Homework 5 M328K
by Mark Lindberg /Marie-Amélie Lawn
Problem 1:
Part 1: 3.5 #6 (⇒) If all of the powers in the prime-power factorization of an integer n are even, then
2km
1 2k2
they can be written in the form n = p2k
1 p2 · · · pm , where p1 , p2 , . . . , pm are the
prime factors, 2k1 , 2k2 , . . . , 2km are the powers, and we know that k1 , k2 , . . . , km
are integers by the definition of even. Then by rules of powers,
k1 k2
km 2
2km
1 2k2
n = p2k
1 p2 · · · pm = (p1 p2 · · · pm ) .
Moreover, we know that the product of primes in an integer, due to the closure of
integers under addition. Thus, n is equal to the square of some integer, and so, by
the definition of a perfect square, n is a perfect square.
(⇐) If n is a perfect square, then there is an integer a such that a2 = n. We can take
the prime-power factorization of a, so a = pk11 pk22 · · · pkmm , and so we get that
2km
1 2k2
n = a2 = (pk11 pk22 · · · pkmm )2 = p2k
1 p2 · · · pm
and so we see that, by the definition of even, in the prime-power factorization of
n, the powers of all of the prime factors are even.
√
Part 2 Proof by Contradiction: Let p be a prime number, and assume that n p is rational.
√
Then there exists non zero integers a and b such that (a, b) = 1 and n p = ab . That is,
√
there is a rational expression for n p in reduced form.
Now, we raise both sides to the nth power, canceling the root to get
a n a n
√
n n
p =p=
= n.
b
b
Multiplying both sides by bn , we get bn p = an .
Then, by the definition of divisors, p | an . Because p is prime, a previous theorem tells
us that if p | an , then p | a and by the definition of divisors, there exists an integer k
such that a = kp.
Then
an = (kp)n = k n pn = bn p.
Now since n ≥ 2, we have that bn = k n pn−1 , and because n ≥ 2, n − 1 ≥ 1, and so
we have at least one factor of p, so we can say that p | bn . By the same reasoning as
above, this gives that p | b. But we assumed that (a, b) = 1, and if p | a and p | b, then
(a, b) ≥ p > 1, which leads to a contradiction. Thus, it must be the case that for all
√
integers n ≥ 2, n p is irrational.
Part 3-1: 3.7 #2
a) 3x + 4y = 7
Using the Euclidean Algorithm we first find gcd(3, 4) and a linear combination
that gives the gcd:
4=3∗1+1
3=1∗3
Hence (3, 4) = 1 = (−1)3 + (1)4, for example.
1
Multiplying this equation by 7, this yields (−7)3 + (7)4 = 7
Subtracting from 3x + 4y = 7, we get
3(x + 7) + 4(y − 7) = 0.
Then 3(x + 7) = −4(y − 7), which gives that −4 | 3(x + 7), and because (4, 3) = 1,
−4 | (x + 7). This means that there is an integer c such that x + 7 = −4c, or
x = −4c − 7. Similarly, 3 | −4(y − 7) ⇒ 3 | (y − 7), so y − 7 = 3c, and y = 3c + 7.
Thus, our solution set is S = {(−4c − 7, 3c + 7), c ∈ Z}.
b) 12x + 18y = 50
For this problem we have
18 = 12 ∗ 1 + 6
12 = 6 ∗ 2 + 0
(12, 18) = 6 but 50 is not a multiple of 6, so there are no solutions.
c) 30x + 47y = −11
We proceed similarly to the previous parts:
47 = 30 ∗ 1 + 17
30 = 17 ∗ 1 + 13
17 = 13 ∗ 1 + 4
13 = 4 ∗ 3 + 1
4=1∗4+1
Hence (30, 47) = 1 = (11)30 + (−7)47, for example.
Then (−121)(30) + (77)(47) = −11
30(x + 121) + 47(y − 77) = 0
30(x + 121) = −47(y − 77)
−47 | 30(x + 121)
−47 | x + 121
x + 121 = −47c
x = −47c − 121
30 | −47(y − 77)
30 | y − 77
y − 77 = 30c
y = 30c + 77
S = {(−47c − 121, 30c + 77), c ∈ Z}
d) 25x + 95y = 970
95 = 25 ∗ 3 + 20
25 = 20 ∗ 1 + 5
20 = 5 ∗ 4 + 0
(25, 95) = 5 = 4 ∗ 25 − 95
970 = 776 ∗ 25 − 194 ∗ 95
25(x − 776) + 95(y + 194) = 0
25(x − 776) = −95(y + 194)
−95 | 25(x − 776)
−19 | 5(x − 776)
−19 | x − 776
−19c = x − 776
x = 776 − 19c
2
25 | 95(y + 194)
5 | 19(y + 194)
5 | y + 194
y + 194 = 5c
y = 5c − 194
S = {(776 − 19c, 5c − 194), c ∈ Z}
e) 102x + 1001y = 1
1001 = 102 ∗ 9 + 83
102 = 83 ∗ 1 + 19
83 = 19 ∗ 4 + 7
19 = 7 ∗ 2 + 5
7=5∗1+2
5=2∗2+1
2=1∗2+0
(102, 1001) = 1 = 422 ∗ 102 − 43 ∗ 1001
102(x − 422) + 1001(y + 43) = 0
102(x − 422) = −1001(y + 43)
−1001 | 102(x − 422)
−1001 | x − 422
x − 422 = −1001c
x = 422 − 1001c
102 | −1001(y + 43)
102 | y + 43
y + 43 = 102c
y = 102c − 43
S = {(422 − 1001c, 102c − 43), c ∈ Z}
Part 3-2: 3.7 #6 We have x plantains in each of the 63 piles, plus the seven extra, and this must equal
some integer y times the 23 travelers. In mathematical notation, 63x + 7 = 23y ⇒
63x − 23y = −7. This is a linear Diophantine equation. We can absorb the negative
sign into the z = −x, meaning that 63z + 23y = 7. Then:
63 = 2 ∗ 23 + 17
23 = 1 ∗ 17 + 6
17 = 2 ∗ 6 + 5
6=1∗5+1
5=5∗1+0
(63, 23) = 1 = −4 ∗ 63 + 11 ∗ 23
−28 ∗ 63 + 77 ∗ 23 = 7
(z0 , y0 ) = (−28, 77)
Then, with x = −z, we have (28, 77) as a solution, so there were 28 plantains in each
pile, and each traveler got 77.
Part 4-1 For $3, we have the setup of
x + 10y + 25z = 300,
where x is the number of cents (commonly called pennies), y is the number of dimes,
and z is the number of quarters, with the restriction that x + y + z = 50. This can be
3
rewritten as
x = 50 − y − z,
and substituting gives
px + 10y + 25z = 50 − y − z + 10y + 25z = 50 + 9y + 24z = 300.
Hence
9y + 24z = 250,
and we want y, z ≥ 0, y + z ≤ 50 to have a realistic solution. Then:
24 = 2 ∗ 9 + 6
9=1∗6+3
6=2∗3+0
(24, 9) = 3 = 9 ∗ 3 + 24(−1)
But 3 - 250, so there are no integers solutions.
For $3, we have an almost identical setup, with
x + 10y + 25z = 200 ⇒ 9y + 24z = 150.
But then 3 ∗ 50 = 150, so we have that 9 ∗ 150 + 24(−50) = 150. Then our solution set,
remembering that x = 50 − y − z is {(−50 + 5c, 150 − 8c, −50 + 3c), c ∈ Z}. Noticing that
we have to have c > 16 so that −50 + 3c ≥ 0, and that c < 19, so that 150 − 8c > 0, we
have solutions for c = 17, 18. Namely, we can have 35 cents, 14 dimes, and 1 quarter
or 40 cents, 6 dimes, and 4 quarters.
Problem 2:
P
P
j
Part 1 Proof by contradiction. P (x) = ni=0 ai xi and Q(x) = m
j=0 bj x . Let P (x)Q(x) =
Pm+n
k
k=0 ck x and assume that gcd(P Q) > 1, hence that there exists a prime factor p
which divides gcd(P Q) (and therefore all coefficients of PQ).
Since gcd(P ) = gcd(Q) = 1, there exists i0 and j0 , such that
For all i < i0 ,
p|ai but p - ai0
For all j < j0 ,
p|bj but p - bj0 .
Now by assumption, we have that
X
p|ci0 +j0 =
i+j=i0 +j0
p|
P
i+j=i0 +j0
i<i0 or j<j0
X
ai bj = ai0 bj0 +
ai bj
i+j=i0 +j0
i<i0 or j<j0
ai bj since in the sum i < i0 or j < j0 , hence p | ai0 bj0 and this is a
contradiction, since this means that either p|ai0 or p|bj0 .
Part 2 We use the previous result. Let P 0 =
Consequently, by part 1), we have
P
gcd(P ) ,
1 = gcd(P 0 ) gcd(Q0 ) = gcd(P 0 Q0 ) = gcd
which proves the statement.
4
Q0 =
Q
gcd(Q) .
Then gcd(P 0 ) = gcd(Q0 ) = 1.
PQ
gcd(P Q)
=
,
gcd(P ) gcd(Q)
gcd(P ) gcd(Q)
Problem 3:
Part 1 Because n! = n(n − 1)(n − 2) · · · 3 · 2 · 1, the product of all positive integers less than
or equal to n, we have that for all primes p ≤ n, p | n!. By a theorem shown in class,
there must be some prime p1 such that p1 | n! − 1, because for n ≥ 2, n! > 1. However,
if p1 | n! − 1, then p1 - n! − 1 + 1 = n!.1 Thus, p1 can not be any of the primes p
that are less than or equal to n, meaning that p1 > n. Since p1 | n! − 1, we have that
p1 ≤ n! − 1 < n, so combining, we have shown that there must be a prime p1 such
that n < p1 < n!, for all n ≥ 2. (Which is a little tighter than the requirements of the
problem, actually.)
Part 2 Proof by contradiction. Assume that there is a greatest prime, pm . Then look at pm .
By part 1, we know that there is a prime pn such that pm < pn < pm !, and so pn > pm .
But this is a contradiction to the assumption that pm is the greatest prime. Thus, there
is no greatest prime, and there are actually infinitely many prime numbers.
Problem 4:
Part 1
a) 3x + 6y + 5z = 7
We first notice that 6y + 5z is a linear combination of 6 and 5 and hence must
be a multiple of (5, 6) = 1 or equivalently 6y + 5z = 1.w. So we can replace it in
the original equation, which is therefore equivalent to 3x + w = 7. This is a linear
Diophantine equation with two variables, and we know how to solve it. We then
first solve this new equation:
(3, 1) = 1 = 3 − 2 ∗ 1
3 ∗ 7 + 1 ∗ (−14) = 7
The set of solutions for (x, w) is {(7 − k, −14 + 3k), k ∈ Z}
Then −14 + 3k = 6y + 5z. For a given integer k, this is again a linear Diophantine
equation with two variables, namely y and z and again we can solve it using the
technic we know. Because 6(1) − 5(1) = 1, 6(−14 + 3k) + 5(14 − 3k) = −14 + 3k.
Then we have that (y, z) = {(−14 + 3k − 5`, 14 − 3k + 6`), ` ∈ Z}.
Therefore, our overall solutions are:
(x, y, z) = {(7 − k, −14 + 3k − 5`, 14 − 3k + 6`), k, ` ∈ Z}
b) 101x + 102y + 103z = 1
(102, 103) = 1 = 103 − 102
102y + 103z = w
101x + w = 1
(101, 1) = 1 = 101 + 1(−100)
(x, w) = {(1 − k, −100 + 101k), k ∈ Z}
102y + 103z = −100 + 101k
102(100 − 101k) + 103(−100 + 101k) = −100 + 101k
(y, z) = {(100 − 101k − 103`, −100 + 101k + 102`), `, k ∈ Z}
(x, y, z) = {(1 − k, 100 − 101k − 103`, −100 + 101k + 102`), `, k ∈ Z}
c) 2x1 + 5x2 + 4x3 + 3x4 = 5
(4, 3) = 1 = 4(1) + 3(−1)
1
Obviously, let p | a, with p prime and a an integer. Assume p | a and p | a + 1. Then p must divide 1, and
hence p = ±1, which is a contradiction since p is prime.
5
4x3 + 3x4 = y1
2x1 + 5x2 + y1 = 5
(5, 1) = 1 = 5(1) + 1(−4)
5x2 + y1 = y2
2x1 + y2 = 5
(2, 1) = 1 = 2(1) + 1(−1)
(x1 , y2 ) = {(1 − k, −1 + 2k), k ∈ Z}
5x2 + y1 = −1 + 2k
5(−1 + 2k) + 1(4 − 8k) = −1 + 2k
(x2 , y1 ) = {(−1 + 2k − `, 4 − 8k + 5`), k, ` ∈ Z}
4x3 + 3x4 = 4 − 8k + 5`
4(4 − 8k + 5`) + 3(−4 + 8k − 5`) = 4 − 8k + 5`
(x3 , x4 ) = {(4 − 8k + 5` − 3m, −4 + 8k − 5` + 4m), k, `, m ∈ Z}
(x1 , x2 , x3 , x4 ) = {(1−k, −1+2k−`, 4−8k+5`−3m, −4+8k−5`+4m), k, `, m ∈ Z}
Part 2 Letting a, b, c be the number of cupcakes of each type we buy, we want to find solutions
to the equation 3a + 5b + 10c = 151, with a, b, c ≥ 1 (Must buy at least one of each.)
(10, 5) = 5 = 10(1) + 5(−1)
w = b + 2c
3a + 5w = 151
(3, 5) = 1 = 5(−1) + 3(2)
3(302) + 5(−151) = 151
(a, w) = {(302 − 5k, −151 + 3k), k ∈ Z}
b + 2c = −151 + 3k
1(−1) + 2(1) = 1
1(151 − 3k) + 2(−151 + 3k) = 151 + 3k
(b, c) = {(151 − 3k − 2`, −151 + 3k + `), k, ` ∈ Z}
(a, b, c) = {(302 − 5k, 151 − 3k − 2`, −151 + 3k + `), k, ` ∈ Z}
Then, we have the combinations:
a 42 42 37 37 37 32 32 32 32 32 27 27 27 27 27 27
b
1
3
2
4
6
1
3
5
7
9
2
4
6
8 10 12
c
2
1
3
2
1
5
4
3
2
1
6
5
4
3
2
1
a 22 22 22 22 22 22 22 22 17 17 17 17 17 17 17 17 17
b
1
3
5
7
9 11 13 15 2
4
6
8 10 12 14 16 18
c
8
7
6
5
4
3
2
1
9
8
7
6
5
4
3
2
1
a 12 12 12 12 12 12 12 12 12 12 12
b
1
3
5
7
9 11 13 15 17 19 21
c 11 10 9
8
7
6
5
4
3
2
1
a
7
7
7 7 7
7
7
7
7
7
7
7
b
2
4
6 8 10 12 14 16 18 20 22 24
c 12 11 10 9 8
7
6
5
4
3
2
1
a
2
2
2
2
2
2
2
2
2
2
2
2
2
2
b
1
3
5
7
9 11 13 15 17 19 21 23 25 27
c 14 13 12 11 10 9
8
7
6
5
4
3
2
1
(Cupcakes.)
6
Problem 5:
Part 1: 4.1 #12 Because n | m, by the definition of divisors, there is an integer k such that m = kn. By
the definition of congruence, a ≡ b (mod m) means that m | (a−b), and by the definition
of divisors, there is an integer q such that a − b = qm. Then, a − b = qm = qkn. We
know that qk is an integer because the integers are closed under multiplication, so, by
the definition of divisors, n | (a − b). Then, by the definition of mod, a ≡ b (mod n).
Part 2: 4.1 #14 By the definition of congruence, the fact that a ≡ b (mod c) means that c | (a − b), or by
the definition of divisors, there is an integer m such that a − b = mc, or a = mc − b and
b = a − mc. By Bézout’s theorem, there are integers p, q such that (a, c) = pa + qc is the
smallest possible positive linear combination of a and c. Similarly, there are integers
x, y such that (b, c) = xb + yc is the smallest possible positive linear combination of
b and c. By substitution, (a, c) = pa + qc = p(mc − b) + qc = c(pm + q) − pb. We
know that pm + q is an integer by the closure of the integers under multiplication and
addition, so (a, c) can be written as a linear combination of b and c, because it is the
smallest positive linear combination of a and c, and because all linear combinations of
b and c are multiples of (b, c), we have that (a, c) | (b, c). Similarly, with substitution,
(b, c) = xb + yc = x(a − mc) + yc = c(y − xm) + ax, and y − xm is an integer because the
integers are closed under addition and multiplication, so by the same reasoning as above,
since (b, c) can be written as a linear combination of (a, c), we have that (b, c) | (a, c).
Because any gcd must be at least 1, by a previous theorem, because (a, c) | (b, c) and
(b, c) | (a, c), we have that (a, c) = (b, c).
Part 3: 4.1 #30 Proof by Induction:
Base Case: n = 1. Then 41 = 4 and 1 + 3(1) = 4, and 4 − 4 = 0 = 0 ∗ 9, so 9 | 4 − 4,
and 4 ≡ 4 (mod 9) by the definitions of divisors and congruences respectively.
Inductive Hypothesis: 4n ≡ 1 + 3n (mod 9).
Induction step: 4n+1 ≡ 1 + 3(n + 1) (mod 9).
By the Inductive Hypothesis, we have that 9 | 4n − 1 − 3n, and so there is some integer k
such that 9k = 4n − (1 + 3n). Then we have that 4n+1 − (1 + 3(n + 1)) = 4n+1 − 4 − 3n =
4n ∗ 4 − 4 − 12n + 9n = 4(4n − 1 − 3n) + 9n = 4(9k) + 9n = 9(4k + n). 4k + n is an
integer by the closure of the integers under multiplication and addition, so we have that
9 | 4n+1 − (1 + 3(n + 1)), and so by the definition of mod, 4n+1 ≡ 4n+1 − (1 + 3(n + 1))
(mod 9), which proves the result.
Problem 6:
Part 1 If there are integers a, b, k such that a2 + b2 = 4k + 3, then a2 + b2 − 3 = 4k, so
4 | a2 + b2 − 3 by the definition of divisors, and this means that, by the definition of
congruence, a2 + b2 ≡ 3 (mod 4). Thus, a2 + b2 ≡ 3 (mod 4) is equivalent to a, b, k such
that a2 + b2 = 4k + 3 existing. Proof that this is impossible: We can see that we have
three cases: Either a and b are even, one of a and b is odd and the other is even, or
both a and b are odd.
Case 1: a and b are even. Then there are integers m and n such that a = 2m and b = 2n
and we have that a2 + b2 = 4m2 + 4n2 = 4(m2 + n2 ), and m2 + n2 is an integer
7
by the closure of the integers under addition and multiplication. Thus, by the
definition of divisors, 4 | a2 + b2 , and so by the definition of congruence, a2 + b2 ≡ 0
(mod 4). Thus in this case a2 + b2 6≡ 3 (mod 4).
Case 2: WLOG, let a be even and b be odd. Then, there are integers m and n such that
a = 2m and b = 2n + 1. Then we have that a2 + b2 = 4m2 + 4n2 + 4n + 1 =
4(m2 + n2 + n) + 1, and m2 + n2 + n is an integer by the closure of the integers
under addition and multiplication. Thus, by the definition of divisors, 4 | a2 +b2 −1,
and so by the definition of mod, a2 + b2 ≡ 1 (mod 4). Thus in this case a2 + b2 6≡ 3
(mod 4).
Case 3: a and b are odd. Then there are integers m and n such that a = 2m + 1 and b =
2n+1. Consequently a2 +b2 = 4m2 +4m+1+4n2 +4n+1 = 4(m2 +m+n2 +n)+2,
and m2 + m + n2 + n is an integer by the closure of the integers under addition
and multiplication. Thus, by the definition of divisors, 4 | a2 + b2 − 2. Hence by
the definition of congruence, a2 + b2 ≡ 2 (mod 4) and a2 + b2 6≡ 3 (mod 4).
Thus, we have shown that for all possible cases, a2 + b2 6≡ 3 (mod 4), and so there are
no integers a, b, k such that a2 + b2 = 4k + 3.
Part 2 If there are integers a, b, c, k such that a2 + b2 + c2 = 8k + 7, then a2 + b2 + c2 − 7 = 8k,
so 8 | a2 + b2 + c2 − 7 by the definition of divisors, and this means that, by the definition
of mod, a2 + b2 + c2 ≡ 7 (mod 8). Thus, a2 + b2 + c2 ≡ 7 (mod 8) is equivalent to
a, b, c, k such that a2 + b2 + c2 = 8k + 7 existing. Proof that this is impossible: There
are four cases for each of the numbers, a ≡ 0 mod 4), a ≡ 1 mod 4), a ≡ 2 mod 4), or
a ≡ 3 mod 4).
Case 1: a ≡ 0 (mod 4). Then there is an integer k such that a = 4k, so a2 = 16k 2 =
8(2k 2 ) ≡ 0 (mod 8).
Case 2: a ≡ 1 (mod 4). Then there is an integer k such that a = 4k + 1, so a2 =
16k 2 + 8k + 1 = 8(2k 2 + k) + 1 ≡ 1 (mod 8).
Case 3: a ≡ 2 (mod 4). Then there is an integer k such that a = 4k + 2, so a2 =
16k 2 + 16k + 4 = 8(2k 2 + 2k) + 4 ≡ 4 (mod 8).
Case 4: a ≡ 3 (mod 4). Then there is an integer k such that a = 4k + 3, so a2 =
16k 2 + 24k + 9 = 8(2k 2 + 3k + 1) + 1 ≡ 1 (mod 8).
Then we can see that we have 12 various combinations for a2 + b2 + c2 , mod 8, arranged
as follows WLOG.
Case 1: a2 ≡ 0 (mod 8), b2 ≡ 0 (mod 4), and c2 ≡ 0 (mod 8). Then a2 +b2 +c2 ≡ 0+0+0 = 0
(mod 8). And since 0 6≡ 7 (mod 8), we have that this does not lead to what we
want.
Case 2: a2 ≡ 0 (mod 8), b2 ≡ 0 (mod 4), and c2 ≡ 1 (mod 8). Then a2 +b2 +c2 ≡ 0+0+1 = 1
(mod 8). And since 1 6≡ 7 (mod 8), we have that this does not lead to what we
want.
Case 3: a2 ≡ 0 (mod 8), b2 ≡ 0 (mod 4), and c2 ≡ 4 (mod 8). Then a2 +b2 +c2 ≡ 0+0+4 = 4
(mod 8). And since 4 6≡ 7 (mod 8), we have that this does not lead to what we
want.
8
Case 4: a2 ≡ 0 (mod 8), b2 ≡ 1 (mod 4), and c2 ≡ 1 (mod 8). Then a2 +b2 +c2 ≡ 0+1+1 = 2
(mod 8). And since 2 6≡ 7 (mod 8), we have that this does not lead to what we
want.
Case 5: a2 ≡ 0 (mod 8), b2 ≡ 1 (mod 4), and c2 ≡ 4 (mod 8). Then a2 +b2 +c2 ≡ 0+1+4 = 5
(mod 8). And since 5 6≡ 7 (mod 8), we have that this does not lead to what we
want.
Case 6: a2 ≡ 0 (mod 8), b2 ≡ 4 (mod 4), and c2 ≡ 4 (mod 8). Then a2 +b2 +c2 ≡ 0+4+4 =
8 ≡ 0 (mod 8). And since 0 6≡ 7 (mod 8), we have that this does not lead to what
we want.
Case 7: a2 ≡ 1 (mod 8), b2 ≡ 1 (mod 4), and c2 ≡ 1 (mod 8). Then a2 +b2 +c2 ≡ 1+1+1 = 3
(mod 8). And since 3 6≡ 7 (mod 8), we have that this does not lead to what we
want.
Case 8: a2 ≡ 1 (mod 8), b2 ≡ 1 (mod 4), and c2 ≡ 4 (mod 8). Then a2 +b2 +c2 ≡ 1+1+4 = 6
(mod 8). And since 6 6≡ 7 (mod 8), we have that this does not lead to what we
want.
Case 9: a2 ≡ 1 (mod 8), b2 ≡ 4 (mod 4), and c2 ≡ 4 (mod 8). Then a2 +b2 +c2 ≡ 1+4+4 =
9 ≡ 1 (mod 8). And since 1 6≡ 7 (mod 8), we have that this does not lead to what
we want.
Case 10: a2 ≡ 4 (mod 8), b2 ≡ 4 (mod 4), and c2 ≡ 4 (mod 8). Then a2 + b2 + c2 ≡
4 + 4 + 4 = 12 ≡ 4 (mod 8). And since 4 6≡ 7 (mod 8), we have that this does not
lead to what we want.
Thus, in all possible cases, we never encounter a number which is ≡ 7 (mod 8), we must
conclude that it is impossible for such a number to exist.
Problem 7: First we prove that 10n ≡ −1 (mod 11), or equivalently, by the definitions of
congruence and divisors, that there exists an integer k such that10n + 1 = 11k when n is odd.
Proof by induction:
X
Base case: n = 1. Then 101 + 1 = 11 = 11 ∗ 1. Base case holds.
Inductive Hypothesis: 102k+1 ≡ −1 (mod 11), with k a positive integer.
Induction step: we want to show that 102k+3 ≡ −1 (mod 11), with 2k + 3 being the next odd
integer.
Proof: By the inductive hypothesis, the definition of mod, and the definition of divisors,
there is an integer m such that 102k+1 + 1 = 11m, or 102k+1 = 11m − 1. Then we have that
102k+3 + 1 = 102 102k+1 + 1 = 100(11m − 1) + 1 = 11(100m) − 99 = 11(100m − 9), and so
11 | 102k+3 + 1 by the definition of divisors, so by the definition of mod, 102k+3 ≡ −1 (mod
11), which proves the result.
Now, we prove similarly that 10n ≡ 1 (mod 11), when n is even. Proof by induction:
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Base case: n = 0. Then 100 − 1 = 0 = 11 ∗ 0. Base case holds.
Inductive Hypothesis: 102k ≡ −1 (mod 11), where k is a positive integer.
Want to show: 102k+2 ≡ 1 (mod 11), with 2k + 2 being the next even integer.
Proof: By the induction hypothesis, there is an integer m such that 102k − 1 = 11m, or
102k = 11m + 1. Then we have that 102k+2 − 1 = 102 102k − 1 = 100(11m + 1) − 1 =
11(100m) + 99 = 11(100m + 9), and so 11 | 102k+2 + 1 by the definition of divisors, so by
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the definition of mod, 102k+2 ≡ 1 (mod 11). Thus, inductive step complete, and proof by
induction complete.
Now, we know that any palindrome with an even number of digits can be written as
a1 a2 . . . ap ap . . . a2 a1 = a1 102p−1 + a2 102p−2 + · · · + ap 10p+1 + ap 10p + · · · + a2 101 + a1 100 .
Then, take our whole number mod 11. If p is even, we get that
a1 102p−1 + a2 102p−2 + · · · + ap 10p+1 + ap 10p + · · · + a2 101 + a1 ∗ 100
≡ a1 (−1) + a2 (1) + · · · + ap (−1) + ap (1) + · · · + a2 (−1) + a1 (1) (mod 11)
≡ 0 (mod 11)
If p is odd, we have
a1 102p−1 + a2 102p−2 + · · · + ap 10p+1 + ap 10p + · · · + a2 101 + a1 ∗ 100
≡ a1 (−1) + a2 (1) + · · · + ap (1) + ap (−1) + · · · + a2 (−1) + a1 (1) (mod 11)
≡ 0 (mod 11)
And so we get that, in either case, our palindrome is a multiple of 11, by the definition of
mod.
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