Factors and Multiples, and How They Differ

A factor of a number is a whole number that divides it with nothing left over. Picture it as a side of a rectangle with that area. Each factor has a partner, the matching side, and the two multiply to the original number.

Problem

A class arranges 24 floor cushions into one full rectangular block, every row the same length, no cushion left over. How many different row counts work?

Show a hint

A number of rows works only if you can split 24 into that many equal rows with nothing left over. Try 1 row, then 2, then 3, and keep going, checking each time whether 24 divides up cleanly.

Show the full solution

Pair each row count with the row length it forces.

1 \times 24, 2 \times 12, 3 \times 8, 4 \times 6

Both numbers in every pair work as a row count, so the full list is

1, 2, 3, 4, 6, 8, 12, 24

, which is

8

of them. These are the factors, or divisors, of 24, the numbers that divide it with nothing left over. They come two at a time because each side of a rectangle fixes the opposite side.

Problem

Find all the factors of 48 by writing factor pairs

1 \times 48, 2 \times 24, \dots

and stopping once the two sides of a pair would cross. How many factors does 48 have?

Show a hint

The moment you find a small side like 3, the matching side is forced, since 3 rows must each be $48 \div 3$ tiles long. So every small factor you find hands you a large one for free.
Watch the two sides of each pair close in on each other. Once the next side you would test is bigger than the partner it would pair with, every grid past that point is just one you have already seen turned on its side.

Show the full solution

Climb from 1, writing each factor beside the partner it forces.

1 \times 48, 2 \times 24, 3 \times 16, 4 \times 12, 6 \times 8

Reading both numbers out of every pair gives

1, 2, 3, 4, 6, 8, 12, 16, 24, 48

, which is

10

factors. You stop after 6 because the next divisor, 8, is already standing on the right of

6 \times 8

. Once the number you are about to test would be larger than its own partner, every rectangle has already turned up, just mirrored.

If $6$ is a factor of $36$ , then $36$ is a multiple of $6$ . Count by sixes, $6, 12, 18, 24, 30, 36$ , and you land on $36$ after six steps. Factor and multiple are one relationship read from its two ends.

Problem

A train starts at stone 0 and rolls forward exactly 7 stones at a time. How many of the four stones 28, 35, 50, 63 does it land on?

Show a hint

Keep adding $7$ . $7, 14, 21, 28, 35, 42, 49, 56, 63$ . Which of the four stones appear on that list?

Show the full solution

Count by sevens.

7, 14, 21, 28, 35, 42, 49, 56, 63

. Stones 28, 35, and 63 all appear, and 50 does not, since the train steps from 49 straight to 56. That is

3

stones. Each landing reads two ways.

7

is a factor of 28 because

28 = 4 \times 7

, and 28 is a multiple of 7 because it sits four 7-steps out from 0.

Problem

For each pair, decide whether the first number is a factor of the second, a multiple of it, both, or neither. (a) 4 and 12 (b) 15 and 5 (c) 8 and 8 (d) 6 and 9. In how many of the four pairs is the first number a factor of the second?

Show a hint

For each pair ask two separate questions. Does the first number divide the second evenly, and is the first number itself a whole number times the second?
$a$ is a factor of $b$ when $b = a \times (whole number)$ . Check the pair where both numbers are equal carefully.

Show the full solution

(a)

12 = 4 \times 3

, so 4 is a factor of 12. (b) 15 does not divide 5, so 15 is a multiple of 5 rather than a factor. (c)

8 = 8 \times 1

, so 8 is both a factor and a multiple of itself. (d) 6 does not divide 9 evenly, so neither. Only (a) and (c) qualify, giving

2

pairs. Parts (a) and (b) are the same fact told from opposite ends, and (c) shows every number is both a factor and a multiple of itself.

Problem

The factors of 36 pair up as

1 \times 36, 2 \times 18, 3 \times 12, 4 \times 9, 6 \times 6

, nine factors in all. How many factors does 49 have?

Show a hint

Every pair so far used two different side lengths. Look hard at $6 \times 6$ , and note that $49 = 7 \times 7$ .
List every whole number that divides 49 evenly.

Show the full solution

The pairs are

1 \times 49

and

7 \times 7

, and the second adds only one new factor because 7 is its own partner. The factors are

1, 7, 49

, so 49 has

3

. Factors normally arrive two at a time, so a count is even unless one factor pairs with itself, which happens exactly for perfect squares. That is why 36 and 49 come out odd while 35, splitting only as

1 \times 35

and

5 \times 7

, has four.

A factor of a number is a way to fill a full rectangle, and the two side lengths form a factor pair. On the left,

3 \times 4 = 12

uses two different partners. On the right, the square

6 \times 6 = 36

uses one number on both sides, so the two partners have become the same factor. That single self-paired side, the gold corner on the diagonal, is the lone unpaired factor that gives a perfect square its odd total.

Problem

Priya lists every factor of 20, and Owen lists every multiple of 20. Only one of them can ever finish. How many numbers are on that finished list?

Show a hint

A factor of 20 is a side of a rectangle with area 20, so no factor can be longer than 20. Owen's list has no such ceiling.
Hunt the factors in pairs, $1 \times 20$ , $2 \times 10$ , $4 \times 5$ .

Show the full solution

Priya is the one who finishes, since no factor of 20 can be bigger than 20 itself. Hunting in pairs gives

1 \times 20

2 \times 10

4 \times 5

, so her list is

1, 2, 4, 5, 10, 20

, which is

6

numbers. Owen never finishes, because from any multiple he can always take one more step of 20. Note that 20 sits on both lists, as its own largest factor and its own smallest positive multiple.

Problem

A lighthouse flashes every 8th second. How many flashes fall in the range 100 through 250 inclusive?

Show a hint

The flashes happen exactly at the multiples of $8$ . Count every multiple of $8$ up to $250$ , then peel off the ones that fall below $100$ .
To count multiples of $8$ up to a number, see how many whole steps of $8$ fit inside it. For $250$ , how many times does $8$ go in?

Show the full solution

The flashes sit at the multiples of 8. Up to 250, the largest is

248 = 31 \times 8

, since

32 \times 8 = 256

overshoots, so 31 multiples fit. Below 100, the largest is

96 = 12 \times 8

, so 12 fall short of the window. Subtracting leaves

31 - 12 = 19

Counting to the top and removing everything under the bottom beats listing them all. Check the endpoints though, since an endpoint that is itself a multiple has to be counted.

Now picture two multiples of $n$ added or subtracted. The shared factor $n$ can be pulled out, so the result is still a multiple of $n$ . But add a non-multiple to a multiple and the result is never a multiple.

Problem

A machine dispenses and swallows tokens only in packs of 9. Starting at 0, a player receives 153, then 72 more, then returns 18. Find the final count, and explain why it must be a multiple of 9.

Show a hint

Every amount the machine moves is a whole number of $9$ s. Write each of $153$ , $72$ , and $18$ as $9$ times something, then collect the $9$ s.
If three amounts are $9 \times 17$ , $9 \times 8$ , and $9 \times 2$ , what happens when you pull the $9$ out in front of the whole calculation?

Show the full solution

The straight count is

153 + 72 - 18 = 207

. Every amount is a whole number of 9s,

153 = 9 \times 17

72 = 9 \times 8

, and

18 = 9 \times 2

, so the total is

9 \times 17 + 9 \times 8 - 9 \times 2 = 9 \times (17 + 8 - 2) = 9 \times 23 = 207 .

The 9 is a shared factor of every piece, so it slides out front and leaves a whole number behind. Adding or subtracting whole numbers of 9s can only ever build another whole number of 9s.

Problem

A turnstile reads 240, a multiple of 12. A glitch adds 17. Explain why 240 + 17 cannot be a multiple of 12, then find the smallest positive number to add to restore a multiple of 12.

Show a hint

Suppose $240 + 17$ were a multiple of $12$ . Then $17$ would be the difference of two multiples of $12$ . What does the previous problem say such a difference has to be?
The new count is $257$ . Find the next multiple of $12$ above it, then see how far away it sits.

Show the full solution

240 + 17 = 257

. Since

12 \times 21 = 252

and

12 \times 22 = 264

, the count 257 sits between two multiples of 12 and is not one, so the next marker is 264 and you must add

264 - 257 = 7 .

It could never have been a multiple, because if it were, the difference

257 - 240 = 17

would have to be a multiple of 12 as well. The stray 17 is 5 past a multiple of 12, which is exactly why 7 more gets you back.

Problem

Envelope

A

holds a multiple of 8 dollars and envelope

B

holds a multiple of 8 dollars, so the two can never total exactly 250 dollars. What is the closest total they can reach?

Show a hint

By closure, what kind of number is $A + B$ forced to be, whatever $A$ and $B$ are?
Compare 250 with $8 \times 31$ and $8 \times 32$ .

Show the full solution

Both envelopes hold multiples of 8, so the total is a multiple of 8 as well. Since

8 \times 31 = 248

and

8 \times 32 = 256

, the total skips straight past 250, and the closest it can come is

248

, for instance with

A = 240

and

B = 8

. Closure settled the question without either amount ever being known.

The number $12$ is built from $3$ and $4$ , since $12 = 4 \times 3$ . So is every multiple of $12$ also a multiple of $3$ , and of $4$ ? And does it work the other way? Is every multiple of $3$ a multiple of $12$ ?

Problem

A workshop packs beeswax sheets in bundles of 12, and

12 = 4 \times 3

, so any whole number of bundles holds a number of sheets that is a multiple of 3 and of 4. The reverse does not follow. Among the multiples of 3 from 3 through 60, how many are not multiples of 12?

Show a hint

Every bundle already splits into groups of 3, so a stack of bundles does too. The question runs the other way. Count 3, 6, 9, 12, and on up to 60, and mark which of those land exactly on a whole number of bundles.
Instead of testing each multiple of 3 one at a time, count how many multiples of 3 sit in the range, then count the multiples of 12 in the same range and take those away. Every multiple of 12 is already a multiple of 3, so each one you remove really was on the first list.

Show the full solution

There are

60 \div 3 = 20

multiples of 3 from 3 through 60. Among them,

60 \div 12 = 5

are multiples of 12, namely 12, 24, 36, 48 and 60. Striking those five out leaves

20 - 5 = 15 .

Chaining runs downward only. Any

n \times 12 = (n \times 4) \times 3

is a multiple of 3 and of 4, but a multiple of 3 such as 6 splits into 3s while falling short of a full bundle of 12.

Problem

A secret number is a multiple of 6, a perfect square, and lies between 100 and 400 inclusive. Find the sum of all possible values.

Show a hint

If a square $k \times k$ is a multiple of $6$ , the side $k$ itself has to supply the factor $2$ and the factor $3$ . What does that force $k$ to be a multiple of?
So every secret number is the square of a multiple of $6$ . Square $6, 12, 18, 24$ in turn, keep the ones that land between $100$ and $400$ , and add those.

Show the full solution

1 2^{2} = 144

and

1 8^{2} = 324

are the only ones that qualify, and

144 + 324 = 468

468

A multiple of 6 carries a factor of 2 and a factor of 3, and in a square

k \times k

those can only come from

k

itself, so

k

is a multiple of 6 and the square is a multiple of

6 \times 6 = 36

. Squaring multiples of 6,

6^{2} = 36

sits below 100,

1 2^{2} = 144

and

1 8^{2} = 324

both land in range, and

2 4^{2} = 576

overshoots.

Practice these ideas

Practice

A school fits 40 lockers into one full rectangular block, every row the same length and no locker left over. Climb the factor pairs of 40 from

1 \times 40

and stop once the two sides of a pair would cross. How many factors does 40 have in all?

Show the solution

Climb the pairs from 1 and let each small side fix its partner.

1 \times 40, 2 \times 20, 4 \times 10, 5 \times 8

Reading both sides of every pair gives

1, 2, 4, 5, 8, 10, 20, 40

, so 40 has

8

factors. The numbers 3 and 6 do not divide 40, so they are skipped, and you stop after 6, because

7 \times 7 = 49

already passes 40, so every factor above 6 is the partner of one you have. Four pairs with no side meeting itself gives an even count, exactly what a non-square should have.

Practice

A ring-toss scoreboard only ever adds 8 points per throw. Five friends claim scores of 24, 30, 48, 56, and 60. How many of these could the game actually show?

Show the solution

A score can appear only if it is a whole number of 8-point throws. Dividing,

24 = 8 \times 3

48 = 8 \times 6

, and

56 = 8 \times 7

all come out clean, while 30 leaves a remainder of 6 and 60 leaves a remainder of 4. So

3

of the five claimed scores are possible. Testing whether something is a multiple of 8 is just one division and a glance at the remainder.

Practice

Decide whether each claim about 61 is true or false. (a) 1 is a factor of 61. (b) 61 is a multiple of 61. (c) 0 is a multiple of 61. (d) 61 is a factor of 0. Enter your four verdicts in order.

Show the solution

(a)

61 = 1 \times 61

, so 1 is a factor of 61. (b)

61 = 61 \times 1

, so 61 is a multiple of itself. (c) and (d) are the single line

0 = 61 \times 0

read from its two ends, which makes 0 a multiple of 61 and 61 a factor of 0. All four hold, so the verdict is

all four true

. The three facts behind this are that 1 divides every number, every number is a multiple of itself, and 0 is a multiple of everything.

Practice

Find the smallest and largest factors of

56

, then add them together.

Show the solution

The smallest factor of

56

1

and the largest is

56

, so the sum is

1 + 56

57

Nothing smaller than

1

divides a whole number evenly, so

1

sits at the bottom of every factor list. And a factor bigger than

56

would need a partner smaller than

1

to multiply out to

56

, which no whole number does, so

56

is its own largest factor. Every factor list is bookended the same way,

1

at the low end and the number itself at the high end.

Practice

Two bookends sit on every factor list.

1

is always the smallest factor, the single full-length row, and the number itself is always the largest, since no side of a rectangle can outrun its own area. So a factor of

n

is always trapped between

1

and

n

, which is why a factor hunt can never wander off forever.

Practice

Which of 72, 81, and 98 has an odd number of factors? Explain through self-pairing.

Show the solution

Of 72, 81, and 98, only

81 = 9 \times 9

is a perfect square, since

8 \times 8 = 64

and

10 \times 10 = 100

bracket the other two. So the odd factor count belongs to

81

, whose factors are

1, 3, 9, 27, 81

, five of them. Factors normally pair off as

d

with

n \div d

, which makes the count even. Only a perfect square owns a factor that pairs with itself, and that lone factor tips the total to odd.

Practice

A ribbon is marked every 23 centimeters. What is the largest multiple of 23 below 500?

Show the solution

Fit as many whole 23-steps under the cap as possible.

500 \div 23 = 21 remainder 17

Those 21 steps land at

21 \times 23 = 483

, and the next mark,

22 \times 23 = 506

, sails past 500. The greatest multiple of

n

below a bound is the bound divided by

n

with the remainder dropped, then multiplied back by

n

Practice

A bakery packs rolls in trays of 26. What is the smallest three-digit multiple of 26?

Show the solution

A three-digit number is 100 or more, and

100 \div 26 \approx 3.8

. You cannot bake 3.8 trays, so round up to 4 whole trays, giving

26 \times 4 = 104

. Rounding down to 3 would land at 78, still two digits. The smallest multiple of

n

at or above a bound comes from dividing by

n

and rounding the quotient up, then multiplying back.

Practice

Two-digit codes lit from inside are multiples of 6 whose digits sum to 6. How many codes light up?

Show the solution

Build the codes from the digit rule first, since it leaves the shorter list. The two-digit numbers whose digits sum to 6 are

15, 24, 33, 42, 51, 60

, and no others, since a tens digit of 7 or more already overshoots. Each has digit sum 6, a multiple of 3, so each is divisible by 3 for free, and the only gate left is being even. That keeps 24, 42, and 60, so

3

codes light up. When two rules overlap, start from whichever one produces the shorter list.

Practice

Find the largest three-digit multiple of 15 whose three digits are all different.

Show the solution

The greatest three-digit multiple of 15 is

990 = 15 \times 66

, but its digits are 9, 9, 0 and the 9s repeat. Stepping down by 15 gives

975 = 15 \times 65

, whose digits 9, 7, 5 are all different, so the answer is

975

. Walking downward from the top means the first survivor you meet is the largest one, so you can stop the moment a candidate passes.

Practice

An amphitheater has 7 rows with 14, 28, 42, … seats. Find the total number of seats.

Show the solution

The seven rows hold

14, 28, 42, 56, 70, 84, 98

, which is

14 \times 1

through

14 \times 7

. Pull the shared 14 out front.

14 \times (1 + 2 + 3 + 4 + 5 + 6 + 7) = 14 \times 28 = 392

Pairing from the ends makes the inner sum quick,

1 + 7 = 8

2 + 6 = 8

3 + 5 = 8

, plus the middle 4, so

3 \times 8 + 4 = 28

. When every term shares a common step, factor that step out and add only the small counting numbers left.

Practice

A runner claps at every 13th metre. How many claps fall strictly between metres 150 and 850?

Show the solution

The claps sit at the multiples of 13. The largest one strictly under 850 is

845 = 13 \times 65

, since

13 \times 66 = 858

overshoots, so 65 multiples sit below the top. The largest at or below 150 is

143 = 13 \times 11

, so 11 sit below the window. Subtracting,

65 - 11 = 54

Neither 150 nor 850 is a multiple of 13, so the word strictly costs nothing here. Count up to the top, subtract what falls below the bottom, and always check whether an endpoint is itself a multiple.

Practice

Two trucks each carry a multiple of 15 apples. Show 400 is impossible, then find the nearest reachable total.

Show the solution

Two multiples of 15 add to

15 a + 15 b = 15 (a + b)

, still a multiple of 15. But

15 \times 26 = 390

and

15 \times 27 = 405

, so 400 sits strictly between two multiples and is not one, which makes the claimed total impossible. Of those two neighbours, 405 is only 5 away while 390 is 10 away, and 405 is reachable as

15 + 390

, so the nearest total is

405

. Once a target fails the membership test for a closed set, the best you can do is the nearest member to it.

Practice

A bead artist works only in kits of 24. For each of 2, 3, 4, 8, 9, and 16, decide whether the bead count must be a multiple of that number. How many of the six are guaranteed?

Show the solution

The factors of 24 are

1, 2, 3, 4, 6, 8, 12, 24

. Of the six sizes asked about, 2, 3, 4, and 8 are on that list, so a bead count of

24 \times k

must be a multiple of each of them, giving

4

guaranteed. Neither 9 nor 16 divides 24, and 24 itself is the counterexample for both. Being a multiple of

n

only forces divisibility by the factors of

n

, so the inheritance chains downward and stops there.

Practice

Find every perfect square that is a multiple of 45 and stays below 2000. How many such numbers are there?

Show the solution

Since

45 = 3^{2} \times 5

, a square multiple of 45 needs a factor of 5, and a prime can only get into

s \times s

by living in

s

, so

s

carries a 5 and likewise a 3. That makes

s

a multiple of 15 and the square a multiple of

1 5^{2} = 225

. Squaring,

1 5^{2} = 225

and

3 0^{2} = 900

stay under 2000, while

4 5^{2} = 2025

overshoots, so there are

2

such numbers. A perfect square needs every prime to an even power, which rounds the requirement up from 45 to 225.

Practice

A perfect number equals the sum of its proper factors:

1 + 2 + 3 = 6

and

1 + 2 + 4 + 7 + 14 = 28

. They are rare, and no one has ever found an odd perfect number or proved none exists.