Download The digits 1,2,3,4,5,6 are written down in some order to form a six

HOMEWORK ASSIGNMENT 7
ACCELERATED PROOFS AND PROBLEM SOLVING [MATH08071]
Each problem will be marked out of 4 points.
Exercise 1 ([1, Exercise 16.6]).
The digits 1, 2, 3, 4, 5, 6 are written down in some order to form a six-digit number.
(a) How many such six-digit numbers are there altogether?
(b) How many such six-digit numbers are even?
(c) How many are divisible by 4.
(d) How many are divisible by 8. (Hint: First show that the remainder on dividing a
six-digit number abcdef is 4d + 2e + f .)
Solution. Let a1 a2 a3 a4 a5 a6 be a six-digit number such that
ai ∈ 1, 2, 3, 4, 5, 6
for every i ∈ {1, 2, 3, 4, 5, 6}. We assume that the six-digit numbers we get do not have
repeating digits (this is what “in some order” means). Then the digits a1 , a2 , a3 , a4 , a5 , a6
must be all different.
It follows from [1, Proposition 16.1] that there are 6! = 720 such six-digit numbers
a1 a2 a3 a4 a5 a6 altogether, because this is the number of rearrangements of the digits
1, 2, 3, 4, 5, 6.
There are 3 × 5! such six-digit numbers a1 a2 a3 a4 a5 a6 that are even. Indeed, once we
fix the last digit a6 , we have 5! such numbers a1 a2 a3 a4 a5 a6 . But the last digit a6 must be
even if the six-digit number is even. So we have exactly 3 possibilities for the last digit
(being 2, 4, or 6). This gives us 3 × 5! = 360 possible combinations for a1 a2 a3 a4 a5 a6 to
be even.
The number a1 a2 a3 a4 a5 a6 is divisible by 4 if and only if a5 a6 is divisible by 4. But
there are exactly 8 choices for a5 a6 to be divisible by 4, which are given by the integers
12, 16, 24, 32, 36, 52, 56, 64. Once we fix a5 a6 , we have exactly 4! possibilities for a1 a2 a3 a4 .
Thus, there are 8 × 4! = 192 such integers that are divisible by 4.
The number a1 a2 a3 a4 a5 a6 is divisible by 8 if and only if a4 a5 a6 is divisible by 8. But
a4 a5 a6 = a4 × 100 + a5 × 10 + a6 ≡ 4a4 + 2a5 + a6 mod 8,
since 100 ≡ 4 mod 8 and 10 ≡ 2 mod 8. Thus, we see that a4 a5 a6 is divisible by 8 if and
only if 4a4 + 2a5 + a6 is divisible by 8. Checking all possible combinations, we see that
there are exactly 14 possibilities for a4 a5 a6 to be divisible by 8, which are given by the
numbers
136, 152, 216, 256, 264, 312, 352, 416, 432, 456, 512, 536, 624, 632,
which implies that there are 14 × 3! = 84 such integers that are divisible by 8, because
there are 3! possibilities for a1 a2 a3 once we fix a4 a5 a6 .
Exercise 2 ([1, Exercise 17.2]). Which of the following statements are true and which are
false? Give proofs or counterexamples.
(a) For any sets A, B, C, we have
A ∪ (B ∩ C) = (A ∪ B) ∩ (A ∪ C).
This assignment is due on Thursday 12ve November 2015.
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(b) For any sets A, B, C, we have
(A − B) − C = A − (B − C).
(c) For any sets A, B, C, we have
(A − B) ∪ (B − C) ∪ (C − A) = A ∪ B ∪ C.
Solution. Let A, B, C be any sets. Let us prove that the statement
A ∪ (B ∩ C) = (A ∪ B) ∩ (A ∪ C)
is true. Let x be an element in A ∪ (B ∩ C). Then either x ∈ A or x ∈ (B ∩ C). If x ∈ A,
then x ∈ A ∪ B and x ∈ A ∪ C, which implies that
x ∈ (A ∪ B) ∩ (A ∪ C),
since x ∈ A ∪ B and x ∈ A ∪ C. If x ∈ B ∩ C, then x ∈ B and x ∈ C, which implies that
x ∈ A ∪ B and x ∈ A ∪ C, which implies (again) that x ∈ (A ∪ B) ∩ (A ∪ C). Thus, we see
that
A ∪ (B ∩ C) ⊆ (A ∪ B) ∩ (A ∪ C),
since we proved that x ∈ (A ∪ B) ∩ (A ∪ C) for any x ∈ A ∪ (B ∩ C). Let y be an element
in (A ∪ B) ∩ (A ∪ C). Then y ∈ A ∪ B and y ∈ A ∪ C. If y ∈ A, then
y ∈ A ∪ (B ∩ C)
obviously. If y 6∈ A, then y ∈ B and y ∈ C, since we know that y ∈ A ∪ B and y ∈ A ∪ C.
Thus, if y 6∈ A, then y ∈ B ∩ C, which implies that y ∈ A ∪ (B ∩ C). Therefore, we proved
that
(A ∪ B) ∩ (A ∪ C) ⊆ A ∪ (B ∩ C),
since we proved that y ∈ A ∪ (B ∩ C) for every y in (A ∪ B) ∩ (A ∪ C). Hence, we see that
(A ∪ B) ∩ (A ∪ C) = A ∪ (B ∩ C).
The statements (A − B) − C = A − (B − C) and (A − B) ∪ (B − C) ∪ (C − A) = A ∪ B ∪ C
are both wrong in general. To see this, put A = B = C = Z. Then
(A − B) − C = ∅ − C = ∅ 6= Z = A = A − ∅ = A − (B − C),
since A − B = Z − Z = ∅ and B − C = Z − Z = ∅. Similarly, we see that
(A − B) ∪ (B − C) ∪ (C − A) = ∅ ∪ ∅ ∪ ∅ = ∅ 6= Z = Z ∪ Z ∪ Z = A ∪ B ∪ C.
Exercise 3 ([1, Exercise 17.4]). Prove that if A, B, C are finite sets, then
|A ∪ B ∪ C| = |A| + |B| + |C| − |A ∩ B| − |A ∩ C| − |B ∩ C| + |A ∩ B ∩ C|.
Solution. Let A, B, C be any finite sets. Applying [1, Proposition 17.2] to (A ∪ B) ∪ C,
we see that
|A ∪ B ∪ C| = |A ∪ B| + |C| − |(A ∪ B) ∩ C| = |A ∪ B| + |C| − |(A ∩ C) ∪ (B ∩ C)|,
because (A∪B)∩C = (A∩C)∪(∩C) by [1, Proposition 17.1]. Applying [1, Proposition 17.2]
to (A ∩ C) ∪ (B ∩ C), we see
|(A ∩ C) ∪ (B ∩ C)| = |A ∩ C| + |B ∩ C| − |(A ∩ C) ∩ (B ∩ C)|,
where (A ∩ C) ∩ (B ∩ C) = A ∩ B ∩ C obviously. Thus, we see that
|(A ∩ C) ∪ (B ∩ C)| = |A ∩ C| + |B ∩ C| − |A ∩ B ∩ C|,
and we can plug in this expression for |(A ∩ C) ∪ (B ∩ C)| in the first equality we got
above. This gives us
|A∪B∪C| = |A∪B|+|C|− |A∩C|+|B∩C|−|A∩B∩C| = |A∪B|+|C|−|A∩C|−|B∩C|+|A∩B∩C|,
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where |A ∪ B| = |A| + |B| − |A ∩ B| by [1, Proposition 17.2] (again). Thus, we finally have
|A ∪ B ∪ C| = |A| + |B| − |A ∩ B| + |C| − |A ∩ C| − |B ∩ C| + |A ∩ B ∩ C|,
which implies that |A ∪ B ∪ C| = |A| + |B| + |C| − |A ∩ B| − |A ∩ C| − |B ∩ C| + |A ∩ B ∩ C|.
Exercise 4 ([1, Exercise 17.6]). How many integers between 1 and 10000 are neither
squares nor cubes?
Solution. For every positive integer i, put
n
o
Ai = n ∈ Z 10000 > n > 1, n = mi for some positive m ∈ Z .
Then |A2 | is the number of integers between 1 and 10000 that are squares, and |A3 | is the
number of integers between 1 and 10000 that are cubes. Then |A2 ∪ A3 | is the number of
integers between 1 and 10000 that are either squares or cubes. Then
10000 − |A2 ∪ A3 |
is the number of integers between 1 and 10000 that are neither squares nor cubes. How
to find |A2 ∪ A3 |? We have
|A2 ∪ A3 | = |A2 | + |A3 | − |A2 ∩ A3 |
by [1, Proposition 17.1]. How to find |A2 | and |A3 |? We have
j√
k
|A2 | =
10000 = 100,
where we denote by bxc the biggest integer that does not exceed x ∈ R. Similarly, we see
that
j√
k
3
|A3 | =
10000 = 21,
√
since 3 10000 ≈ 21.544. How to find |A2 ∩ A3 |? It follows from [1, Excercise 11.8(a)] that
|A2 ∩ A3 | = |A6 |,
where |A6 | is the number of integers between 1 and 10000 that are sixth powers. Then
j√
k
6
|A6 | =
10000 = 4,
√
since 3 10000 ≈ 4.642. Thus, we see that
|A2 ∪ A3 | = |A2 | + |A3 | − |A2 ∩ A3 | = 100 + 21 − 4 = 117,
which implies that there are 10000 − |A2 ∪ A3 | = 9883 integers between 1 and 10000 that
are neither squares nor cubes.
Exercise 5 ([1, Exercise 17.8]). (a) Find the number of integers between 1 and 5000 that
are divisible by neither 3 nor 4.
(b) Find the number of integers between 1 and 5000 that are divisible by none of the
numbers 3, 4 and 5.
(c) Find the number of integers between 1 and 5000 that are divisible by one or more of
the numbers 4, 5 and 6.
Solution. For every positive integer i, put
n
o
Ai = n ∈ Z 5000 > n > 1, i divides n .
Then |Ai | is the number of integers between 1 and 5000 that are divisible by i. Then
$
%
5000
|Ai | =
i
3
for every positive integer i.
Let us first find the number of integers between 1 and 5000 that are divisible by neither
3 nor 4. Since |A3 | is the number integers between 1 and 5000 that are divisible by 3,
and |A4 | is the number integers between 1 and 5000 that are divisible by 4, we see that
|A3 ∪ A4 | is the number of integers between 1 and 5000 that are divisible by either 3 or 4.
Then
5000 − |A3 ∪ A4 |
is the number of integers between 1 and 5000 that are divisible by neither 3 nor 4. But
|A3 ∪ A4 | = |A3 | + |A4 | − |A3 ∩ A4 |
by [1, Proposition 17.1]. Then
$
% $
%
5000
5000
|A3 ∪ A4 | =
+
− |A3 ∩ A4 | = 1666 + 1250 − |A3 ∩ A4 | = 2916 − |A3 ∩ A4 |,
3
4
and we must find |A3 ∩ A4 | now. But
|A3 ∩ A4 | = |A12 |
because a positive integer is divisible by 3 and 4 if and only if it is divisible by 12. Then
$
%
5000
|A3 ∪ A4 | = 2916 − |A3 ∩ A4 | = 2916 −
= 2916 − 416 = 2500,
12
which implies that there are 5000 − 2500 = 2500 integers between 1 and 5000 that are
divisible by neither 3 nor 4.
Now let us find find the number of integers between 1 and 5000 that are divisible by
none of the numbers 3, 4 and 5. This number is
5000 − |A3 ∪ A4 ∪ A5 |,
since |A3 ∪ A4 ∪ A5 | is the number of integers between 1 and 5000 that are divisible by at
least one number among 3, 4 and 5. But
|A3 ∪ A4 ∪ A5 | = |A3 | + |A4 | + |A5 | − |A3 ∩ A4 | − |A3 ∩ A5 | − |A4 ∩ A5 | + |A3 ∩ A4 ∩ A5 |
by Exercise 3. We already know that |A3 | = 1666, |A4 | = 1250, and |A3 ∩ A4 | = 416.
Similarly, we see that
%
$
%
$
%
$
5000
5000
5000
= 1000, |A3 ∩A5 | = |A15 | =
= 333, |A4 ∩A5 | = |A20 | =
= 250,
|A5 | =
5
15
20
and |A3 ∩ A4 ∩ A5 | = |A60 | = 83. Thus, we have
|A3 ∪ A4 ∪ A5 | = 1666 + 1250 + 1000 − 416 − 333 − 250 + 83 = 3000,
which implies that there are 5000 − 3000 = 2000 integers between 1 and 5000 that are
divisible by none of the numbers 3, 4 and 5.
Finally, let us find the number of integers between 1 and 5000 that are divisible by one
or more of the numbers 4, 5 and 6. This number is |A4 ∪ A5 ∪ A6 |. We have
|A4 ∪ A5 ∪ A6 | = |A4 | + |A5 | + |A6 | − |A4 ∩ A5 | − |A4 ∩ A6 | − |A5 ∩ A6 | + |A4 ∩ A5 ∩ A6 |,
by [1, Proposition 17.1]. We know that |A4 | = 1250, |A5 | = 1000, and |A4 ∩ A5 | = 250.
Similarly, we have |A6 | = 833. Similarly, we see that
$
%
5000
|A5 ∩ A6 | = |A30 | =
= 166,
30
because a positive integer is divisible by 5 and 6 if and only if it is divisible by 30. And
|A4 ∩ A6 | = |A12 | = 416
4
because a positive integer is divisible by 4 and 6 if and only if it is divisible by 12. Finally,
we have
$
%
5000
|A4 ∩ A5 ∩ A6 | = |A60 | =
= 83
60
because a positive integer is divisible by 4, 5 and 6 if and only if it is divisible by 60. Thus,
there are
|A4 ∪ A5 ∪ A6 | = 1250 + 1000 + 833 − 250 − 416 − 166 + 83 = 2334
integers between 1 and 5000 that are divisible by one or more of the numbers 4, 5 and 6.
References
[1] M. Liebeck, A concise introduction to pure mathematics
Third edition (2010), CRC Press
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