 \times 25 = 100\), and \(8 \times 125 = 1000\), and scan for one before you compute.
Problem
All in your head: $$16 \times 125 \times 5.$$
Show a hint
- $16=2\times8$, and $8$ partners with $125$.
Show the full solution
$16\times125=2\times(8\times125)=2\times1000=2000$, then $2000\times5=10000$.
Problem
Compute $$8 \times 625 \times 125.$$
Show a hint
- 8 and 125 are partners that make 1000.
Show the full solution
$8\times125=1000$, so $1000\times625=625000$. Partner-hunting collapses the whole product.
Problem
In your head: $$2 \times 17 \times 5 \times 4 \times 25.$$
Show a hint
- Two partner pairs sit in the list.
- $2\times5=10$ and $4\times25=100$.
Show the full solution
Pair them off: $(2\times5)\times(4\times25)\times17=10\times100\times17=1000\times17=17000$.
So far multiplication has only borrowed addition’s tricks. The new rule shows up when the two operations meet in one problem, either when a sum has to be multiplied or when two products share a factor.
Problem
Split the multiply: $$6 \times 12.$$
Show a hint
- Think of 12 as $10+2$.
Show the full solution
$6\times(10+2)=60+12=72$. Breaking a number into easy parts turns one hard multiply into two simple ones.
Problem
Use the bridge: $$5 \times 21.$$
Show a hint
- Write 21 as $20+1$.
Show the full solution
$5\times(20+1)=100+5=105$.
Problem
A trail crew is building $9$ identical fence panels. Each panel needs $20$ long boards and $7$ short ones. Count the boards two different ways. First find what one panel uses and multiply by 9. Then count all the long boards and all the short boards separately and add. Both counts should match, so what total do they give?
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One panel uses \(20 + 7 = 27\) boards, so nine panels use \(9 \times 27 = \boxed{243}\). Counting by type gives the same total, \(9 \times 20 = 180\) long boards and \(9 \times 7 = 63\) short ones, which add to 243.$$9 \times (20 + 7) = (9 \times 20) + (9 \times 7).$$Multiplying a sum is the same as multiplying each part and adding. That is the distributive property, the bridge between \(\times\) and \(+\).
Problem
Split the awkward factor: $$8 \times 43.$$
Show a hint
- $43=40+3$.
Show the full solution
$8\times(40+3)=320+24=344$.
Problem
The bridge spans subtractions too. Compute $$4 \times 998$$ with nothing written down.
Show a hint
- 998 is unfriendly, but 1000 is a dream, and \(998 = 1000 - 2\).
Show the full solution
998 is 2 short of 1000, so multiply by 1000 and take the extra back.$$4 \times 998 = 4 \times (1000 - 2) = 4000 - 8 = \boxed{3992}.$$Distributing works over subtraction too, which handles any factor sitting just below a round number.
Problem
No pencil: $$7 \times 1003.$$
Show a hint
- $1003=1000+3$.
Show the full solution
$7\times(1000+3)=7000+21=7021$. Split across the round number, then add the small piece.
Problem
Now run the bridge in reverse: compute $$86 \times 14 - 76 \times 14.$$
Show a hint
- Both products contain a 14. Say the expression in words: “eighty-six 14s minus seventy-six 14s leaves …”
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Read it as eighty-six 14s minus seventy-six 14s, which leaves ten 14s.$$86 \times 14 - 76 \times 14 = (86 - 76) \times 14 = 10 \times 14 = \boxed{140}.$$This is the distributive property read right to left. When two products share a factor, subtract the counts and multiply once.
Problem
Collapse it: $$53 \times 28 + 47 \times 28.$$
Show a hint
- Both products share a 28, so how many 28s in all?
Show the full solution
Read it as fifty-three 28s plus forty-seven 28s, which is one hundred 28s: $(53+47)\times28=100\times28=2800$.
One case is still open. The last lesson handled negative numbers under addition and subtraction. What happens when a negative number turns up in a multiplication?
Problem
Compute $$(-5) \times 3.$$
Show a hint
- One negative factor, so the answer is negative.
Show the full solution
$5\times3=15$, and one minus sign flips it across zero: $-15$.
Problem
A research submarine descends steadily, changing its height by $-8$ metres every minute. It starts at the surface (height 0) and descends for $5$ minutes. Use repeated addition to find its final height, that is, find $$5 \times (-8).$$
Show the full solution
Five copies of \(-8\) stack up.$$(-8) + (-8) + (-8) + (-8) + (-8) = \boxed{-40}.$$Repeated addition covers a positive count of a negative quantity. A negative count of copies has no meaning yet, and that needs a different tool.
So one strange case remains. What could (−1) × (−8) possibly mean? “Add −8 to itself −1 times” is gibberish. When the old meaning runs out, mathematicians let patterns decide. Study the ladder below.
Problem
The ladder forces \((-1) \times (-8) = 8\), since any other answer would break the pattern. Extend the ladder one more rung. What must $$(-2) \times (-8)$$ be?
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Each step down the ladder drops the left factor by 1 and raises the product by 8. From \((-1) \times (-8) = 8\), one more step gives$$(-2) \times (-8) = 8 + 8 = \boxed{16}.$$Repeated addition has no meaning here, so the pattern decides, and keeping it unbroken forces negative times negative to land positive.
Problem
Settle the score from Lesson 1.1: compute $$(-4) \times (-6),$$ and know why it is not \(-24\).
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Start from \(4 \times 6 = 24\). One minus sign takes the opposite,$$(-4) \times 6 = -24,$$and the second takes it back.$$(-4) \times (-6) = \boxed{24}.$$Each negative factor is one flip across zero, so a product is negative only when its negative factors come in an odd count.
Problem
Compute $$(-2) \times (-6).$$
Show a hint
- Two negatives cancel, so the answer is positive.
Show the full solution
The two minus signs flip back to positive: $2\times6=12$.
Problem
Count the signs: $$(-1) \times (-2) \times (-5).$$
Show a hint
- Three negatives, so is that odd or even?
Show the full solution
Three negative factors is an odd count, so the product is negative: $1\times2\times5=10$ gives $-10$.
Problem
Count first, then compute: $$(-5) \times (-7) \times (-4).$$
Show a hint
- Three negative factors, so odd or even?
Show the full solution
An odd count of negatives makes the product negative: $5\times7\times4=140$, so $-140$.
Problem
Five signs, one decision: $$(-2) \times (-2) \times (-2) \times (-2) \times (-2).$$
Show a hint
- Count the minus signs first.
Show the full solution
Five negatives is an odd count, so the product is negative. $2\times2\times2\times2\times2=32$, giving $-32$.
One last stretch before you go. These closing problems fold the whole lesson together, partner-hunting, the distributive bridge, and the sign rules, all at once.
Problem
Compute $$(-4) \times (-25) \times (-1) \times (-1).$$
Show a hint
- Count the minus signs. Even or odd?
Show the full solution
Four negative factors is an even count, so the product is positive: $4\times25\times1\times1=100$.
Problem
Lesson 1.1 promised you would one day do this in your head: $$(2025 \times 2030) - (2020 \times 2025).$$ Today is that day.
Show a hint
- Both giant products share the factor 2025.
- \(2025 \times 2030\) is 2030 copies of 2025, and \(2020 \times 2025\) is 2020 copies of it. How many copies survive the subtraction?
Show the full solution
A 2025 sits inside each product, so read it as 2030 copies of 2025 minus 2020 copies, and the subtraction leaves ten copies.$$2025 \times (2030 - 2020) = 2025 \times 10 = \boxed{20250}.$$When look-alike products are subtracted, factor out what they share instead of multiplying.
Practice these ideas
Practice
Compute $$2 \times 87 \times 5.$$
Show the solution
Partner the 2 with the 5: \(2 \times 5 = 10\), so the product is \(87 \times 10 = \boxed{870}\).
Practice
Compute $$4 \times 6 \times 250 \times 5 \times 2.$$
Show the solution
Partner-hunt twice: \(4 \times 250 = 1000\) and \(5 \times 2 = 10\). What remains is \(1000 \times 10 \times 6 = \boxed{60000}\).
Practice
Compute $$5 \times 7 \times 5 \times 2 \times 2.$$
Show the solution
Pair each 5 with a 2: \((5 \times 2) \times (5 \times 2) \times 7 = 10 \times 10 \times 7 = \boxed{700}\).
Practice
Compute $$4 \times 25 \times 4 \times 25 \times 4.$$
Show the solution
Two pairs of partners, \((4 \times 25) \times (4 \times 25) = 100 \times 100 = 10000\), with one spare 4 left over, so \(4 \times 10000 = \boxed{40000}\).
Practice
Compute $$1 \times 1 \times 487 \times 1 \times 1.$$
Show the solution
Multiplying by 1 changes nothing, so strip the 1s away and the product is just \(\boxed{487}\). For \(\times\), 1 does what 0 does for \(+\).
Practice
A licence plate reads $37094$. Multiply its five digits together, the fast way.
Show the solution
The third digit is a 0, and a single 0 makes the whole product 0, so \(3 \times 7 \times 0 \times 9 \times 4 = \boxed{0}\). No other digit matters.
Practice
Compute $$(61 \times 0 \times 88) + 425.$$
Show the solution
Resist working out \(61 \times 88\), since the 0 makes that whole product 0. The sum is \(0 + 425 = \boxed{425}\). Spotting the zero saves all the work.
Practice
Compute $$125 + 125 + 125 + 125 + 125 + 125 + 125 + 125.$$
Show the solution
Eight copies of 125 is \(8 \times 125\), an old partner pair, so the sum is \(\boxed{1000}\). When a sum repeats a single number, switch to \(\times\).
Practice
Compute $$7 \times 1004.$$
Show the solution
Split across the sum: \(7 \times (1000 + 4) = 7000 + 28 = \boxed{7028}\).
Practice
Compute $$6 \times 997.$$
Show the solution
\(997 = 1000 - 3\), so \(6 \times 997 = 6000 - 18 = \boxed{5982}\).
Practice
Compute $$13 \times 9 + 9 \times 87$$, minding where each 9 sits.
Show the solution
One 9 sits to the right of its \(\times\), the other to the left, but order never matters, so both products share the 9: \(9 \times (13 + 87) = 9 \times 100 = \boxed{900}\).
Practice
Compute $$23 \times 9 + 23.$$
Show the solution
The spare 23 at the end is really \(23 \times 1\). So the sum is \(23 \times (9 + 1) = 23 \times 10 = \boxed{230}\).
Practice
Compute $$64 \times 35 - 54 \times 35.$$
Show the solution
Sixty-four 35s minus fifty-four 35s leaves ten of them: \((64 - 54) \times 35 = 10 \times 35 = \boxed{350}\).
Practice
Multiplication distributes over addition, but does it distribute over itself? Test the claim \(3 \times (10 \times 4) = (3 \times 10) \times (3 \times 4)\) by working out both sides, then enter the right side minus the left side.
Show the solution
The left side is \(3 \times (10 \times 4) = 3 \times 40 = 120\). The right side is \((3 \times 10) \times (3 \times 4) = 30 \times 12 = 360\). They differ by \(\boxed{240}\), so the claim is false, and the distributive bridge runs between \(\times\) and \(+\) only. Tripling a product once is not the same as tripling both of its factors.
Practice
Compute $$(-6) \times 7.$$
Show the solution
One negative factor means one flip: \(6 \times 7 = 42\), flipped to \(\boxed{-42}\).
Practice
Compute $$(-9) \times (-12).$$
Show the solution
Two negative factors mean two flips, back to positive: \(9 \times 12 = \boxed{108}\).
Practice
Compute $$(-1) \times (-3) \times (-10)$$, and count the negative factors before you multiply.
Show the solution
Three negative factors mean three flips, \(1 \times 3 \times 10 = 30\) lands on \(\boxed{-30}\). An odd count of negative factors leaves a product negative, and an even count leaves it positive.
Practice
Compute $$15 \times 12 + 15 \times 38 + 25 \times 12 + 25 \times 38.$$
Show the solution
Collapse in stages: \(15 \times (12 + 38) = 15 \times 50\), and \(25 \times (12 + 38) = 25 \times 50\). Then \(15 \times 50 + 25 \times 50 = (15 + 25) \times 50 = 40 \times 50 = \boxed{2000}\). Four multiplications collapsed into one.
Practice
For thousands of years, everyone multiplied big numbers the same way you do, digit by digit, carrying as you go. In 1956 the great Russian mathematician Andrey Kolmogorov conjectured that nothing fundamentally faster was possible, and in 1960 he said so in his seminar at Moscow State University. In the audience sat a 23-year-old student named Anatoly Karatsuba. Within about a week, he found a genuinely faster method, and humanity had been overlooking a multiplication shortcut for four thousand years. Computers still use his idea today when they multiply enormous numbers, and an even faster method was discovered as recently as 2019. Even the “simplest” operations still hold secrets, and maybe the next shortcut will have your name on it. 🚀