Download The Chinese Restaurant Approach to Integer

The Chinese Restaurant Approach
to Integer Representation Problems
by Harold Reiter
University of North Carolina Charlotte
We are given a set G of generators and a process P for producing integers from
the members of G. Actually we’ll let G be a set, a multiset, or a sequence. G
can be finite or infinite. Each choice of G and P gives rise to a set R of results.
We’ll see how varying G and P gives rise to many different types of problems.
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Some examples of G’s are:
G1 = {100 , 101 , 102 , . . .}, the powers of 10.
G2 = {2, 3, 5, 7, . . .}, the primes.
G3 = {1, 2, 3, 4, . . .}, the positive integers.
G4 = {1, 2, 4, 8, . . .}, the non-negative integral powers of 2.
G5 = {1, 3, 9, 27, . . .}, the non-negative integral powers of 3.
G6 = {1, −2, 4, −8, . . .}, the integral powers of −2.
G7 = {p, q}, any two distinct primes (or relative primes).
G8 = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11}, the integers from 1 to 11.
G9 = {1, 2, 3, 5, 8, 13, . . .}, the Fibonacci numbers.
G10 = {1, 5, 10, 25, 50, 100}, the values in cents of US coins.
G11 = {1, 4, 9, 16, 25, · · ·}, the perfect squares.
G12 = {1, 2, 6, 24, . . .}, the factorials of positive integers.
We will also allow G to be a multiset, that is, an object which can have
multiple membership, but which is unchanged by a rearrangement of its
members. For example {1, 1, 2, 2, 3, 3, 3} is a multiset which is not the same
as {1, 2, 3}, but which is the same as {1, 1, 3, 2, 3, 3, 2}. In this case the three
instances of the ‘3’ are considered distinct. We can match these generating
sets Gi with generating processes Pj . Some examples we will use are:
P1 :
r = x1 + x2 + · · · + xn ,
where n is arbitrary and the xi are all distinct members of G. In other words,
the set R of results of applying P to G is given by R = {r = x1 +x2 +· · ·+xn :
n is arbitrary and the xi are all distinct members of G}
P2 :
r = d1 · x1 + d2 · x2 + · · · + dn · xn ,
where n is arbitrary, the xi are all distinct members of G, and the di are
decimal digits.
P3 :
r = a 1 · x1 + a 2 · x2 + · · · + a n · xn ,
where n is arbitrary, the xi are all distinct members of G, and the ai are
arbitrary positive integers.
For example, we might insist that the ai belong to a given set C.
P4 :
r = a 1 · x1 + a 2 · x2 + · · · + a n · xn ,
where n is arbitrary, the xi are all distinct members of G, and the ai all
belong to some prescribed subset C of the positive integers. This generalizes
all the examples below.
For the next process, we no longer take integer conbinations to get members of R. We allow instead any arithmetic operations.
P5 :
r = x1 21 x2 22 · · · 2n−1 xn ,
where n is arbitrary, 2i ∈ {+, −, ÷, ×} for each i, the xi are all distinct
members of G, and parentheses are inserted so that the expression is unambiguous.
For all the examples above, note that the number of summands is arbitrary. For those below, we fix n and require that the expression have exactly
n members of G. In this case it makes a difference whether 0 is included
in the set C of allowed coefficients. If 0 is included, then expressions could
effectively have fewer than n members of G.
P6 (n) :
r = x1 ± x2 ± · · · ± xn ,
where n is fixed and the xi are all distinct members of G.
P7 (n) :
r = x1 + x2 + · · · + xn ,
where n is fixed and the xi are all distinct members of G.
Of course, these Pi could be generalized in the same way that we did for
those with the arbitrary n, simply by requiring the coefficients be members
of some set C.
Then there is the n-fixed analog of P5 .
P8 (n) :
r = x1 21 x2 22 · · · 2n−1 xn ,
where n is fixed, 2i ∈ {+, −, ÷, ×} for each i, the xi are all distinct members
of G, and parentheses are inserted so that the expression is unambiguous.
What questions do we ask about these G’s and P ’s? Is the set R of
numbers generated finite or infinite?
1. Suppose R is infinite.
(a) Is R the set of all positive integers? If not, what is the smallest
omitted (positive integer) value?
(b) Is each member of R represented uniquely? If not, how many ways
can the integer n be represented?
(c) Is there a nice algorithm for finding the nth one?
(d) Is the complement of R finite? If so, what is the largest integer
not in R?
2. Suppose R is finite.
(a) What is the cardinality of R?
(b) Is representation unique?
(c) What is the largest member of R?
(d) What is the smallest positive integer not in R?
There is another type of question which fits very nicely in this framework.
Suppose we are given the P and the R. We can ask questions about the
sets, lists, or multisets G which result in R when P is applied. For example,
problems 4 and 7 below are of this type. Here are some sample problems:
1. Problem 7, 1986 AIME.
The increasing sequence 1, 3, 4, 9, 10, 12, 13, . . . consists of those integers
which are powers of three or sums of distinct powers of 3. Find the
100th term of the sequence (where 1 is the 1st term, 3 is the 2nd term,
and so on).
2. NC-SC ARML Training Session, 1994.
How many numbers can be expressed as a sum of four distinct members
of the set {17, 21, 25, 29, 33, 37, 41}?
3. The Postage Stamp Problem
An unlimited supply of postage stamps of denominations p and q are
available, where p and q are relatively prime. What is the largest
amount of postage that cannot be produced using these stamps?
4. The Balance Pan Problem.
Given a finite set {w1 , w2 , w3 , . . . , wn }, and a two-pan balance, what
is the largest number of weights which can be determined by the wi ?
How should the values w1 , w2 , w3 , . . . , wn be selected to minimize the
number of duplicated values?
5. McNuggets Problem.
Chicken McNuggets come in packages of size 6, 9 and 20. What is the
largest number of McNuggets which cannot be purchased?
6. How many integers can be obtained as a sum of two or more of the
numbers 1, 3, 5, 10, 20, 50, 82? (Alan Tucker’s Applied Combinatorics,
problem 40, page 179.)
7. If pairs of distinct elements of the set S are added, the following ten
numbers are obtained:
1967, 1972, 1973, 1974, 1975, 1980, 1983, 1984, 1989, 1991.
What are the elements of S? (USAMTS, 1995)
8. Use each of the nine digits 1, 2, 3, 4, 5, 6, 7, 8, and 9 exactly twice to
form distinct prime numbers whose sum is as small as possible.
9. How many numbers can be obtained as the product of two or more
of the numbers 3, 4, 4, 5, 5, 6, 7, 7, 7? (Alan Tucker’s Applied Combinatorics, problem 34, page 179.)
10. Brackets can be inserted into the expression 1 ÷ 2 ÷ 3 ÷ 4 in various
ways. For example, (1 ÷ 2) ÷ (3 ÷ 4) equals 2/3 whereas 1 ÷ ((2 ÷ 3) ÷ 4)
equals the whole number 6. Similarly, brackets can be inserted into
1 ÷ 2 ÷ 3 ÷ 4 ÷ 5 ÷ 6 ÷ 7 ÷ 8 ÷ 9 ÷ 10 ÷ 11 to produce a large collection
of answers, many of which are integers. What do you get if you divide
the largest of these integers by the smallest of these integers? (London
Times Brainteaser #1696).
11. How many of the first 100 positive integers are expressible as a sum of
three or fewer members of the set {30 , 31 , 32 , 33 , 34 } if we are allowed to
use the same power more than once. For example, 5 can be represented,
but 8 cannot. (1991 State Math Contest of North Carolina.)
12. How many integers can be expressed as a sum of two or more different
members of the set {0, 1, 2, 4, 8, 16, 31}? (1994 UNC Charlotte Contest.)
13. In a standard ‘infix’ calculator, unparenthesized expressions are evaluated as follows:
(a) Multiplications and divisions are done first, starting at the left
and moving to the right.
(b) Additions and subtractions are done, again moving from left to
right.
For example, 1 − 2 ÷ 3 − 4 is −3 32 . How many different values are
possible if each ∗ is replaced by one of +, −, ×, or ÷ in the expression
1 ∗ 1 ∗ 1 ∗ 1 ∗ 1 ∗ 1 ∗ 1 ∗ 1? (1987 State Math Contest of North Carolina.)
14. The number 113 can be expressed as a sum of multiples of powers of
−3. In fact, there exist integers a0 , a1 , a2 , a3 , a4 , such that 0 ≤ ai ≤ 2
and 113 = a4 (−3)4 + a3 (−3)3 + a2 (−3)2 + a1 (−3)1 + a0 (−3)0 . Find
the value of a0 + a1 + a2 + a3 + a4 . (1990 State Math Contest of North
Carolina.)
15. John has 2 pennies, 3 nickels, 2 dimes, 3 quarters, and 8 dollars. For
how many different amounts can John make an exact purchase?
16. Show that 1997 can be written in the form
±12 ± 22 ± 32 · · · ± k 2
for some positive integer k. Next show that every positive integer can
be written in this way. (Paul Erdös)
17. Do there exist positive integers N and M such that every positive
ingerger larger than M can be written as the sum of at most N distinct
Fibonacci numbers? (Leo Schneider)
SOLUTIONS
1. Problem 7, 1986 AIME.
The increasing sequence 1, 3, 4, 9, 10, 12, 13, . . . consists of those integers
which are powers of three or sums of distinct powers of 3. Find the
100th term of the sequence (where 1 is the 1st term, 3 is the 2st term,
and so on).
Solution. Let G be the powers of 3. Let P be the process which allows
addition of distinct members of G. The elements of R are representable
in ternary using just the digits 0 and 1. So they all look like binary
representations. The ordering of the elements of R is the same as they
would be ordered in binary. Therefore the 100th one is the one whose
ternary representation is the binary representation of 100. Since
100 = 64 + 32 + 4
= 2 6 + 25 + 22
= 1 · 26 + 1 · 25 + 0 · 2 4 + 0 · 2 3 + 1 · 2 2 + 0 · 2 1 + 0 · 2 0
= 11001002 ,
it follows that the 100th member of R is 36 +35 +32 = 729+243+9 = 981.
2. NC-SC ARML Training Session, 1994. How many numbers can be
expressed as a sum of four distinct members of the set
{17, 21, 25, 29, 33, 37, 41}?
Solutions. Let G = {17, 21, 25, 29, 33, 37, 41} and let P be P7 (4).
• Each member of G is one more than a multiple of four. Therefore,
any sum of four of them is a multiple of 4. The smallest such
number is 17+21+25+29 = 92 and the largest is 29+33+37+41 =
140 and all the multiples of 4 between them are obtainable. There
are 140−92
+ 1 = 48
+ 1 = 13 such numbers.
4
4
• We transform the problem into a simpler one. Because 17 =
4 · 4 + 1, 21 = 4 · 5 + 1, . . . , 41 = 4 · 10 + 1, it makes sense to set up
a correspondance between R and the set of numbers generated by
{4, 5, 6, 7, 8, 9, 10} using P7 (4) OR between R and those numbers
generated by {−3, −2, −1, 0, 1, 2, 3} using P7 (4). The set R in this
case is just {−6, −5, −4, −3, −2, −1, 0, 1, 2, 3, 4, 5, 6}, which has 13
members.
• There are 74 = 35 ways to choose four elements from the sevenmember set G. But there is some duplication. The table below
shows the number of ways Tn to write n, (alternatively 116n + 4n)
as a sum of four members of {−3, −2, −1, 0, 1, 2, 3} (alternatively,
{17, 21, 25, 29, 33, 37, 41}).
n -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6
Tn 1 1 2 3 4 4 5 4 4 3 2 1 1
3. The Postage Stamp Problem An unlimited supply of postage stamps
of denominations p and q are available, where p and q are relatively
prime. What is the largest amount of postage that cannot be produced
using these stamps.
Solution. Let G = {p, q} and let P be P3 . The largest amount not
representable is pq − p − q. If pq − p − q were so representable, then
pq − p − q = (M − 1)p + (N − 1)q where M and N are POSITIVE
integers. However, since pq = M p + N q and (p, q) = 1 it follows that
p|N (since p|N q). Similarly q|M . But then M p + N q must be of the
form Kpq + Lpq where K and L are both POSITIVE integers which
contradicts pq = M p + N q = Kpq + Lpq. So we know that pq − p − q
is not representable.
To show that this is the largest such number, observe that any positive
integer k can be written in the form
(∗)
k = Ap + Bq
where A and B are (not necessarily positive) integers. It follows from
(*) that k > 0 can be written in the form k = M p + N q where N is
positive and −q < M <= 0. Now, consider pq + k = pq + M p + N q =
(q + M )p + N q. We see that (q + M ) and N are both positive. Thus, we
have proved that any integer greater than pq can be written in the form
M p + N q where M and N are positive. Finally then, if k > pq − p − q,
we have k + p + q = M p + N q where M and N are positive so that
k = (M − 1)p + (N − 1)q as desired.
4. The Balance Pan Problem. Given a finite set {w1 , w2 , w3 , . . . , wn }, and
a two-pan balance, what is the largest number of weights which can
be determined by the wi ? How should the values w1 , w2 , w3 , . . . , wn be
selected to minimize the number of duplicated values?
Solution. There are three things we can do with each weight: a) put it
on the left pan, b) put it on the right pan, or c) leave it out. Thus there
are 3n ways to distribute the weights. But one of these is the one with
no weights on either pan, and half the rest are obtainable from the other
half by switching the weights in the two pans. For each configuration of
weights, (U, V ), where U represents the set of weights on the left pan
P
P
and V those on the right, there corresponds the sum i∈V − j∈U .
Zero corresponds to the configuration (φ, φ), and the transformation
P
P
P
P
i∈V −
j∈U →
i∈U −
j∈V represents a one-to-one correspondance
between positive sums and negative sums. Hence the set R of results
n
has cardinality at most 3 2−1 . This occurs when no two configurations
P
give the same sum. Note that the set { ai wi |ai ∈ {0, −1, 1}} has 3n
P
members if and only if { (ai + 1)wi |ai ∈ {0, −1, 1}} has 3n members.
We have the following theorem the proof of which is omitted:Theorem
n
The set R of weighable values is the set {1, 2, 3, . . . 3 2−1 } if and only of
the set of weights is 1, 3, 9, . . . , 3n−1 . Thus this problem is obtainable
by choosing G as the powers of 3, and P as the process P4 with A =
{−1, 0, 1}.
5. McNuggets Problem. Chicken McNuggets come in packages of size 6,
9 and 20. What is the largest number of McNuggets which cannot be
purchased?
Solution. Since 6 and 9 have a gcd of 3, and 20 is congruent to 2 modulo
3, it makes sense to distinguish three sets, S0 , S1 , and S2 , defined as
follows:
S0 = {n|n = 6a + 9b for some non-negative integers a and b}
S1 = {n|n = 6a + 9b + 20 for some non-negative integers a and b}
and
S2 = {n|n = 6a + 9b + 40 for some non-negative integers a and b}.
Thus,
S0 = {6, 9, 12, 15, 18, 21, . . .}
S1 = {20, 26, 29, 32, 35, 38, 41, 44, . . .}
and
S2 = {40, 46, 49, 52, . . .}.
Therefore, by inspection we see that 43 is the largest number not in
S
S
S0 S1 S2 .
6. How many integers can be obtained as a sum of two or more of the
numbers 1, 3, 5, 10, 20, 50, 82? (Alan Tucker’s Applied Combinatorics,
problem 40, page 179.)
Solution. Each member of G = {1, 3, 5, 10, 20, 50, 82} is larger than
the sum of the smaller members of G except 82. Therefore the only
duplicated sums involve 82. The minimal duplicated sums are 82 + 1 =
50 + 20 + 10 + 3 and 82 + 3 = 50 + 20 + 10 + 5. The other duplicates
is 88 and 86. Thus there are
!
7
2 −
− 1 − 4 = 116
1
7
different sums, since we don’t want to count the
empty sum, and the four duplicated values.
7
1
= 7 singletons, the
7. If pairs of distinct elements of the set S are added, the following ten
numbers are obtained:
1967, 1972, 1973, 1974, 1975, 1980, 1983, 1984, 1989, 1991.
What are the elements of S? (USAMTS, 1995)
Solution. Since 52 = 10 it makes sense to look for a five element
set. (In fact there are six element sets which have exactly 10 distinct
pairwise sums. Can you find one?) Suppose the elements of the set are
a < b < c < d < e. Then the ten pairwise sums are {a+b, a+c, b+c, a+
d, b+d, a+e, b+e, c+d, c+e, d+e} but they are not necessarily in order
from smallest to largest. We can say that the smallest, a + b = 1967,
the largest d + e = 1991, and the sum of them all is a + b + c + d + e =
1
(1967 + 1972 + 1973 + 1974 + 1975 + 1980 + ... + 1991) = 4947, because
4
each of the numbers a, b, c, d, e appears in exactly four sums. It follows
that c = 989. Since a + c is the second largest in the list, it follows
that a = a + c − c = 1972 − 989 = 983 and from this it follows that
b = 1967 − 983 = 984. Reasoning similarly results in e = 1000 and
d = 991.
8. Use each of the nine digits 1, 2, 3, 4, 5, 6, 7, 8, and 9 exactly twice to
form distinct prime numbers whose sum is as small as possible.(USAMTS,
1991)
Solution. From M&IQ, No. 3, 1992. Clearly 4, 6, and 8 cannot occur
as the units digit. Similarly, 2 and 5 may occur as units digit at most
once. Thus the sum must be at least
(2(4 + 6 + 8) + (2 + 5))10 + (2 + 5) + 2(1 + 3 + 7 + 9) = 477.
To see that 477 can be attained, we first note that the only two primes
in the 80’s are 83 and 89, and that the only two in the 60’s are 61 and
67. This leads to the solutions
2 + 5 + 23 + 41 + 47 + +59 + 61 + 67 + 83 + 89 = 477
and
2 + 5 + 29 + 41 + 47 + 53 + 61 + 67 + 83 + 89 = 477.
Can you solve this problem when each digit must be used exactly
once?. . . exactly thrice?
9. How many numbers can be obtained as the product of two or more
of the numbers 3, 4, 4, 5, 5, 6, 7, 7, 7? (Alan Tucker’s Applied Combinatorics, problem 34, page 179.)
Solution. Take G as the multiset {3, 4, 4, 5, 5, 6, 7, 7, 7}, and P as the
process P1 with the modification that we must use at least two members
of G and we multiply instead of add. Note that each member n of
R uniquely determines the subset Sn of G whose product it is. We
claim that each product in R uniquely determines its factors among
the multiset. Factor the product n of members of G into primes to get
something of the form n = 2i 3j 5k 7l . The exponent i is odd if and only
if the 6 appears in the product. The number of 5’s and 7’s in Sn is just
k and l respectively and the number of 4’s is b 2i c. The number of 6’s
is i − 2b 2i c, and the number of 3’s is j minus the number of 6’s. Thus
the number of members of R is the number alternative ways to treat
the various values. We can include the 3 or not, include the 6 or not,
include 0, 1, or 2 of the 4’s, 0,1, or 2 of the 5’s, and 0,1,2, or 3 of the
7’s. This number is
2 · 2 · 3 · 3 · 4 − 1 − 5 = 138.
10. Brackets can be inserted into the expression 1 ÷ 2 ÷ 3 ÷ 4 in various
ways. For example, (1 ÷ 2) ÷ (3 ÷ 4) equals 2/3 whereas 1 ÷ ((2 ÷ 3) ÷ 4)
equals the whole number 6. Similarly, brackets can be inserted into
1 ÷ 2 ÷ 3 ÷ 4 ÷ 5 ÷ 6 ÷ 7 ÷ 8 ÷ 9 ÷ 10 ÷ 11 to produce a large collection
of answers, many of which are integers. What do you get if you divide
the largest of these integers by the smallest of these integers? (London
Times Brainteaser #1696).
Solution. This problem is related to the choices
G = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11}
and P = P5 , with the requirement that only ÷’s appear in the expression and the members of G appear in numerical order. The numbers
produced are precisely
1αβγ . . .
,
2abc . . .
where each of the numbers 1, 2, 3, 4, . . . , 11 appears exactly once. Another way to write these is as
11!
,
(2abc . . .)2
where a, b, c, . . . belong to {3, 4, . . . , 11}. The largest integer value is
11!
= 9979200. Since 11! = 11 · 7 · 5 · 5 · 3 · 3 · 3 · 3 · 2 · 2 · 2 · 2 · 2 · 2 · 2 · 2 =
22
11 · 7(2 · 5 · 8 · 9)2 , it follows that the smallest integer is
11!
= 77.
(2 · 5 · 8 · 9)2
Therefore the quotient in question is 9979200 ÷ 77 = 129600. In fact
1 ÷ (((((((((2 ÷ 3) ÷ 4) ÷ 5) ÷ 6) ÷ 7) ÷ 8) ÷ 9) ÷ 10) ÷ 11) =
3 · 4 · 5 · 6 · 7 · 8 · 9 · 10 · 11
= 9, 979, 200.
2
On the other hand,
((((1 ÷ 2) ÷ 3) ÷ 4) ÷ 5) ÷ (((((6 ÷ 7) ÷ 8) ÷ 9) ÷ 10) ÷ 11)
=
7 · 8 · 9 · 10 · 11
= 77.
2·3·4·5·6
11. How many of the first 100 positive integers are expressible as a sum of
three or fewer members of the set {30 , 31 , 32 , 33 , 34 } if we are allowed to
use the same power more than once. For example, 5 can be represented,
but 8 cannot. (1991 State Math Contest of North Carolina.)
Solution. The number of powers of 3 used is just the sum of the ternary
digits. It is useful therefore to consider the numbers from 1 to 26, 27
to 53, 54 to 80, and 81 to 100. Numbers in the range 1 to 26 have
ternary representation of the form (a2 a1 a0 )3 . How many of these satisfy
a2 + a1 + a0 ≤ 3? There are 16 such numbers. Those in the range 27 to
53 all have the form (1a2 a1 a0 )3 . There are 10 for which a2 +a1 +a0 ≤ 2.
The number in the range 54 to 80 have the form (2a2 a1 a0 )3 . Only 4
of these satisfy a2 + a1 + a0 ≤ 1. The numbers from 81 to 100 all
have the form (1a3 a2 a1 a0 )3 . We want to know how many of that form
are less than 100 and satisfy 1 + a3 + a2 + a1 + a0 ≤ 3. There are 10
numbers in this range which satisfy the conditions. Hence there are
16 + 10 + 4 + 10 = 40 such numbers altogether.
12. How many integers can be expressed as a sum of two or more different
members of the set {0, 1, 2, 4, 8, 16, 31}? (1994 UNC Charlotte Comprehensive Exam.)
Solutions. Take G as the set {1, 2, 4, 8, 16, 31}, and P as P1 .
• The 0 in the set means that we should count the individual elements among the sums. If the 31 was a 32, we’d have unique
(binary) representation. Hence the number of sums is just 26 minus 1(the empty sum) minus the duplicated values, of which there
is just one. Hence there are 64 − 2 = 62 members of R.
• Each of the numbers 1, 2, 3, 4, 5, . . . , 62 is achievable, and 62 is
certainly the largest member of R.
13. In a standard ‘infix’ calculator, unparenthesized expressions are evaluated as follows:
(a) Multiplications and divisions are done first, starting at the left
and moving to the right.
(b) Additions and subtractions are done, again moving from left to
right.
For example, 1 − 2 + 3 − 4 is −3 23 . How many different values are
possible if each ∗ is replaced by one of +, −, ×, or ÷ in the expression
1 ∗ 1 ∗ 1 ∗ 1 ∗ 1 ∗ 1 ∗ 1 ∗ 1? (1987 State Math Contest of North Carolina.)
Solution. Inserting a pluses, b minuses, c times and d divs results in an
expression with value 1 + a − b where a + b ≤ 7. Thus −7 ≤ a − b ≤ 7
implies that there are 15 different values.
14. The number 113 can be expressed as a sum of multiples of powers of
−3. In fact, there exist integers a0 , a1 , a2 , a3 , a4 , such that 0 ≤ ai ≤ 2
and 113 = a4 (−3)4 + a3 (−3)3 + a2 (−3)2 + a1 (−3)1 + a0 (−3)0 . Find
the value of a0 + a1 + a2 + a3 + a4 . (1990 State Math Contest of North
Carolina.)
Use the standard repeated division algorithm to find the base −3 representation of 113, being careful to obtain a remainder of 0, 1 or 2 to
get 113 = 22122−3 .
15. John has 2 pennies, 3 nickels, 2 dimes, 3 quarters, and 8 dollars. For
how many different amounts can John make an exact purchase?
Solution. We’ll count achievable amounts less than $1, and multiply by
9, then add in the 9 values 9.00, 9.01, 9.02, 9.05, 9.06, 9.07, 9.10, 9.11, 9.12.
They are {0, 1, 2, 5, 6, 7, 10, 11, 12, . . .}, exactly three of every five consecutive values. So, counting 0, there are 53 (100) = 60 such values.
Hence there are 9 × 60 − 1 = 539, since we don’t count the value 0.
Now adding in the nine uncounted values, we get 539 + 9 = 548.
16. Show that 1997 can be written in the form
±12 ± 22 ± 32 · · · ± k 2
for some positive integer k. Next show that every positive integer can
be written in this way. (Paul Erdös)
Hint: k 2 − (k + 1)2 − (k + 2)2 + (k + 3)2 = 4.