MAS105
SCHOOL OF MATHEMATICS AND STATISTICS
Autumn Semester
2010–2011
Numbers and Proofs (solutions)
2 hours
Attempt all the questions. The allocation of marks is shown in brackets.
{B} means bookwork, {S} means seen similar, {U} means unseen
MAS105
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Turn Over
MAS105
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(i)
(ii)
[S]
(a)
kappa and mu.[1]
(1 mark)
(b)
Q.[1]
(1 mark)
[S]
(a)
Contrapositive:
If a2 is not divisible by 216, then a is not divisible by 12.[1]
Converse:
If a2 is divisible by 216, then a is divisible by 12.[1]
(2 marks)
(iii)
(b)
The contrapositive is false[1]: take a = 12, then a2 = 144 is not
divisible by 216, but a is divisible by 12[1].
(2 marks)
(c)
The converse is true[1]. For a2 to be divisible by 216 = 23 33 , we
need a to be divisible by 22 [1] and by 32 [1]; in particular, a will be
divisible by 36 and so by 12[1].
(4 marks)
if it is not prime.[1]
Since 2010 = 2 × 3 × 5 × 67[1], there are 3 ways[1] of writing it as
the product of composite numbers:
[B/U]
n (> 1) is
composite
(2 × 3) × (5 × 67);
(2 × 5) × (3 × 67);
i.e., 6 × 335, 10 × 201, 15 × 134.
(iv)
(3 marks)
The highest common factor h of a and b is the integer h such that: h|a,
h|b (so h is a common factor) and if c|a, c|b, then c ≤ h (so h is at least as
[B]
big as any other common factor)[2].
(v)
(2 × 67) × (3 × 5),
(2 marks)
[B]
(a)
a and b are coprime if (a, b) = 1 (i.e., there is no common factor
bigger than 1)[1]. Equivalently, if there exist integers s and t with
as + bt = 1.[1]
(2 marks)
(b)
If (a, bc) = 1, then there are integers s and t with as + (bc)t = 1.
But then as + bt′ = 1 where t′ = ct, so a and b are coprime also.[2]
(2 marks)
(vi)
MAS105
We square 0, . . . , 7 and nd that the possible remainders are 0, 1, 4.[1]
There is no way to add two of 0, 1, 4 to make 6, so we can't add two squares
to get anything with remainder 6 after dividing by 8.[2]
(3 marks)
[S]
2
Continued
MAS105
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(i)
[S]
We use the Euclidean algorithm:
2010
1695
315
120
75
45
30
=
=
=
=
=
=
=
1
5
2
1
1
1
2
× 1695 + 315
× 315 + 120
× 120 + 75
×
75 + 45
×
45 + 30
×
30 + 15
×
15 +
0
so the hcf is 15.[4] Working backwards,
15 = 45 = 30 = 45 − (75 − 45) = 2 × 45 − 75 = 2(120 − 75) − 75
= 2 × 120 − 3(315 − 2 × 120) = 8 × (1695 − 5 × 315) − 3 × 315
= 8 × 1695 − 43(2010 − 1695) = 51 × 1695 − 43 × 2010.[4]
(8 marks)
(ii)
[S]
Let's solve 2010s + 1695t = 78000.[1] We therefore solve
2010s + 1695t = 5200 × 15 = 5200(51 × 1695 − 43 × 2010), [1]
or
2010(s + 5200 × 43) = 1695(5200 × 51 − t).[1]
Divide through by 15[1]:
134(s + 223600) = 113(265200 − t).
So 134|265200 − t, meaning 265200 − t = 134k (and s + 223600 = 113k ) for
some k , giving the general solution
(s, t) = (113k − 223600, 265200 − 134k).[2]
We need a solution with s and t non-negative. For s ≥ 0, we need k ≥ 1979
and for t ≥ 0, we need k ≤ 1979. Thus k = 1979, and (s, t) = (27, 14).[2]
(8 marks)
(iii)
As (2010, 1695) = 15, and 15 - 85, the second has no solution.[1] To
solve 1695x ≡ 75 (mod 2010), we write it as
[S]
1695x = 2010y + 75.
Using the earlier result, we need
1695x − 2010y = 5 × 15 = 5(51 × 1695 − 43 × 2010).[1]
Then
and
1695(x − 255) = 2010(y − 215),
113(x − 255) = 134(y − 215).[2]
So 134|x − 255, so the solution is x = 255 + 134k , i.e., x ≡ 255 (mod 134),
or x ≡ 121 (mod 134).[2]
(6 marks)
MAS105
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Turn Over
MAS105
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(i)
(a)
ϕ(n) is the number of integers between 1 and n which are coprime
to n[1]. We have ϕ(209) = ϕ(11)ϕ(19) = 10 × 18 = 180[1].
[S]
(2 marks)
(b)
The Fermat-Euler Theorem states that if m ≥ 2 is a natural
number, and (a, m) = 1, then aϕ(m) ≡ 1 (mod m).[3]
[B]
When m = p, a prime, we have ϕ(p) = p − 1, and the FermatEuler theorem would become: if p is prime, and p - a, then ap−1 ≡
1 (mod p), which is exactly Fermat's Little Theorem.[2]
(5 marks)
(c)
(ii)
Let's compute the LHS mod 11 and mod 19.[1] Modulo 11, we
know 510 ≡ 1 (mod 11), so 52011 ≡ 5 (mod 11).[2] Mod 19, we know
518 ≡ 1 (mod 19), so 52011 ≡ 513 (mod 19), and it is easy to calculate
513 ≡ 17 (mod 19). So 52011 ≡ 17 (mod 19).[2] Combining these, we
get 52011 ≡ 93 (mod 209).[1]
(6 marks)
[S]
When n = 4, the assertion is that 24 > 16, which is
[S] Initial step.
true.[1]
Suppose that the assertion is true for some natural
number n = k ≥ 4. Then
Induction step.
(k + 1)! = (k + 1).k! > (k + 1).2k > 2.2k = 2k+1 , [3]
and so the assertion also holds for n = k + 1.
(iii)
k+1
k
(4 marks)
k
23 = 23.3 = (23 )3 = (3k u − 1)3 = (33k u3 − 3.32k u2 + 3.3k u) − 1 =
k+1
3 v − 1, where v = 32k−1 u3 − 3k u2 + u.[2]
[S/U]
For k = 1, 23 = 8 = 31 u − 1 where u = 3.[1] Now we see by
induction that for all n = 3k , we have 2n = nu − 1 for some u (depending
on n) using the rst part, and so n|2n + 1 for n = 3k .[2]
(5 marks)
1
MAS105
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Continued
MAS105
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√
(i)
Suppose not. Then we can write 12 = m/n for coprime integers m
and n.[1] That is, m2 = 12n2 , so 3|m2 , and so 3|m.[1] Write m = 3r, and
substitute in; 9r2 = 12n2 , i.e., 3r2 = 4n2 , so n2 and therefore n are divisible
by 3.[1] This contradicts the coprimality of m and n.[1]
(4 marks)
(ii)
(a)
[B]
(b)
[S]
[S]
When we divide m.000 . . . by n, the remainders at each step must
be between 0 and n − 1, so there are only nitely many possibilities.
As soon as we get two remainders which are the same, the process
repeats.[3]
(3 marks)
Put x = 0.10̇5̇; then 10x = 1.0̇5̇ and 1000x = 105.0̇5̇.[1] Sub-
tracting, we have 990x = 104, so x =
(c)
104
52
=
.[1]
990
495
(2 marks)
The decimal expansion is got by dividing 1.000 . . . by d; to have
period 5 we need that the remainder on dividing 1.00000 by d is the
same as the remainder on dividing 1 by d; i.e., that 105 ≡ 1 (mod d),
or that d|99999.[1] The divisors of 99999 are: 1, 3 and 9 (for these
the period of 1/d is 1) and 41, 123 = 3 × 41, 369 = 32 × 41; 271,
813 = 3×271, 2439 = 32 ×271; 11111 = 41×271, 33333 = 3×11111,
and 99999 itself. So the list is
[U]
41, 123, 271, 369, 813, 2439, 11111, 33333, 99999.[2]
(3 marks)
(iii)
We need f (a) = f ′ (a) = 0. Equivalently, (x − a) is a common factor of
f (x) and f ′ (x).[1] For the given polynomial, f ′ (x) = 4x3 + 9x2 + 2x − 3.
We observe f (1) = f (−1) = 0 and f ′ (−1) = 0, so −1 is a repeated root of
the given polynomial[2], and 1 is a simple root.[1] As the constant term is
the product of the roots, we see −2 is the other root, so
[S]
x4 + 3x3 + x2 − 3x − 2 = (x + 1)2 (x − 1)(x + 2).[1]
(5 marks)
(iv)
The division algorithm: Let f and g ∈ R[x], with g non-zero. Then
there are polynomials q and r in R[x] with f = qg + r and where r = 0 or
has degree strictly less than the degree of g .[2]
So
[S]
x4 + 3x3 + x2 − 3x − 2 = q(x)(x2 − 2x) + (ax + b), [1]
as the remainder is linear. Put x = 0 to get −2 = q(0).0 + b, so b = −2.[1]
Then put x = 2 to nd 36 = q(2).0 + (2a + b), so a = 19 and the remainder
is 19x − 2.[1]
(5 marks)
End of Question Paper
MAS105
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