Download MATH10040: Numbers and Functions Homework 4: Solutions

MATH10040: Numbers and Functions
Homework 4: Solutions
1. (a) Find s, t ∈ Z satisfying 323s + 143t = 1.
(b) Find s < 0 and t > 0 satisfying 323s + 143t = 1.
(c) Find s, t ∈ Z satisfying 323s + 143t = 43.
Solution:
(a) We have
323
143
37
32
5
=
=
=
=
=
2 · 143 + 37
3 · 37 + 32
32 + 5
6·5+2
2 · 2 + 1.
Thus
1=5−2·2
= 13 · 5 − 2 · 32
= 13 · 37 − 15 · 32
= 58 · 37 − 15 · 143
=
=
=
=
=
5 − 2 · (32 − 6 · 5)
13 · (37 − 32) − 2 · 32
13 · 37 − 15 · (143 − 3 · 37)
58 · (323 − 2 · 143) − 15 · 143
58 · 323 − 131 · 143.
So we can take s = 58 and t = −131.
(b) By the solution to Problem 9 on Homework 3, for any m ∈ Z,
S = 58 − 143m and T = 323m − 131 will also work. Taking
m = 1, gives S = −85 and T = 192.
(c) We multiply both sides of the identity
58 · 323 − 131 · 143 = 1
by 43, to get
(43 · 58) · 323 + (−131 · 43) · 143 = 43.
2. (a) How many solutions are there to the equation
x + y + z + w = 10
where x, y, z, w are non-negative integers? [Hint: This is equivalent to one of the problems in homework 3.]
(b) How many solutions are there to the equation
x + y + z + w + u = 12
where x, y, z, u, w are non-negative integers?
(c) How many solutions are there to the equation
x + y + z + w + u = 12
where x, y, z, u, w are positive integers?
Solution:
(a) This is just a re-phrasing of Problem 5 on Homework 3 (a solution
(x, y, z, w) gives
us a distribution of sugar lumps). Thus the
answer is 13
=
286.
3
(b) Same
problem with 10 → 12 and 3 → 4. So the answer is
16
= 1820.
4
(c) Let X = x − 1, Y = y − 1, Z = z − 1, W = w − 1 and U = u − 1.
Then
x + y + z + w + u = 12 ⇐⇒ X + Y + Z + W + U = 12 − 5 = 7
and x, y, z, w, u are positive if and only if X, Y, Z, W, U are nonnegative.
So we just need to count the number of nonnegative solutions of
X + Y + Z + W + U = 7,
which is 7+4
= 11
= 330 by the method of parts (a) and (b).
4
4
3. Suppose that a, b ∈ Z \ {0} with (a, b) = 1. Prove that (an , bm ) = 1
for any n, m ≥ 1.
Solution: We will prove the equivalent statement that if (an , bm ) > 1,
then (a, b) > 1 for any n, m.
So suppose (an , bm ) > 1. Then there is a prime p dividing an and bm .
Since p is prime, p|an =⇒ p|a and p|bm =⇒ p|b. Then p is a common
divisor of a and b and thus (a, b) > 1.
4. Let a, b ∈ Z and let g = (a, b). Let ` = ab/g.
(a) Show that a|` and b|`.
?
(b) Show that if m ∈ Z and if a|m and b|m then `|m. (` is called
the least common multiple of a and b.)
Solution:
(a) Since g|a, a = a0 g for some a0 ∈ Z. Similarly b = b0 g for some
b0 ∈ Z. Thus
(a0 g)(b0 g)
`=
= a0 b0 g = ab0 =⇒ a|`
g
and
` = a0 b0 g = ba0 =⇒ b|`.
(b) We have (a0 , b0 ) = 1, by a result from class, and ` = a0 b0 g.
Suppose that a|m and b|m. Then m = as = a0 gs and m = bt =
b0 gt for some s, t ∈ Z. It follows (cancelling g) that a0 s = b0 t. So
a0 |b0 t. Since (a0 , b0 ) = 1 it follows (from a key Proposition from
class) that a0 |t. Thus t = a0 r for some r ∈ Z and hence
m = b0 gt = b0 ga0 r = `r =⇒ `|m.
5. Prove that every integer of the form 8n + 1 (n ≥ 1) is composite.
Solution: Recall
x3 + 1 = (x + 1)(x2 − x + 1).
So
8n + 1 = (2n )3 + 1 = (2n + 1)(22n − 2n + 1) for all n ≥ 1
and thus 2n + 1 divides 8n + 1 for all n ≥ 1, so that these numbers
can never be prime.
6. Show that (7n + 16, 3n + 7) = 1 for every n ≥ 1.
Solution: We apply the method of Euclid’s algorithm to the pair of
numbers, in order to find their greatest common divisor:
7n + 16 = 2 · (3n + 7) + (n + 2)
3n + 7 = 3 · (n + 2) + 1.
So (7n + 16, 3n + 7) = (3n + 7, n + 2) = (n + 2, 1) = 1 for all n ≥ 1.
7. Find the remainder of 355 + 553 on division by 7.
Solution: We have
27 = 33 ≡ −1
(mod 7) =⇒ 36 = (33 )2 ≡ (−1)2 ≡ 1
(mod 7).
Thus
355 ≡ 31
(mod 7)
since
55 ≡ 1
(mod 6).
Also
(mod 7) =⇒ 553 ≡ (−1)3 ≡ −1
55 ≡ −1
(mod 7).
So
355 + 553 ≡ 3 + (−1) ≡ 2
(mod 7).
The answer is: 2
8. Show that 999912 leaves remainder 1 on division by 28.
Solution: Since 28 = 4 · 7 and (4, 7) = 1, it’s enough to show that
999912 ≡ 1 (mod 4) and 999912 ≡ 1 (mod 7).
Now,
9999 ≡ −1
(mod 4) =⇒ 999912 ≡ (−1)12 ≡ 1
(mod 4),
while
9999 ≡ 3
(mod 7)
and 33 ≡ −1
(mod 7)
gives
999912 ≡ 312 ≡ (−1)4 ≡ 1
(mod 7).
9. What are the last two digits of 357358 ?
Solution: Let A = 357358 . The question amounts to asking for the
remainder of A on division by 100; i.e. we must find the unique number
r with 0 ≤ r < 100 and A ≡ r (mod 100).
Now 100 = 22 · 52 = 4 · 25, so we first calculate A (mod 4) and A
(mod 25):
357 ≡ 1
(mod 4) =⇒ A = 357358 ≡ 1358 ≡ 1
(mod 4).
On the other hand, since
357 ≡ 7
(mod 25)
and 72 ≡ −1
(mod 25)
we get
357358 ≡ 7358 ≡ (−1)179 ≡ −1
(mod 25).
So A ≡ 1 (mod 4) and A ≡ −1 (mod 25). Thus r is the unique
number smaller than 100 satisfying r ≡ 1 (mod 4) and r ≡ −1
(mod 25). This latter congruence means that r is of the form 25t − 1.
So r = 24, 49, 74 or 99. 49 is the only number in this list congruent
to 1 (mod 4).
So r = 49 and the last two digits of A are 49.