Download MATH3221: Number Theory Philipp BRAUN Homework until Test #2

MATH3221: Number Theory
Homework until Test #2
Philipp BRAUN
Section 3.1
page 43, 1. It has been conjectured that there are infinitely many primes of the form n2 − 2.
Exhibit five such primes.
52
Solution. Five such primes are e.g. 2 = 4 − 2 = 22 − 2, 7 = 9 − 2 = 32 − 2, 23 = 25 − 2 =
− 2, 47 = 49 − 2 = 72 − 2 and 79 = 81 − 2 = 92 − 2.
2. Give an example to show that the following conjecture is not true: Every positive integer can
be written in the form p + a2 , where p is either a prime or 1, and a ≥ 0.
Counterexample: 25. Since none of the numbers 25, 25 − 12 = 24, 25 − 22 = 21, 25 − 32 =
16, 25 − 42 = 9 are primes and a has to be less than 5 since 52 = 25 which forced p to be
zero, there is neither a prime p such that there is a such that p + a2 = 25, nor there is a
such that 1 + a2 = 25.
3. Prove each of the assertions below:
(a) Any prime of the form 3n + 1 is of the form 6m + 1
(b) Each integer of the form 3n + 2 has a prime factor of this form.
(c) The only prime of the form n3 − 1 is 7.
(d) The only prime p for which 3p + 1 is a perfect square is p = 5.
(e) The only prime of the form n2 − 4 is 5.
Proof.
(a) Let p = 3n + 1 be any prime. Since p is a prime by assumption, 3n has to be even
(since otherwise 3n + 1 would be even and therefore divisible by 2, which contradicts
the assumption that p is prime), i.e. 2 | 3n. Since 3 - 2, 2 | n by Theorem 3.1 (since
2 is a prime). That means that there is an integer m such that n = 2m. Therefore
3n + 1 = 3(2m) + 1 = 6m + 1, as we wanted to show.
(b) Let m = 3n + 2 be any integer. If m is prime, we are done. Assume m is not a prime.
Therefore there are integers a, b such that |a|, |b| > 1 and ab = 3n + 2. Consider the
following cases: Case I : a = 3k, b = 3l for integers k, l. Then ab = 3k · 3l = 3(3kl),
which is of the form 3z and therefore cannot equal 3n + 2. Case II : a = 3k, b = 3l + 1
(or vice versa) for integers k, l. Then ab = 3k(3l + 1) = 3(3kl + k), which is also of the
form 3z. Case III : a = 3k + 1, b = 3l + 1 for integers k, l. Then ab = (3k + 1)(3l + 1) =
3(3kl + k + l) + 1, which is of the form 3z + 1.
1
So none of these cases can occur, so at least one of the factors a, b has to be 3z + 2 for
an integer z. Either this factor is a prime, or we can use the same argument to show
that 3z + 2 = (3x + 2)y with integers x, y. Since |a|, |b| > 1, 3z + 2 < 3n + 2. Thus
in every step the absolute value of the factor of the form 3a + 2 will be smaller, so the
method has to terminate with a prime of that form.
(c) Since n3 − 1 = (n2 + n + 1)(n − 1), (n − 1) | (n3 − 1). So n must be less than or equal
to 2. For n = 1 we have 13 − 1 = 0, which is not the prime. So the only prime of the
form n3 − 1 is 23 − 1 = 7.
(d) Assume (3p+1) = n2 for a prime p and an integer n. Then p = 13 (n2 −1) = 31 (n+1)(n−
1). Since p is a prime, 31 (n − 1) = 1, which implies n = 4 and therefore p = 13 · 5 · 3 = 5.
(e) Since n2 − 4 = (n + 2)(n − 2), n − 2 has to equal 1 (because otherwise n2 − 4 is no prime
since (n − 2) | (n2 − 4)), i.e. n = 3, so the only prime of that form is 32 − 4 = 5.
4. If p ≥ 5 is a prime number, show that p2 + 2 is composite.
Proof. p2 + 2 = (p + 1)(p − 1) + 3. Since p is a prime bigger than 3, p is not divisible by
3. But since one of three consecutive numbers is divisible by three, so is either (p − 1) or
(p + 1). Since 3 | 3, by Theorem 2.2 (vii) 3 | (p + 1)(p − 1) + 3, so p2 + 2 is divisible by 3
and therefore composite.
6. Establish each of the following statements:
(a) Every integer of the form n4 + 4, with n > 1, is composite.
(b) If n > 4 is composite, then n divides (n − 1)!.
(c) Any integer of the form 8n + 1, where n ≥ 1, is composite.
(d) Each integer n > 11 can be written as the sum of two composite numbers.
Proof.
(a) n4 + 4 = (n2 + 2n + 2)(n2 − 2n + 2), and for n > 1, each of these factors is bigger than
1, hence n4 + 4 is composite.
√
(b) If n > 4 is composite, then it is a square number (then 2 n < n since n > 4, and
√
√
therefore n | 1·2·. . .· n·. . .·2 n·. . .·(n−1) = (n−1)!) or it can be written as a product of
√
√
a prime 1 < p < n and an integer n < m < n. Since (n−1)! = 1·2·p·. . .·m·. . .·(n−1),
pm | (n − 1)!, i.e. n | (n − 1)!.
(c) 8n + 1=(2n )3 + 1 = (2n + 1)((2n )2 − (2n ) + 1). Since both factors are bigger than 1 for
n ≥ 1, 8n + 1 is not prime and therefore composite.
(d) In case n > 11 is even, there is k > 5 such that n = 2k. Therefore n = 6 + 2(k − 3),
where 6 = 2 · 3 is a composite number and 2 · (k − 3) is a composite number (since
k − 3 > 1 since k > 5). In case n is odd, there is k > 5 such that n = 2k + 1. Therefore
n = 9 + 2(k − 4), where 9 = 3 · 3 and 2(k − 4) are composite numbers (since k − 4 > 1
because k > 5).
2
Section 3.2
page 49, 1. Determine whether the integer 701 is prime by testing all primes p ≤
divisors. Do the same for integer 1009.
√
701 as possible
√
Solution. 701 ≈ 26.47, i.e. we have to test the primes 2, 3, 5, 7, 11, 13, 17, 19 and 23 as
possible divisors of 701:
701/2 = 350.5 ∈
/ Z ⇒ 2 - 701.
701/3 = 233.6 ∈
/ Z ⇒ 3 - 701.
701/5 = 140.2 ∈
/ Z ⇒ 5 - 701.
701/7 ≈ 100.14 ∈
/ Z ⇒ 7 - 701.
701/11 = 63.72 ∈
/ Z ⇒ 11 - 701.
701/13 ≈ 53.92 ∈
/ Z ⇒ 13 - 701.
701/17 ≈ 41.24 ∈
/ Z ⇒ 17 - 701.
701/19 ≈ 36.89 ∈
/ Z ⇒ 19 - 701.
701/23 ≈ 30.48 ∈
/ Z ⇒ 23 - 701. Hence 701 is a prime.
√
Since 1009 ≈ 31.76, we have to test the primes 2, 3, 5, 7, 11, 13, 17, 19, 23, 29 and 31 as
possible divisors of 1009:
1009/2 = 504.5 ∈
/ Z ⇒ 2 - 1009.
1009/3 = 336.3 ∈
/ Z ⇒ 3 - 1009.
1009/5 = 201.8 ∈
/ Z ⇒ 5 - 1009.
1009/7 ≈ 144.14 ∈
/ Z ⇒ 7 - 1009.
1009/11 = 91.72 ∈
/ Z ⇒ 11 - 1009.
1009/13 ≈ 77.62 ∈
/ Z ⇒ 13 - 1009.
1009/17 ≈ 59.35 ∈
/ Z ⇒ 17 - 1009.
1009/19 ≈ 53.11 ∈
/ Z ⇒ 19 - 1009.
1009/23 ≈ 43.87 ∈
/ Z ⇒ 23 - 1009.
1009/29 ≈ 34.79 ∈
/ Z ⇒ 29 - 1009.
1009/31 ≈ 32.54 ∈
/ Z ⇒ 31 - 1009. Hence 1009 is a prime.
2. Employing the Sieve of Eratosthenes, obtain all the primes between 100 and 200.
Solution.
101 103 107 109
100
102
104
105
106
108
113 H
110
111
112
114
115
116
117
118
119
H
H
120 121
122 123 124 125 126 127 128 129
H
131 H
137 139
130
132
133
134
135
136
138
H
H
149
140
141
142
143
144
145
146
147
148
H
151 157 150
152
153
154
155
156
158
159
H
H
160 161
162 163 164 165 166 167 168 169
H
H
170 171 172 173 174 175 176 177 178 179
3
H
181 189
186
188
785
187
184
183
182
180
H
191 193 197 199
190
192
194
195
196
198
(x means x is divisible by 2, 3, or 5, xA means that x is divisible by 7, 11, or 13.)
Thus the primes between 100 and 200 are 101, 103, 107, 109, 113, 127, 131, 137, 139, 149, 151,
157, 163, 167, 173, 179, 181, 191, 193, 197 and 199.
3. Given that p - n for all primes p ≤
two primes.
√
3
n, show that n > 1 is either a prime or the product of
Proof. If n was the product of three or more primes, at least one of them had to be less
√
√
than or equal to 3 n. Since by assumption none of the primes p ≤ 3 n divides n, this is a
contradiction. Hence n is a prime or a product of two primes.
4. Establish the following facts:
√
(a) p is irrational for any prime p.
√
√
(b) If a > 0 and n a is rational, then n a must be an integer.
√
(c) For n ≥ 2, n n is irrational.
√
√
Proof. (a) Assume p is rational. Then there are integers a, b with p = ab and
gcd(a, b) = 1. Square the equation and multiply both sides with b to get b2 p = a2 . That
means p | a2 and since p is a prime p | a by Theorem 3.1. Therefore there is an integer c such
that a = cp. Applying that to our first equation, we have b2 p = (cp)2 = c2 p2 , or b2 = c2 p.
That means p | b2 , and again by Theorem 3.1 p | b, so p | gcd(a, b), a contradiction.
√
√
n
(b) Assume n a is rational, i.e. n a = pq for relatively prime integers p, q. Then a = pqn .
Since p and q are relatively prime, pn and q n are relatively prime, too. That a is an integer
√
forces q n to be 1, so q = 1, i.e. n a = p, an integer.
√
√
√
√
(c) Since 2n > n for n ≥ 2, n 2n > n n. But n 2n = 2, so n n < 2. On the other hand
√
√
√
n
n > 1 since 1n = 1 < n. By part (b), n n is either an integer or irrational, so n n has to
be irrational as there is no integer between 1 and 2.
5. Show that any composite three-digit number must have a prime factor less than or equal to
31.
Proof. Let c be any composite three-digit number. Since c is composite, there are
integers a, b such that c = ab. Assume a and b are both greater than 31, i.e. at least 32.
Then ab ≥ 32 · 32 = 25 · 25 = 210 = 1024, which contradicts to the fact that c is a three-digit
number by assumption.
4
Section 3.3
page 57, 1. Verify that the integers 1949 and 1951 are twin primes.
√
Solution. Since 1951 ≈ 44.17, we have to check if 1949 or 1951 are divisible by
2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, or 43. Obviously, both numbers are odd and therefore not divisible by 2. Since the sum of the digits of 1949, 1+9+4+9=23 and the sum of
the digits of 1951, 1+9+5+1=16 are both not divisible by 3, both numbers are not divisible
by 3. Since both numbers do not end with “0” or “5”, they are not divisible by 5. Using
the same method as in exercise 1 on page 49, we have
1949/7 ≈ 278.43 ∈
/ Z ⇒ 7 - 1949, 1951/7 ≈ 278.71 ∈
/ Z ⇒ 7 - 1951
1949/11 = 177.18 ∈
/ Z ⇒ 11 - 1949, 1951/11 = 177.36 ∈
/ Z ⇒ 11 - 1951
1949/13 ≈ 149.92 ∈
/ Z ⇒ 13 - 1949, 1951/13 ≈ 150.08 ∈
/ Z ⇒ 13 - 1951
1949/17 ≈ 114.65 ∈
/ Z ⇒ 17 - 1949, 1951/17 ≈ 114.76 ∈
/ Z ⇒ 17 - 1951
1949/19 ≈ 102.58 ∈
/ Z ⇒ 19 - 1949, 1951/19 ≈ 102.68 ∈
/ Z ⇒ 19 - 1951
1949/23 ≈ 84.74 ∈
/ Z ⇒ 23 - 1949, 1951/23 ≈ 84.83 ∈
/ Z ⇒ 23 - 1951
1949/29 ≈ 67.21 ∈
/ Z ⇒ 29 - 1949, 1951/29 ≈ 67.28 ∈
/ Z ⇒ 29 - 1951
1949/31 ≈ 62.87 ∈
/ Z ⇒ 31 - 1949, 1951/31 ≈ 62.94 ∈
/ Z ⇒ 31 - 1951
1949/37 = 52.675 ∈
/ Z ⇒ 37 - 1949, 1951/37 = 52.729 ∈
/ Z ⇒ 37 - 1951
/ Z ⇒ 41 - 1949, 1951/41 = 47.58536 ∈
/ Z ⇒ 41 - 1951
1949/41 = 47.53658 ∈
1949/43 ≈ 45.33 ∈
/ Z ⇒ 43 - 1949, 1951/43 ≈ 45.37 ∈
/ Z ⇒ 43 - 1951. Therefore 1949 and
1951 both are prime. Since 1951 − 1949 = 2, they are twin primes.
2. (a) If 1 is added to a product of twin primes, prove that a perfect square is always obtained.
(b) Show that the sum of twin primes p and p + 2 is divisible by 12, provided that p > 3.
Proof. (a) Assume p and p + 2 are twin primes. Then their product is p(p + 2) = p2 + 2p.
If we add 1, we have p2 + 2p + 1, which is equal to (p + 1)2 , which obviously is a perfect
square.
(b) The sum of the twin primes p and p + 2 is p + (p + 2) = 2p + 2. Since p is a prime
greater than 3, p is of the form 3k + 2 for an integer k (if it was of the form 3k, it would
be divisible by 3, which contradicts the assumption that p is a prime, and if it was of the
form 3k + 1, p + 2 = 3k + 1 + 2 = 3(k + 1) would not be a prime). So 2p + 2 is of the form
2(3k + 2) + 2 = 6k + 4 + 2 = 6k + 6 = 6k + 6 = 6(k + 1). Since p is a prime, k is odd
(because if k is even, then so is 3k + 2, which contradicts that p is a prime greater than
3). Hence k + 1 is even, i.e. there is an integer n such that k + 1 = 2n. That means that
2p + 2 = 6(k + 1) = 6 · 2 · n = 12n, so the sum of the primes p and p + 2 is divisible by 12.
3. Find all pairs of primes p and q satisfying p − q = 3.
Solution. Since the difference should be 3, one of the primes has to be even and the
5
other has to be odd. Since the only even prime is 2, the only of these pairs is p = 5 and
q = 2.
5. In 1752, Goldbach submitted the following conjecture to Euler: Every odd integer can be
written in the form p + 2a2 , were p is either a prime or 1 and a ≥ 0. Show that the integer 5777
refutes this conjecture.
Solution. a has to be less than or equal to 53, since 2 · 542 = 5832 > 5777. The
set 5777 − 2a2 |0 ≤ a ≤ 53 = {5777 = 53 · 109, 5775 = 5 · 1155, 5769 = 3 · 1923, 5759 =
13·443, 5745 = 5·1149, 5727 = 3·1909, 5705 = 5·1141, 5679 = 3·1893, 5649 = 3·1883, 5615 =
5·1123, 5577 = 3·1859, 5535 = 5·1107, 5489 = 11·499, 5439 = 3·1813, 5385 = 5·1077, 5327 =
7·761, 5265 = 5·1053, 5199 = 3·1733, 5129 = 23·223, 5055 = 5·1011, 4977 = 3·1659, 4895 =
5 · 979, 4809 = 3 · 1603, 4719 = 3 · 1573, 4625 = 5 · 925, 4527 = 3 · 1509, 4425 = 5 · 885, 4319 =
7 · 617, 4209 = 3 · 1403, 4095 = 5 · 819, 3977 = 41 · 97, 3855 = 5 · 771, 3729 = 3 · 1243, 3599 =
59 · 61, 3465 = 5 · 693, 3327 = 3 · 1109, 3185 = 5 · 637, 3039 = 3 · 1013, 2889 = 3 · 963, 2735 =
5 · 547, 2577 = 3 · 859, 2415 = 5 · 483, 2249 = 13 · 173, 2079 = 3 · 693, 1905 = 5 · 381, 1727 =
11 · 157, 1545 = 5 · 309, 1359 = 3 · 453, 1169 = 7 · 167, 975 = 5 · 195, 777 = 7 · 111, 575 =
5 · 115, 369 = 3 · 123, 159 = 3 · 53} contains neither a prime nor 1, so 5777 cannot be written
in the wanted form.
7. A conjecture of Lagrange (1775) asserts that every odd integer greater than 5 can be written
as a sum p1 + 2p2 , where p1 , p2 are both primes. Confirm this for all odd integers through 75.
Solution. 7 = 3 + 2 · 2, 9 = 5 + 2 · 2, 11 = 7 + 2 · 2, 13 = 7 + 2 · 3, 15 = 5 + 2 · 5, 17 =
7 + 2 · 5, 19 = 5 + 2 · 7, 21 = 7 + 2 · 7, 23 = 13 + 2 · 5, 25 = 3 + 2 · 11, 27 = 5 + 2 · 11, 29 =
7 + 2 · 11, 31 = 5 + 2 · 13, 33 = 7 + 2 · 13, 35 = 13 + 2 · 11, 37 = 11 + 2 · 13, 39 = 13 + 2 · 13, 41 =
7 + 2 · 17, 43 = 5 + 2 · 19, 45 = 7 + 2 · 19, 47 = 13 + 2 · 17, 49 = 3 + 2 · 23, 51 = 5 + 2 · 23, 53 =
7 + 2 · 23, 55 = 17 + 2 · 19, 57 = 19 + 2 · 19, 59 = 37 + 2 · 11, 61 = 3 + 2 · 29, 63 = 5 + 2 · 29, 65 =
7 + 2 · 29, 67 = 5 + 2 · 31, 69 = 7 + 2 · 31, 71 = 13 + 2 · 29, 73 = 47 + 2 · 13, 75 = 13 + 2 · 31.
6
Sections 4.1 & 4.2
page 67, 1. Prove each of the following assertions:
(a) If a ≡ b (mod n) and m | n, then a ≡ b (mod m)
(b) If a ≡ b (mod n) and c > 0, then ca ≡ cb (mod cn)
(c) If a ≡ b (mod n) and the integers a, b, n are all divisible by d > 0, then a/d ≡ b/d (mod n/d)
Proof. (a) a ≡ b (mod n) ⇒ n | (a − b). Since m | n, by Theorem 2.2 (iv) m | (a − b), or
in other words a ≡ b (mod m).
(b) a ≡ b (mod n) ⇒ n | (a − b). Thus there is an integer d such that nd = a − b. Multiply
both sides with c to get cnd = c(a − b) = ca − cb. Since c > 0 and n > 0, cn > 0, and since
cn | (ca − cb), ca ≡ cb (mod cn).
(c) a ≡ b (mod n) ⇒ n | (a − b). Therefore there is an integer c such that cn = a − b. If
a, b and n all are divisible by d, then cn/d = (a − b)/d = a/d − b/d, where cn/d, a/d, b/d
all are integers. So n/d | (a/d − b/d). Since d > 0, n/d > 0 and therefore a/d ≡ b/d (mod
n/d).
2. Give an example to show that a2 ≡ b2 (mod n) need not imply a ≡ b (mod n).
Solution. Let a = 2, b = 3, n = 5. Then a2 = 4, b2 = 9, and 4 ≡ 9 (mod 5) (because
9 − 4 = 5, where 5 is obviously a multiple of 5), but 2 6≡ 3 (mod 5) (because 3 − 2 = 1, and
1 is not a multiple of 5).
3. If a ≡ b (mod n), prove that gcd(a, n) = gcd(b, n).
Proof. a ≡ b (mod n) ⇒ n | (a − b) ⇒ there is an integer c such that cn = a − b.
Let d be any divisor of both a and n. Then there are integers k, l such that dk = a and
dl = n, which means cdl = dk − b, or in other words b = dk − cdl = d(k − cl), so d divides
b. Especially gcd(a, n) | gcd(b, n). Similarly we can show that gcd(b, n) | gcd(a, n), which
forces the gcd’s to be equal.
4. (a) Find the remainders when 250 and 4165 are divided by 7.
(b) What is the remainder when the following sum is divided by 4? 15 + 25 + 35 + · · · + 995 + 1005
10
Solution. (a) 250 = 25 , and 25 ≡ 32 ≡ 5 (mod 7). By Theorem 4.2 (f), 250 ≡ 510
5
(mod 7). 510 = 52 , and 52 ≡ 25 ≡ 4 (mod 7). Again by Theorem 4.2 (f), 250 ≡ 45 (mod
7). 45 = (25 )2 , and 25 ≡ 5 (mod 7), as we already saw. Hence 250 ≡ 52 ≡ 25 ≡ 4 (mod 7),
i.e. the remainder when dividing 250 by 7 is 4.
Since 41 ≡ (−1) (mod 7), 4165 ≡ (−1)65 (mod 7) by Theorem 4.2 (f). Since (−1)65 = −1,
4165 ≡ −1 ≡ 6 (mod 7), so the remainder when dividing 4156 by 7 is 6.
(b) The even numbers to the fifth power have a factor 25 and therefore are divisible by
4 (i.e. congruent 0 (mod 4)). The other numbers are alternating either congruent 1 or
congruent -1, and so are their fifth powers. Since the sum of numbers is congruent to the
sum of any numbers their summands are congruent to by Theorem 4.2 (d), 15 + 25 + 35 +
7
· · · + 995 + 1005 ≡ 1 + 0 − 1 + · · · − 1 + 0 ≡ 0 (mod 4), i.e. the remainder when dividing
the sum by 4 is 0.
5. Prove that the integer 53103 + 10353 is divisible by 39, and that 111333 + 333111 is divisible by
7.
Proof. 53103 + 10353 ≡ (−1)103 + 153 ≡ −1 + 1 ≡ 0 (mod 3) (by Theorem 4.2 (f)), and
53103 + 10353 ≡ 1103 + (−1)53 (mod 13), so both 3 and 13 divide 53103 + 10353 . Since 3 and
13 are relatively prime (in fact, they are both prime numbers), the fist statement is true
by Collary 2 of Theorem 2.4.
For the second statement, note that 111333 + 333111 ≡ (−1)333 + 4111 ≡ −1 + (43 )37 ≡
−1 + (64)37 = −1 + 137 ≡ −1 + 1 ≡ 0 (mod 7).
6. For n ≥ 1, use congruence theory to establish each of the following divisibility statements:
(a) 7 | 52n + 3 · 25n−2 .
(b) 13 | 3n+2 + 42n+1 .
(c) 27 | 25n+1 + 5n+2 .
(d) 43 | 6n+2 + 72n+1 .
Proof. (a) For n = 1, the statement is true since 7 | 49 = 25 + 24 = 25 + 3 · 8 = 52 + 3 · 23 .
Assume the statement is true for n = k. Then 52k + 3 · 25k−2 ≡ 0 (mod 7). 52(k+1) + 3 ·
25(k+1)−2 ≡ 52 · 52k + 25 · 3 · 25k−2 ≡ 25 · 52k + 32 · 3 · 25k−2 ≡ 25(52k + 3 · 25k−2 ) + 7 · 3 · 25k−2
(mod 7). Since 52k + 3 · 25k−2 ≡ 0 (mod 7), 25(52k + 3 · 25k−2 ) ≡ 0 (mod 7) by Theorem
4.2 (e). Since 7 is a factor, 7 · 3 · 25k−2 ≡ 0 (mod 7). Therefore the statement is true for
n = k + 1 by Theorem 4.2 (d).
(b) For n = 1, we have 13 | 91 = 27 + 64 = 33 + 43 , which is true since 13 · 7 = 91.
Assume the statement is true for n = k. Then 3(k+1)+2 + 42(k+1)+1 ≡ 3 · 3k+2 + 42 · 42k+1 ≡
3(3k+2 + 42k+1 ) + 13 · 42k+1 (mod 13). By induction assumption, 3k+2 + 42k+1 ≡ 0 (mod
13), so 3(3k+2 + 42k+1 ) ≡ 0 (mod 13) by Theorem 4.2(e). Since 13 is a factor of 13 · 42k+1 ,
13 · 42k+1 ≡ 0 (mod 13). By Theorem 4.2 (d), the statement is true for n = k + 1.
(c) For n = 1, it is stated that 27 | 189 = 64 + 125 = 26 + 53 , which is true since
27 · 7 = 189. Assume the statement is true for n = k. Then 25(k+1)+1 + 5(k+1)+2 ≡
25 · 25k+1 + 5 · 5k+2 ≡ 32 · 25k+1 + 5 · 5k+2 ≡ 5(25k+1 + 5k+2 ) + 27 · 25k+1 (mod 27). By
assumption and since 27 is a factor of the second summand, for the same reasons as in
parts (a) and (b), 5(25k+1 + 5k+2 ) + 27 · 25k+1 ≡ 0 (mod 27), hence the statement is true
for n = k + 1.
(d) Since 43 | 559 = 216 + 343 = 63 + 73 because 43 · 13 = 559, the statement is true
for n = 1. Assume the statement is true for n = k. Similar to parts (a), (b) and (c),
6(k+1)+2 +72(k+1)+1 ≡ 6·6k+2 +72 ·72k+1 ≡ 6·6k+2 +49·72k+1 ≡ 6(·6k+2 +72k+1 )+43·72k+1 ≡
0 + 0 ≡ 0 (mod 43), thus the statement is true for n = k + 1.
8
7. For n ≥ 1, show that (−13)n+1 ≡ (−13)n + (−13)n−1 (mod 181).
Proof. For n = 1, the statement is true because (−13)2 ≡ 169 ≡ −12 ≡ −131 +
(−13)0 (mod 181). Assume the statement is true for n = k. Then (−13)(k+1)+1 ≡
(−13) (−13)k + (−13)k−1 ≡ (−13)k+1 + (−13)(k+1)−1 (mod 181).
9
Section 4.3
page 73, 1. Use the binary exponentiation algorithm to compute both 1953 (mod 503) and 14147
(mod 1537).
Solution. 1953 = 1932 ·1916·194 ·19, so 1953 ≡ 1932 ·1916 ·130321·19 ≡ 1932 ·(194 )4 ·44·19 ≡
1932 · (44)4 · 44 · 19 ≡ 1932 · 3, 748, 096 · 44 · 19 ≡ (1916 )2 · 243 · 44 · 19 ≡ 2432 · 243 · 44 · 19 ≡
59049 · 243 · 441̇9 ≡ (198 · 243) · (44 · 19) ≡ 48114 · 836 ≡ 329 · 333 ≡ 406 (mod 503)
14147 = 14132 ·1418 ·1414 ·1412 ·141, so 14147 ≡ 14132 ·1418 ·(1412 )2 ·19881·141 ≡ 14132 ·1418 ·
(1437)2 · 1437 · 141 ≡ 14132 · (1414 )2 · 2, 064, 969 · 1437 · 141 ≡ 14132 · (778)2 · 778 · 1437 · 141 ≡
(1418 )4 ·605284·778·1437·141 ≡ (1243)4 ·1243·778·1437·141 ≡ (1, 545, 049)2 ·1243·778·1437·
141 ≡ 3642 ·1243·778·1437·141 ≡ 132496·1243·778·1437·141 ≡ (314·1243)·(778·1437)·141 ≡
390302 · 1, 117, 986 · 141 ≡ 1441 · 587 · 141 ≡ 845867 · 141 ≡ 517 · 141 ≡ 72897 · 658 (mod
1537).
2. Prove the following statements:
(a) For any integer a, the units digit of a2 is 0, 1, 4, 5, 6, or 9.
(b) Any one of the integers 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 can occur as units digit of a3 .
(c) For any integer a, the units digit of a4 is 0, 1, 5, or 6.
(d) The units digit of a triangular number is 0, 1, 3, 5, 6, or 8.
Proof. Let a be an integer.
(a) By Division Algorithm, a is of the form 10k + r, where k is an integer and r ∈
{0, 1, · · · , 9}. a2 = (10k + r)2 = 102 k 2 + 2 · 10 · r + r2 , and since 10 is a factor of both 102 k 2
and 2 · 10 · r, a2 ≡ r2 (mod 10). Considering the values r can have, we get:
in case r = 0: a2 ≡ 02 ≡ 0 (mod 10),
in case r = 1: a2 ≡ 12 ≡ 1 (mod 10),
in case r = 2: a2 ≡ 22 ≡ 4 (mod 10),
in case r = 3: a2 ≡ 32 ≡ 9 (mod 10),
in case r = 4: a2 ≡ 42 ≡ 16 ≡ 6 (mod 10),
in case r = 5: a2 ≡ 52 ≡ 25 ≡ 5 (mod 10),
in case r = 6: a2 ≡ 62 ≡ 36 ≡ 6 (mod 10),
in case r = 7: a2 ≡ 72 ≡ 49 ≡ 9 (mod 10),
in case r = 8: a2 ≡ 82 ≡ 64 ≡ 4 (mod 10), and finally
in case r = 9: a2 ≡ 92 ≡ 81 ≡ 1 (mod 10). So we see that in each case a2 is congruent to
either 0, 1, 4, 5, 6, or 9 (mod 10), which means that one of these is the units digit of a2 .
(b) Using the same method as in (a), we have a3 ≡ r3 (mod 10), because by Binomial
Theorem the other summands of the binomial extension of (10k + r)3 have a factor 10.
Considering the cases, we see that
in case r = 0: a3 ≡ 03 ≡ 0 (mod 10),
in case r = 1: a3 ≡ 13 ≡ 1 (mod 10),
10
in case r = 2: a3 ≡ 23 ≡ 8 (mod 10),
in case r = 3: a3 ≡ 33 ≡ 27 ≡ 7 (mod 10),
in case r = 4: a3 ≡ 43 ≡ 64 ≡ 4 (mod 10),
in case r = 5: a3 ≡ 53 ≡ 125 ≡ 5 (mod 10),
in case r = 6: a3 ≡ 63 ≡ 216 ≡ 6 (mod 10),
in case r = 7: a3 ≡ 73 ≡ 343 ≡ 3 (mod 10),
in case r = 8: a3 ≡ 83 ≡ 512 ≡ 2 (mod 10), and finally
in case r = 9: a3 ≡ 93 ≡ 729 ≡ 9 (mod 10). So we see that each of the numbers
0, 1, 2, 3, 4, 5, 6, 7, 8, and 9 occur as units digit of a3 .
(c) We use again the same method to obtain
in case r = 0: a4 ≡ 04 ≡ 0 (mod 10),
in case r = 1: a4 ≡ 14 ≡ 1 (mod 10),
in case r = 2: a4 ≡ 24 ≡ 16 ≡ 6 (mod 10),
in case r = 3: a4 ≡ 34 ≡ 81 ≡ 1 (mod 10),
in case r = 4: a4 ≡ 44 ≡ 256 ≡ 6 (mod 10),
in case r = 5: a4 ≡ 54 ≡ 625 ≡ 5 (mod 10),
in case r = 6: a4 ≡ 64 ≡ 1296 ≡ 6 (mod 10),
in case r = 7: a4 ≡ 74 ≡ 2401 ≡ 1 (mod 10),
in case r = 8: a4 ≡ 84 ≡ 4096 ≡ 6 (mod 10), and finally
in case r = 9: a4 ≡ 94 ≡ 6561 ≡ 1 (mod 10). So the only possibilities for the units digit of
a4 are 0, 1, 5, and 6.
(d) Let a be a triangular number. By problem 1.(a) of section 2.1, there is an integer n
2
such that a = n(n+1)
= n 2+n . By Division Algorithm, n is of the form 10k + r, where k is
2
2
2 2
2
+10k+r
. Not that a
an integer and r ∈ {0, 1, · · · , 9}. So a = (10k+r)2+10k+r = 10 k +2·10·r+r
2
multiple of 5 ends either with 0 or with 5, so 5k ≡ 0 (mod 10) or 5k ≡ 5 (mod 10).
2
In case r = 0: a ≡ 0 +0+10k
≡ 0 + 5k (mod 10), so a ∈ {[0], [5]}
2
2
in case r = 1: a ≡ 1 +1+10k
≡
1 + 5k (mod 10), so a ∈ {[1], [6]}
2
22 +2+10k
in case r = 2: a ≡
≡ 3 + 5k (mod 10), so a ∈ {[3], [8]}
2
32 +3+10k
in case r = 3: a ≡
≡ 6 + 5k (mod 10), so a ∈ {[6], [1]}
2
42 +4+10k
in case r = 4: a ≡
≡ 10 + 5k ≡ 0 + 5k (mod 10), so a ∈ {[0], [5]}
2
2
in case r = 5: a ≡ 5 +5+10k
≡ 15 + 5k ≡ 5 + 5k (mod 10), so a ∈ {[5], [0]}
2
62 +6+10k
in case r = 6: a ≡
≡ 21 + 5k ≡ 1 + 5k (mod 10), so a ∈ {[1], [6]}
2
72 +7+10k
in case r = 7: a ≡
≡ 28 + 5k ≡ 8 + 5k (mod 10), so a ∈ {[8], [3]}
2
2
in case r = 8: a ≡ 8 +8+10k
≡ 36 + 5k ≡ 6 + 5k (mod 10), so a ∈ {[6], [1]}, and finally
2
92 +9+10k
in case r = 9: a ≡
≡ 45 + 5k ≡ 5 + 5k (mod 10), so a ∈ {[0], [5]}. In every case a
2
is element of one of the congruence classes [0], [1], [3], [5], [6], and [8] (mod 10). This implies
the statement.
11
9
3. Find the last two digits of the number 99 .
Solution. 99 ≡ (92 )4 · 9 ≡ 814 · 9 ≡ 14 · 9 ≡ 9 (mod 10), i.e. 99 = 10k + 9 for an integer
9
9
k, and therefore 99 = 910k+9 . So 99 ≡ 910k+9 ≡ 910k · 99 ≡ (910 )k · 99 ≡ (99 · 9)k · 99 (mod
100). Since 99 ≡ 93 · 93 · 93 ≡ 729 · 729 · 729 ≡ 29 · 29 · 29 ≡ 841 · 29 ≡ 41 · 29 ≡ 1189 ≡ 89
(mod 100), (99 · 9)k · 99 ≡ (89 · 9)k · 89 ≡ (801)k · 89 ≡ 1k · 89 ≡ 89 (mod 100). Therefore
9
the last two digits of 99 are 89.
4. Without performing the divisions, determine whether the integers 176,521,221 and 149,235,678
are divisible by 9 or 11.
Solution. Using Theorem 4.5, we only have to check if the sum of the digits of the
numbers are divisible by 9 to check divisibility by 9. 1 + 7 + 6 + 5 + 2 + 1 + 2 + 2 + 1 = 27,
which is divisible by 9, so 176,521,221 is divisible by 9. 1 + 4 + 9 + 2 + 3 + 5 + 6 + 7 + 8 = 45,
which is also divisible by 9, so 149,235,678 is divisible by 9, as well.
To check divisibility by 11, we use Theorem 4.6: 1 − 7 + 6 − 5 + 2 − 1 + 2 − 2 + 1 = −3, which
is not divisible by 11, so 176,521,221 is not divisible by 11. 1−4+9−2+3−5+6−7+8 = 9,
which is also not divisible by 11, so 149,235,678 is not divisible by 11.
7. Establish the following divisibility criteria:
(a) An integer is divisible by 2 if and only if its units digit is 0, 2, 4, 6, or 8.
(b) An integer is divisible by 3 if and only if the sum of its digits is divisible by 3.
(c) An integer is divisible by 4 if and only if the number formed by its tens and units digits is
divisible by 4.
(d) An integer is divisible by 5 if and only if its units digit is 0 or 5.
Proof. (a) An integer n is divisible by 2 if and only if there is an integer k such that
n = 2k. Let l be that number out of k, k − 1, k − 2, k − 3, and k − 4 that is divisible by
5. Since 2l is divisible by 10, it is congruent 0 (mod 10). k is equal to one of the numbers
l, l + 1, l + 2, l + 3, and l + 4. In case k = l, n ≡ 2k ≡ 2l ≡ 0 (mod 10), in case k = l + 1,
n ≡ 2k ≡ 2(l + 1) ≡ 2l + 2 ≡ 2 (mod 10), in case k = l + 2, n ≡ 2k ≡ 2(l + 2) ≡ 2l + 4 ≡ 4
(mod 10), in case k = l + 3, n ≡ 2k ≡ 2(l + 3) ≡ 2l + 6 ≡ 6 (mod 10), and finally in case
k = l + 4, n ≡ 2k ≡ 2(l + 4) ≡ 2l + 8 ≡ 8 (mod 10), so the units digit of n is one of the
numbers 0, 2, 4, 6, and 8. Trivially, if n ends with 0, 2, 4, 6, or 8, it is even and therefore
divisible by 2.
(b) Let n be an integer. There are m ∈ Z, a0 , a1 , . . . , am ∈ {0, 1, . . . , 9} such that n =
m
m
P
P
10m am + · · · + 10a1 + a0 . Let P (x) =
ak xk . Then P (10) = n and P (1) =
am , the
k=0
k=0
sum of the digits of n. Since 1 ≡ 10 (mod 3), P (1) ≡ P (10) (mod 3) by Theorem 4.4,
which implies the statement.
(c) Let n be an integer. There are m ∈ Z, a0 , a1 , . . . , am ∈ {0, 1, . . . , 9} such that n =
10m am + · · · + 10a1 + a0 . Since 100 = 4 · 25, 100 ≡ 4 · 25 ≡ 0 · 25 ≡ 0 (mod 4). Therefore
12
100 · 10m ≡ 0 · 10m ≡ 0 (mod 4) for any m ≥ 0; in other words 10k ≡ 0 (mod 4), for
k = m + 2. Therefore 10m am + · · · + 10a1 + a0 ≡ 10a1 + a0 (mod 4). That is why n ≡ 0
(mod 4) ⇔ 10a1 + a0 ≡ 0 (mod 4). Note that 10a1 + a0 is the number formed by the tens
and units digits of n to obtain the statement.
(d) Let n be an integer. There are m ∈ Z, a0 , a1 , . . . , am ∈ {0, 1, . . . , 9} such that n =
10m am + · · · + 10a1 + a0 . Since 10 = 2 · 5, 10k ≡ 2k · 5k ≡ 0 (mod 5), if k > 0. Hence
10m am + · · · + 10a1 + a0 ≡ a0 (mod 5). The statement is true since a0 ≡ 0 (mod 5) if and
only if a0 = 0 or a0 = 5.
8. For any integer a, show that a2 − a + 7 ends in one of the digits 3, 7, or 9.
Proof. By Division Algorithm, a = 10k + r, where k is an integer and r ∈ {0, 1, . . . , 9}.
Therefore a2 − a + 7 = 102 k 2 + 2 · 10 · r + r2 − 10k − r + 7, which is congruent r2 − r + 7
(mod 10). Consider the following cases:
in case r = 0: a2 − a + 7 ≡ 02 − 0 + 7 ≡ 7 (mod 10),
in case r = 1: a2 − a + 7 ≡ 12 − 1 + 7 ≡ 7 (mod 10),
in case r = 2: a2 − a + 7 ≡ 22 − 2 + 7 ≡ 9 (mod 10),
in case r = 3: a2 − a + 7 ≡ 32 − 3 + 7 ≡ 13 ≡ 3 (mod 10),
in case r = 4: a2 − a + 7 ≡ 42 − 4 + 7 ≡ 19 ≡ 9 (mod 10),
in case r = 5: a2 − a + 7 ≡ 52 − 5 + 7 ≡ 27 ≡ 7 (mod 10),
in case r = 6: a2 − a + 7 ≡ 62 − 6 + 7 ≡ 37 ≡ 7 (mod 10),
in case r = 7: a2 − a + 7 ≡ 72 − 7 + 7 ≡ 49 ≡ 9 (mod 10),
in case r = 8: a2 − a + 7 ≡ 82 − 8 + 7 ≡ 63 ≡ 3 (mod 10), and finally
in case r = 9: a2 − a + 7 ≡ 92 − 9 + 7 ≡ 79 ≡ 9 (mod 10). So we see that in each case
a2 − a + 7 is in one of the congruence classes [3], [7], [9] (mod 10), and therefore ends in
3, 7, or 9.
13
Section 4.4
1. Solve the following linear congruences:
(a) 25x ≡ 15 (mod 29).
(b) 5x ≡ 2 (mod 26).
(c) 6x ≡ 15 (mod 21).
(d) 36x ≡ 8 (mod 102).
(e) 34x ≡ 60 (mod 98).
(f) 140x ≡ 133 (mod 301).
Solution. (a) By Theorem 4.7, 25x ≡ 15 (mod 29) has a solution since gcd(25, 29) =
1 | 15. The solution is [18] (mod 29), since 25 · 18 ≡ 450 ≡ 15 (mod 29). The solution is
unique by Theorem 4.7.
(b) Since gcd(5, 26) = 1, 5x ≡ 2 (mod 26) has a unique solution, namely [20] (mod 26).
(c) Since gcd(6, 21) = 3 | 15, 6x ≡ 15 (mod 21) has three solutions. By trial-and-error we
get [6] (mod 21) as a solution, so by Theorem 4.7 the others are [13] (mod 21) and [20]
(mod 21).
(d) Since gcd(36, 102) = 6, which does not divide 8, there are no solutions.
(e) Since gcd(34, 98) = 2, 34x ≡ 60 (mod 98) has two solutions. By trial-and-error we find
[45] (mod 98) (because 34 · 45 ≡ 1530 ≡ 60 (mod 60)), so the other one is [94] (mod 98).
(f) Since gcd(140, 301) = 7 by the hint, which divides 133, there are seven congruence
classes which elements solve 140x ≡ 133 (mod 301). By trial-and-error, we find [16] (mod
301), because 140 · 16 ≡ 2240 ≡ 133 (mod 301). By Theorem 4.7, the other solutions are
[59] (mod 301), [102] (mod 301), [145] (mod 301), [188] (mod 301), [231] (mod 301), and
[274] (mod 301).
2. Using congruences, solve the following Diophantine equations below:
(a) 4x + 51y = 9.
(b) 12x + 25y = 331.
(c) 5x − 53y = 17.
Solution. (a) 4x + 51y = 9 ⇔ 4x ≡ 9 (mod 51). The solution of this linear congruence is
[15] (mod 51), and with x = 15, we have y = 9−4·15
= −1 we have a solution of 4x+51y = 9.
51
The general solution therefore is x = 15 + 51t, y = −1 − 4t, where t is any integer.
(b) 12x + 25y = 331 ⇔ 12x ≡ 331 ≡ 6 (mod 25). The solution of this linear congruence
is [13] (mod 25). For x = 13, we have y = 331−12·13
= 7, so the general solution is
25
x = 13 + 25t, y = 7 − 12t for any integer t.
(c) 5x − 53y = 17 ⇔ 5x ≡ 17 (mod 53). The solution of this linear congruence is (by
trial-and-error) [14] (mod 53). With x = 14, we get y = 17−5·14
= 1, which leads to the
−53
general solution x = 14 + 53t, y = 1 + 5t, where t is an arbitrary integer.
14
4. Solve each of the following sets of simultaneous congruences:
(a) x ≡ 1 (mod 3), x ≡ 2 (mod 5), x ≡ 3 (mod 7).
(b) x ≡ 5 (mod 11), x ≡ 14 (mod 29), x ≡ 15 (mod 31).
(c) x ≡ 5 (mod 6), x ≡ 4 (mod 11), x ≡ 3 (mod 17).
(d) 2x ≡ 1 (mod 5), 3x ≡ 9 (mod 6), 4x ≡ 1 (mod 7), 5x ≡ 9 (mod 11).
Solution. (a) By Chinese Remainder Theorem, we have to solve 35x1 ≡ 1 (mod 3),
21x2 ≡ 1 (mod 5), 15x3 ≡ 1 (mod 7). Trial-and-error gives us x1 = 2, x2 = 1, x3 = 1,
therefore x̄ = 1 · 35 · 2 + 2 · 21 · 1 + 3 · 15 · 1 = 157 is a solution, so the solution is [52] (mod
105).
(b) Solving 899x1 ≡ 1 (mod 11), 341x2 ≡ 1 (mod 29), 319x3 ≡ 1 (mod 31), we get
x1 = 7, x2 = 4, x3 = 7, so we have x̄ = 5 · 899 · 7 + 14 · 341 · 4 + 15 · 319 · 7 = 31465 + 19096 +
33495 = 84056, so the solution is [4944] (mod 9889).
(c) Solving 187x1 ≡ 1 (mod 6), 102x2 ≡ 1 (mod 11), 66x3 ≡ 1 (mod 17), we get x1 =
1, x2 = 4, x3 = 8, so we have x̄ = 5 · 187 · 1 + 4 · 102 · 4 + 3 · 66 · 8 = 935 + 1632 + 1584 = 4151,
so the general solution is [785] (mod 1122)
(d) Solving 462x1 ≡ 1 (mod 5), 385x2 ≡ 1 (mod 6), 330x3 ≡ 1 (mod 7), 210x4 ≡ 1 (mod
11), we get x01 = 3, x02 = 1, x03 = 1, x04 = 1, considering that x1 must be a multiple of 2, we
have x1 = 8, considering that x3 must be a multiple of 4, we have x3 = 8, considering that
x4 must be a multiple of 5, we have x4 = 45, so we have x̄ = 21 1 · 462 · 8 + 13 9 · 385 · 1 + 41 1 ·
330 · 8 + 15 9 · 210 · 45 = 1848 + 1155 + 660 + 17010 = 20673, so the solution is [2193] (mod
2310).
5. Solve the linear congruence 17x ≡ 3 (mod 2 · 3 · 5 · 7) by solving the system 17x ≡ 3 (mod 2),
17x ≡ 3 (mod 3), 17x ≡ 3 (mod 5), 17x ≡ 3 (mod 7).
Solution. By Chinese Remainder Theorem, we have to solve 105x1 ≡ 1 (mod 2), 70x2 ≡ 1
(mod 3), 42x3 ≡ 1 (mod 5), and 30x1 ≡ 1 (mod 7). By trial-and-error, one solution is
x01 = 1, x02 = 1, x03 = 3, x04 = 4. Considering that xi has to be divisible by 17, we get
x1 = 17, x2 = 34, x3 = 68, x4 = 102, from which we get x̄ = 3 · 105 · 17 + 3 · 70 · 34 + 3 · 42 ·
1
68 + 3 · 30 · 102 = 17
3(1785 + 2380 + 2856 + 3060) = 1779. Hence the solution is [99] (mod
210).
6. Find the smallest integer a > 2 such that 2 | a, 3 | a + 1, 4 | a + 2, 5 | a + 3, 6 | a + 4.
Solution. Since 4 | a + 2 implies 2 | a and 6 | a + 4 implies 3 | a + 1, we do not need these
conditions. So we only have the conditions a ≡ 2 (mod 4), a ≡ 2 (mod 5), a ≡ 2 (mod 6).
Since 4 and 6 are not relatively prime, we cannot use the Chinese Remainder Theorem.
But by trial-and-error, we get 62 as a solution. This is the smallest solution, because a ≡ 2
(mod 6) implies that if 2 < a < 62, a ∈ {8, 14, 20, 26, 32, 38, 44, 50, 56}, but 8 6≡ 2 (mod 5),
14 6≡ 2 (mod 5), 20 6≡ 2 (mod 5), 26 6≡ 2 (mod 5), 32 6≡ 2 (mod 4), 38 6≡ 2 (mod 5), 44 6≡ 2
(mod 5), 50 6≡ 2 (mod 5), and 56 6≡ 2 (mod 5).
15
7. (a) Obtain three consecutive integers, each having a square factor.
(b) Obtain three consecutive integers, the first of which is divisible by a square, the second by a
cube, and the third by a fourth power.
Solution. (a) The condition holds if a ≡ 0 (mod 22 ), a ≡ 8 (mod 32 ), and a ≡ 23
(mod 52 ). Using the Chinese Remainder Theorem, we have x1 = 1, x2 = 1, x3 = 16 as
a solution of 225x1 ≡ 1 (mod 4), 100x2 ≡ 1 (mod 9), 36x3 ≡ 1 (mod 25). Therefore
x̄ = 8 · 100 · 1 + 23 · 36 · 16 = 800 + 13248 = 14048 is a solution, i.e. [548] (mod 900) is the
general solution, therefore 548 = 22 · 137, 549 = 32 · 61, 550 = 52 · 22 are three such integers.
(b) Similarly to part (a), we get x1 = 18, x2 = 16, x3 = 11 as a solution of 432x1 ≡ 1 (mod
25), 400x2 ≡ 1 (mod 27), 675x3 ≡ 1 (mod 16). Therefore x̄ = 26 · 400 · 16 + 14 · 675 · 11 =
166400 + 103950 = 270350, and the general the solution is [350] (mod 10800), and the three
integers could e.g. be 350 = 52 · 14, 351 = 33 · 39, 352 = 24 · 22.
8. (Brahmagupta, 7th century A.D.) When eggs in a basket are removed 2, 3, 4, 5, 6 at a time
there remain, respectively, 1, 2, 3, 4, 5 eggs. When they are taken out 7 at a time, none are left over.
Find the smallest number of eggs that could have been contained in the basket.
Solution. Let x be the number off eggs. We know that x ≡ 1 (mod 2), x ≡ 2 (mod 3),
x ≡ 3 (mod 4), x ≡ 4 (mod 5), x ≡ 5 (mod 6), x ≡ 0 (mod 7), which is equal to x ≡ 4 (mod
5), x ≡ 11 (mod 12), x ≡ 0 (mod 7). To solve this, we use the Chinese Remainder Theorem:
84x1 ≡ 1 (mod 5), 35x2 ≡ 1 (mod 12), 60x3 ≡ 1 (mod 7), so x1 = 4, x2 = 11, x3 = 2, thus
x̄ = 4 · 84 · 4 + 11 · 35 · 11 = 1344 + 4235 = 5579, and the general solution is [119] (mod
420). The smallest number of eggs is 119.
16