Math 6

Perfect, abundant, deficient and amicable numbers

Add up the factors of a number, leaving out the number itself. Does the total fall short, overshoot, or land exactly on the number, and can two numbers each produce the other?

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Factors and multiplesGrade 4

Where you are 10 in Divisibility and primes

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Perfect, abundant, deficient and amicable numbers

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Why this exists

The factors of 6 other than 6 itself are 1, 2 and 3, and 1 + 2 + 3 = 6. The factors give back the number they came from. Try the same with 8 and you get 1 + 2 + 4 = 7, one short. Try 18 and you get 1 + 2 + 3 + 6 + 9 = 21, three too many.

Greek mathematicians studied numbers like 6 more than two thousand years ago, and Greek writers gave the three kinds the names still used: perfect, abundant and deficient. Euclid’s Elements (Book IX, Proposition 36) gives a rule that builds perfect numbers, and some questions it raised are still open today.

There is a stranger pattern too. The factors of 220, leaving out 220, add to 284, and the factors of 284, leaving out 284, add straight back to 220. Each number is made from the other. That pair has been known since ancient times; later Greek writers credited it to the followers of Pythagoras. Pairs like it are called amicable, from the Latin for friendly.

None of this is needed to do arithmetic. It is here because it is a good reason to get fast and careful at finding all the factors of a number, and because it shows how a simple question about whole numbers can stay unanswered for centuries.

Definition

Everything in this lesson is about whole numbers from 1 up.

The proper factors of a number are its factors other than the number itself. The proper factors of 12 are 1, 2, 3, 4 and 6; the number 12 is a factor of 12 but not a proper one.

We write s(n) for the sum of the proper factors of n. So s(6) = 6, s(8) = 7 and s(18) = 21.

A number n is

perfect when its proper factors add to exactly n, that is s(n) = n;
abundant when they add to more than n, that is s(n) > n;
deficient when they add to less than n, that is s(n) < n.

Every number from 1 up is exactly one of the three, because a total is either more than n, less than n, or equal to it.

Two different numbers a and b are an amicable pair when the proper factors of a add to b and the proper factors of b add to a:

s(a) = b \qquad\text{and}\qquad s(b) = a

Intuition

Lay the proper factors end to end and compare the row with the number itself.

Figure 1The proper factors laid end to end fall short of 8, land exactly on 6 and run past 18.

Which way the row goes depends on how many factors the number has, and especially on its small ones. Factors come in pairs that multiply to the number: for 18 the pairs are 1 \times 18, 2 \times 9 and 3 \times 6. Each small factor brings a big partner with it. A factor of 2 brings n \div 2, which is half the number on its own; a factor of 3 brings a third, and a factor of 6 a sixth. A multiple of 6 bigger than 6 has a half, a third and a sixth of itself in its row, which already make the whole number, and the 1 pushes it past. A prime has no small factors at all except 1, so its row is just 1.

That is the rough picture: many small factors, abundant; few factors, deficient. Perfect numbers sit on the knife edge between the two, which is why they are rare.

Results

Primes are deficient. If p is prime, its only proper factor is 1, so s(p) = 1.

Powers of 2 fall short by exactly one. For 2, 4, 8, 16, \ldots

s(2^k) = 2^k - 1

Multiples of perfect and abundant numbers are abundant. If n is perfect or abundant, then 2n, 3n, 4n and every larger multiple of n is abundant.

Euclid’s rule for perfect numbers. If 2^k - 1 is prime, then

2^{k-1} \times (2^k - 1)

is perfect. With k = 2, 3, 5, 7 it gives 6, 28, 496 and 8128. Leonhard Euler later proved that every even perfect number comes from this rule. Nobody knows whether an odd perfect number exists: none has ever been found, and nobody has proved there are none.

The smallest amicable pair is 220 and 284.

Why it is true

Primes. A prime p has exactly two factors, 1 and p. Leaving out p leaves 1, and 1 < p.

Powers of 2. The proper factors of 2^k are 1, 2, 4, \ldots, 2^{k-1}, because by unique factorization a factor of 2^k can contain no prime but 2. Add them one at a time: 1 = 2 - 1, then 1 + 2 = 3 = 4 - 1, then 3 + 4 = 7 = 8 - 1. Each time the total is one less than the next power of 2, and adding that power doubles it and keeps it one short. So the total stops at 2^k - 1.

Multiples. Let n be perfect or abundant, so s(n) \ge n, and let m = c \times n with c \ge 2. Take each proper factor d of n and multiply it by c. Each c \times d divides m, and each is less than m, so each is a proper factor of m, and they add to c \times s(n) \ge c \times n = m. The factor 1 of m is not on that list, since every c \times d is at least 2. So s(m) \ge m + 1, and m is abundant.

Euclid’s rule. Write q = 2^k - 1, a prime, and n = 2^{k-1} \times q. By unique factorization the factors of n are the powers of 2 up to 2^{k-1}, and those same powers times q. Leave out n itself, which is the last power of 2 times q. The powers of 2 add to 2^k - 1 = q, as just shown. The powers of 2 up to 2^{k-2}, times q, add to q \times (2^{k-1} - 1). Together:

\begin{aligned} s(n) &= q + q \times (2^{k-1} - 1) \\ &= q \times 2^{k-1} = n \end{aligned}

Euler’s proof that no other even perfect number exists is harder and is left for later study.

Worked examples

Example 1

Sorting three numbers

Decide whether 28, 40 and 45 are perfect, abundant or deficient.

Solution

List each number’s factors in pairs that multiply to it, so none is missed, then add all but the number itself.

28 = 1 \times 28 = 2 \times 14 = 4 \times 7, and 3, 5, 6 do not divide 28. The proper factors are 1, 2, 4, 7, 14:

1 + 2 + 4 + 7 + 14 = 28

so 28 is perfect.

40 = 1 \times 40 = 2 \times 20 = 4 \times 10 = 5 \times 8, and 3 and 6 do not divide 40. The proper factors add to

1 + 2 + 4 + 5 + 8 + 10 + 20 = 50

and 50 > 40, so 40 is abundant.

45 = 1 \times 45 = 3 \times 15 = 5 \times 9. The proper factors add to 1 + 3 + 5 + 9 + 15 = 33, and 33 < 45, so 45 is deficient.

Example 2

Listing factors from a factorization

Decide whether 196 is perfect, abundant or deficient.

Solution

Its prime factorization is 196 = 2^2 \times 7^2. By unique factorization every factor of 196 is a power of 2 up to 2^2 times a power of 7 up to 7^2, so a grid lists them all:

	1	7	49
1	1	7	49
2	2	14	98
4	4	28	196

Leave out 196 and add the other eight:

1 + 7 + 49 + 2 + 14 + 98 + 4 + 28 = 203

203 > 196, so 196 is abundant. This was predictable: 196 = 7 \times 28 is a multiple of the perfect number 28.

Example 3

The friendly pair

Show that 220 and 284 are an amicable pair.

Solution

220 = 2^2 \times 5 \times 11. Its factors are the powers 1, 2, 4 times each of 1, 5, 11, 55, which gives twelve: 1, 2, 4, 5, 10, 20, 11, 22, 44, 55, 110 and 220.

Leave out 220 and add, grouping to keep the sums easy:

\begin{aligned} s(220) &= (1 + 2 + 4) + (5 + 10 + 20) \\ &\quad + (11 + 22 + 44) + (55 + 110) \\ &= 7 + 35 + 77 + 165 = 284 \end{aligned}

Now 284 = 2^2 \times 71, and 71 is prime. Its factors are 1, 2, 4, 71, 142, 284, so

s(284) = 1 + 2 + 4 + 71 + 142 = 220

Each sum is the other number, and 220 \ne 284, so the pair is amicable.

Example 4

The next perfect number

Use Euclid’s rule to find the first perfect number after 28, and check it by adding.

Solution

The rule needs 2^k - 1 to be prime. k = 2 and k = 3 give 3 and 7, which produce 6 and 28. k = 4 gives 15 = 3 \times 5, not prime, so the rule says nothing. k = 5 gives 31, which is prime, so

2^4 \times 31 = 16 \times 31 = 496

is perfect. Check: the factors of 496 are 1, 2, 4, 8, 16 and those times 31, that is 31, 62, 124, 248, 496. Leaving out 496:

\begin{aligned} s(496) &= (1 + 2 + 4 + 8 + 16) \\ &\quad + (31 + 62 + 124 + 248) \\ &= 31 + 465 = 496 \end{aligned}

Example 5

Abundant without the full list

Show that 84 is abundant, then find s(84).

Solution

84 = 3 \times 28, and 28 is perfect. Multiply each proper factor of 28 by 3: 3, 6, 12, 21, 42. Each divides 84 and is less than 84, and they add to 3 \times 28 = 84 already. The factor 1 is not among them, so s(84) is at least 85, and 84 is abundant.

The full list of proper factors is 1, 2, 3, 4, 6, 7, 12, 14, 21, 28 and 42. Adding them in two groups, 1 + 2 + 3 + 4 + 6 + 7 = 23 and 12 + 14 + 21 + 28 + 42 = 117, so

s(84) = 23 + 117 = 140

Example 6

One direction is not enough

Dario finds that s(20) = 22 and says 20 and 22 are an amicable pair. Is he right?

Solution

He has checked only half. The proper factors of 20 are 1, 2, 4, 5, 10, which do add to 22. But 22 = 2 \times 11 has proper factors 1, 2, 11, so

s(22) = 14

and 14 \ne 20. The sum from 22 does not lead back, so the pair is not amicable.

Example 7

Following the sums

Start at a number, find s of it, then s of that, and keep going. Follow the chain from 95 and from 562.

Solution

95 = 5 \times 19, so s(95) = 1 + 5 + 19 = 25. Then s(25) = 1 + 5 = 6, and s(6) = 6. The chain lands on a perfect number and stays there for ever:

95,\ 25,\ 6,\ 6,\ 6,\ \ldots

562 = 2 \times 281, and 281 is prime (no prime up to 16 divides it, and 17 \times 17 > 281). So s(562) = 1 + 2 + 281 = 284. From the friendly pair, s(284) = 220 and s(220) = 284:

562,\ 284,\ 220,\ 284,\ 220,\ \ldots

This chain falls into the amicable pair and goes round it for ever. A perfect number is a loop of length one, an amicable pair a loop of length two.

Boundaries

Only whole numbers from 1 up. Every number from 1 up divides 0, so 0 has no finite list of factors and is left out. The number 1 has no proper factors at all, so s(1) = 0 and 1 is deficient.

Amicable means two different numbers. A perfect number satisfies s(n) = n, so it would pair with itself; the definition asks for a \ne b to rule that out.

Both directions must hold. s(a) = b alone says nothing about s(b), as the pair 20 and 22 shows.

Euclid’s rule needs a prime. When 2^k - 1 is not prime, 2^{k-1} \times (2^k - 1) is not perfect. And the rule makes only even numbers, so it says nothing about odd ones.

Small factors are only a rough guide. Most abundant numbers are even, but not all: 945 is odd and abundant.

Where this goes next

Adding the number itself back gives the sum of all its factors, and a perfect number is one whose factors add to exactly twice it. That sum can be worked out straight from the prime factorization without listing a single factor, a formula met in number theory.

Euclid’s rule ties perfect numbers to primes of the form 2^k - 1, and the search for very large primes of that form is still going on. In the ninth century Thābit ibn Qurra gave a rule that produces some amicable pairs, 220 and 284 among them, and Euler found dozens more in the eighteenth century. The second smallest pair, 1184 and 1210, was missed by them all and found in the 1860s by a sixteen-year-old Italian student, Nicolò Paganini (not the famous violinist). Whether there are infinitely many amicable pairs is not known. Chains that loop back after more than two steps also exist.

The same factor lists feed straight into the next tools, the greatest common factor and least common multiple.

Exercises

Proper factors and their sum

List the proper factors of 50 and find s(50).
Find s(64).
Find s(97).
Find s(100).

Perfect, abundant or deficient

Decide whether each number is perfect, abundant or deficient, and show the sum.

21
36
49
70
Find every abundant number from 1 to 30.
945 = 3^3 \times 5 \times 7. Show that 945 is abundant.

Amicable pairs

1184 = 2^5 \times 37 and 1210 = 2 \times 5 \times 11^2. Show that they are an amicable pair.
Shirin works out s(2620) = 2924. Given 2924 = 2^2 \times 17 \times 43, find s(2924) and decide whether 2620 and 2924 are amicable.
Leona notices s(24) = 36. Are 24 and 36 an amicable pair?

Following the sums

Follow the chain from 18 until it reaches 0.
Follow the chain from 58 until it reaches 0.
Find every number below 30 whose proper factors add to 6.

Using the results

127 is prime and 127 = 2^7 - 1. Which perfect number does Euclid’s rule give?
2^6 - 1 = 63 is not prime, so Euclid’s rule says nothing about 2^5 \times 63 = 2016. Without listing its factors, decide whether 2016 is perfect, abundant or deficient.
Without listing its factors, explain why 280 is abundant.

Reasoning

Explain why every prime is deficient.
A number n has s(n) = 1. Explain why n must be prime.
Tobias says 6 is amicable with itself. Explain why the definition asks for two different numbers.
Ama says every even number is abundant. Is she right? Explain.
Ren doubles a perfect number and expects another perfect number. Explain what he always gets instead.

Answers

1, 2, 5, 10, 25; s(50) = 43.
s(64) = 63: the proper factors are 1, 2, 4, 8, 16, 32, and a power of 2 falls short by one.
s(97) = 1, since 97 is prime.
s(100) = 117: the proper factors 1, 2, 4, 5, 10, 20, 25 and 50 add to 117.
Deficient: 1 + 3 + 7 = 11 < 21.
Abundant: the proper factors 1, 2, 3, 4, 6, 9, 12 and 18 add to 55, and 55 > 36.
Deficient: 1 + 7 = 8 < 49.
Abundant: the proper factors 1, 2, 5, 7, 10, 14 and 35 add to 74, and 74 > 70.
12, 18, 20, 24, 30. Their sums are 16, 21, 22, 36, 42; every other number up to 30 is deficient except the perfect 6 and 28.
The factors are 1, 3, 9, 27 times each of 1, 5, 7, 35. The four groups add to 40, 200, 280 and 1400, which is 1920 in all. Leaving out 945, s(945) = 975 > 945, so 945 is abundant.
The proper factors of 1184 are 1, 2, 4, 8, 16, 32 (adding to 63) and 37, 74, 148, 296, 592 (adding to 1147), so s(1184) = 63 + 1147 = 1210. Those of 1210 are 1, 2, 5, 10 (adding to 18), 11, 22, 55, 110 (adding to 198) and 121, 242, 605 (adding to 968), so s(1210) = 18 + 198 + 968 = 1184. Each sum is the other number: amicable.
s(2924) = 2620. The factors are 1, 2, 4 times each of 1, 17, 43, 731; leaving out 2924, the groups add to 7 + 119 + 301 + 2193 = 2620. So 2620 and 2924 are amicable.
No. s(36) = 55, not 24.
18, 21, 11, 1, 0: s(18) = 21, s(21) = 11, s(11) = 1, s(1) = 0.
58, 32, 31, 1, 0: 58 = 2 \times 29 gives 1 + 2 + 29 = 32, then s(32) = 31, s(31) = 1, s(1) = 0.
6 and 25. s(6) = 1 + 2 + 3 = 6 and s(25) = 1 + 5 = 6; checking the rest below 30 gives no other sum of 6.
2^6 \times 127 = 64 \times 127 = 8128.
Abundant. 2016 = 6 \times 336 is a multiple of the perfect number 6. (In fact s(2016) = 4536.)
280 = 10 \times 28 is a multiple of the perfect number 28, so it is abundant (s(280) = 440).
A prime p has only the factors 1 and p, so s(p) = 1, and 1 < p because p \ge 2.
1 divides every number, so if s(n) = 1 the only proper factor is 1, and n \ne 1 because s(1) = 0. So n has exactly two factors, 1 and n, which is what being prime means.
Every perfect number satisfies s(n) = n, so allowing a = b would make each perfect number “amicable” and the word would add nothing. Asking a \ne b keeps the two ideas apart.
No. 8 is even and s(8) = 7 < 8. So are 2, 4, 10, 14 and 16.
An abundant number. 2n is a multiple of the perfect number n, so s(2n) \ge 2n + 1. For example s(12) = 16 and s(56) = 64.