QuanticaPrealgebraOpen in the courseA factor of a number is a side of a rectangle with that area, every row the same length, nothing left over. Factors travel in pairs, since each side has a matching partner on the opposite edge whose product is 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. $$1\times24,\quad 2\times12,\quad 3\times8,\quad 4\times6$$ 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 \(\boxed{8}\) 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 \(1\times48, 2\times24, \ldots\) 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 \(48\div3\) 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. $$1\times48,\quad 2\times24,\quad 3\times16,\quad 4\times12,\quad 6\times8$$ Reading both numbers out of every pair gives \(1,\ 2,\ 3,\ 4,\ 6,\ 8,\ 12,\ 16,\ 24,\ 48\), which is \(\boxed{10}\) factors. You stop after 6 because the next divisor, 8, is already standing on the right of \(6\times8\). 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 \(\boxed{3}\) stones. Each landing reads two ways. \(7\) is a factor of 28 because \(28=4\times7\), 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 \(b=a\times(\text{whole number})\). Check the pair where both numbers are equal carefully.
Show the full solution
(a) \(12=4\times3\), 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) \(8=8\times1\), 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 \(\boxed{2}\) 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 \(1\times36,\ 2\times18,\ 3\times12,\ 4\times9,\ 6\times6\), 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 \(6\times6\), and note that \(49=7\times7\).
- List every whole number that divides 49 evenly.
Show the full solution
The pairs are \(1\times49\) and \(7\times7\), and the second adds only one new factor because 7 is its own partner. The factors are \(1,\ 7,\ 49\), so 49 has \(\boxed{3}\). 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 \(1\times35\) and \(5\times7\), has four.
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, \(1\times20\), \(2\times10\), \(4\times5\).
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Priya is the one who finishes, since no factor of 20 can be bigger than 20 itself. Hunting in pairs gives \(1\times20\), \(2\times10\), \(4\times5\), so her list is \(1,\ 2,\ 4,\ 5,\ 10,\ 20\), which is \(\boxed{6}\) 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 $$31-12=\boxed{19}$$ 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 \(9\times17\), \(9\times8\), and \(9\times2\), 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, \(153=9\times17\), \(72=9\times8\), and \(18=9\times2\), so the total is $$9\times17+9\times8-9\times2=9\times(17+8-2)=9\times23=\boxed{207}.$$ 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.
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\(240+17=257\). Since \(12\times21=252\) and \(12\times22=264\), the count 257 sits between two multiples of 12 and is not one, so the next marker is 264 and you must add $$264-257=\boxed{7}.$$ 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 \(8\times31\) and \(8\times32\).
Show the full solution
Both envelopes hold multiples of 8, so the total is a multiple of 8 as well. Since \(8\times31=248\) and \(8\times32=256\), the total skips straight past 250, and the closest it can come is \(\boxed{248}\), 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
Bundles of 12 beeswax sheets, with \(12 = 4 \times 3\). (1) Must a whole number of bundles be a multiple of 3? (2) Of 4? (3) Is every multiple of 3 a multiple of 12? Give a counterexample.
Show a hint
- For (1) and (2), picture one bundle as \(3\) rows of \(4\) (or \(4\) rows of \(3\)). If every single bundle already splits into a whole number of threes, what happens when you have several bundles stacked together?
- For (3), look for a heap that you can split into groups of \(3\) with none left over, yet that is too small or too lopsided to fill whole bundles of \(12\). Try counting up \(3, 6, 9, \ldots\) and check each against \(12\).
Show the full solution
(1) Yes. A delivery of \(n\) bundles is \(n\times12 = n\times(4\times3) = (n\times4)\times3\), which is 3 times a whole number. (2) Yes, the same way, \(n\times12 = n\times(3\times4) = (n\times3)\times4\), which is 4 times a whole number. (3) No. A pile of \(\boxed{6}\) sheets is \(3\times2\), so it is a multiple of 3, but it is smaller than one full bundle, so it is not a multiple of 12. This chaining only runs downward, from 12 to the numbers built into it, never back up.
Problem
A secret number is a multiple of 6, a perfect square, and lies between 100 and 400 inclusive. Find every possible value.
Show a hint
- If a square \(k\times k\) is a multiple of \(6\), the side \(k\) itself must supply the factor \(2\) and the factor \(3\). What does that force \(k\) to be a multiple of?
- So the secret number is the square of a multiple of \(6\). Which multiples of \(6\) have squares between \(100\) and \(400\)?
Show the full solution
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 \(6\times6=36\). Squaring multiples of 6, \(6^2=36\) is below 100, \(12^2=144\) and \(18^2=324\) both land in range, and \(24^2=576\) overshoots. $$\boxed{144 \text{ and } 324}$$ Asking for a square that is also a multiple of 6 quietly forces a multiple of 36.
Practice these ideas
Practice
A school fits 40 lockers into one full rectangular block. List all factor pairs of 40 and report how many factors there are in all.
Show the solution
Climb from 1 and let each small factor fix its partner. $$1\times 40,\qquad 2\times 20,\qquad 4\times 10,\qquad 5\times 8$$ The numbers 3, 6, and 7 do not divide 40, so they are skipped. Reading both sides of every pair and sorting gives \(1,\ 2,\ 4,\ 5,\ 8,\ 10,\ 20,\ 40\), which is \(\boxed{8}\) factors. You stop after 5 because the next try, 6, would pair with a partner smaller than itself, and every rectangle down there has already appeared mirrored. Four clean 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 \(\boxed{3}\) 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 \(\boxed{\text{all four true}}\). 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, and give a one-line reason for each.
Show the solution
The smallest factor is 1, since \(56 = 1\times 56\) and nothing smaller than 1 is a positive whole number. The largest is 56 itself, since a factor bigger than 56 would need a partner smaller than 1, and no such whole number exists. $$\boxed{1\ \text{and}\ 56}$$ Every factor list is bookended the same way, 1 at the low end and the number 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 \(\boxed{81}\), 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. $$500 \div 23 = 21 \text{ remainder } 17$$ Those 21 steps land at \(21\times 23 = \boxed{483}\), 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 \(26 \times 4 = \boxed{104}\). 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 \(\boxed{3}\) 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 \(\boxed{975}\). 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. $$14\times(1+2+3+4+5+6+7) = 14\times 28 = \boxed{392}$$ 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, $$65 - 11 = \boxed{54}$$ 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 \(15a + 15b = 15(a+b)\), 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 \(\boxed{405}\). 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 \(\boxed{4}\) 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 \(15^2 = 225\). Squaring, \(15^2 = 225\) and \(30^2 = 900\) stay under 2000, while \(45^2 = 2025\) overshoots, so there are \(\boxed{2}\) 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.