Raji, Exercises 2.4:
1. Find the least common multiple of 14 and 15.
As gcd (14; 15) = 1, we have lcm (14; 15) = (14 15) =1 = 210. Alternatively, 14 = 21 30 50 71 and 15 = 20 31 51 70 , so taking the maximum of the
exponents of each prime we get lcm (14; 15) = 21 31 51 71 = 2 3 5 7 = 210.
2. Find the least common multiple of 240 and 610.
Using the Euclidean algorithm to compute gcd (240; 610) we get the sequence 610, 240, 130, 110, 20, 10 = gcd (240; 610). So lcm (240; 610) =
(240 610) =10 = 24 610 = 14 640. Alternatively, 240 = 24 3 5 and
610 = 2 5 61, so lcm (240; 610) = 24 3 5 61 = 14 640.
3. Find the least common multiple of 25 56 72 11 and 23 58 72 13.
Taking the maximum of the exponents of each prime, we get 25 58 72 13.
My word processor tells me that 25 58 72 13 = 7962 500 000.
4. Show that every common multiple of two positive integers a and b is divisible by the least common multiple of a and b.
This is part 3 of Theorem 18. Raji, in his hint at the end of the proof of
Theorem 18, suggests doing this by doing Exercise 6 …rst. I suspect that
Raji misstated Exercise 6. To make it consistent with the hint, Exercise
6 should say that ab= (a; b) divides m if m is any common multiple of a
and b, not just the least common multiple, which Raji denotes by ha; bi.
After all, Raji has proved in part 2 of Theorem 18 that ha; bi = ab= (a; b),
so the revised Exercise 6 is just what you need for Exercise 4. I proved
this revised Exercise 6 in class, and prove it again below.
5. Show that if a and b are positive integers, then the greatest common divisor of a and b divides their least common multiple. When are the least
common multiple and the greatest common divisor equal to each other ?
As a is a multiple of gcd (a; b), we can write a = gcd (a; b) i. As lcm (a; b)
is a multiple of a, we can write lcm (a; b) = aj. it follows that lcm (a; b) =
gcd (a; b) ij. For the second part, if lcm (a; b) = gcd (a; b), then ij = 1, so
i and j are 1, so lcm (a; b) = a = gcd (a; b). By symmetry, lcm (a; b) = b =
gcd (a; b). So a = b.
6. The revised version reads: ab= (a; b) divides m if m is any common multiple of a and b.
We want to show that ab= (a; b) divides m. Equivalently, we want to show
that ab divides (a; b) m. Bezout’s equation says that sa + tb = (a; b)
for some integers s and t (which Blankinship’s algorithm will compute).
Multiply this equation by m to get sam + tbm = (a; b) m. But b divides
m, so ab divides am, and a divides m, so ab divides bm. Thus ab divides
sam + tbm which is equal to (a; b) m. Yay!
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Raji, Exercises 3.1:
1. Determine whether 3 and 99 are congruent modulo 7.
By de…nition, we need to see if 7 divides 3
14 7 + 2, so the answer is no.
2. Show that if x is an odd integer, then x2
99 =
96. But
96 =
1 (mod 8).
If x is an odd integer, then x = 2n + 1 for some integer n. So x2 =
2
(2n) + 2 (2n) + 1 = 4 n2 + n + 1. As n2 + n = n (n + 1), it is even
(either n or n + 1 is even). So x2 1 = 4 n2 + n is divisible by 8, that
is, x2 1 (mod 8).
3. Show that if a, b, m and n are integers such that m and n are positive,
n j m and a b (mod m), then a b (mod n).
If a b (mod m), then m j (a
a b (mod n).
b). If also n j m, then n j (a
b), that is
4. Show that if ai
bi (mod m) for i = 1; 2; : : : ; n, where m P
is a posin
tive
integer
and
a
,
b
are
integers
for
i
=
1;
2;
:
:
:
;
n,
then
i
i
i=1 ai
Pn
i=1 bi (mod m).
As m divides ai bi for
m divides
P
Pneach i, the
Pndistributive
Pn law shows thatP
n
n
(a
b
).
But
a
b
=
(a
b
),
so
i
i
i
i
i
i
i=1
i=1
i=1
i=1
i=1 ai
Pn
i=1 bi (mod m).
5. For which n does the expression 1 + 2 +
+ (n
1)
0 (mod n) hold ?
Probably the quickest way to do this is to use the fact that
1+2+
+ (n
1) =
(n
1) n
2
To verify this fact, in case you don’t know it, let s = 1 + 2 +
+
(n 2) + (n 1), and note that s = (n 1) + (n 2) +
+ 2 + 1.
Putting these two expressions for s one above the other, and adding, shows
that 2s = n + n +
+ n where there are n 1 terms in that sum. So
2s = (n 1) n.
Finally, note that
(n
1) n
2
1 is even. So the answer to question is:
is divisible by n exactly when n
for those n that are odd.
Clark, Exercises
15.1 Prove that for all m > 0 and all a:
a
a mod m (mod m) .
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What does this mean? A common use of the expression a mod m, especially in some computer languages, is to denote the remainder when you
divide a by m, according to the division algorithm. Clark introduced this
usage earlier in a part I did not assign. So a mod m = r if a = mq + r for
some q and r, where 0 r < m. Clearly if a mod m = r, then a r = mq
is divisible by m, that is, a
r (mod m). So we have proved what we
wanted to prove.
15.2 Using De…nition 15.1 show that the following congruences are true:
385
322 (mod 3)
385
322 (mod 3)
1
17 (mod 3)
33
0 (mod 3)
For the …rst, 385 322 = 63 = 3 21. For the second, 385 ( 322) =
63 = 3 ( 21). For the third, 1 ( 17) = 18 = 3 6, and for the fourth,
33 0 = 33 = 3 11.
15.3 Use Theorem 15.1 to show that the congruences in Exercise 15.2 are valid.
Theorem 15.1 says that a b (mod m) if and only if a mod m = b mod m.
For the …rst congruence, note that 385 = 128 3+1 and 322 = 107 3+1, so
385 mod 3 = 1 = 322 mod 3. For the second, note that 385 = ( 129) 3+2
and 322 = ( 108) 3 + 2, so ( 385) mod 3 = 2 = ( 322) mod 3. For the
third, 1 = 0 3 + 1 and 17 = ( 6) 3 + 1, so 1 mod 3 = 1 = ( 17) mod 3,
For the fourth, 33 = 11 3 + 0 and 0 = 0 3 + 0, so 33 mod 3 = 0 = 0 mod 3.
15.4 (a) Show that a is even , a 0 (mod 2) and a is odd , a 1 (mod 2).
a is even , 2 divides a , 2 divides a 0 , a 0 (mod 2).
a is odd , 2 divides a 1 , a 1 (mod 2).
(b) Show that a is even , a mod 2 = 0 and a is odd , a mod 2 = 1.
a is even , a = 2n + 0 for some n , a mod 2 = 0.
a is odd , a = 2n + 1 for some n , a mod 2 = 1.
15.5 Show that if m > 0 and a is any integer, there is a unique integer r 2
f0; 1; 2; : : : ; m 1g such that a r (mod m).
The division algorithm tells us that there are unique integers q and r such
that a = qm + r and r 2 f0; 1; 2; : : : ; m 1g. But a = qm + r for some q
exactly when a r is divisible by m, which means that a r (mod m).
15.6 Find integers a and b such that 0 < a < 15, 0 < b < 15, and ab
0 (mod 15).
We need ab to be divisible by 15, so take a = 3 and b = 5. We could also
take a = 12 and b = 10.
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15.7 Find integers a and b such that 1 < a < 15, 1 < b < 15, and ab
1 (mod 15).
This is a little trickier, but there are many candidates so it’s easy to …nd
one pair. For example 2 8
1, 4 4
1, 7 13
1, 11 11
1, and
14 14 1.
15.10 Find the value of each of the following.
(a) 232 mod 7.
22 = 4, 23 = 8
1. So 232 = 23 10+2 = 23
10
22
110 4 = 4.
(b) 1035 mod 7.
10 3 (mod 7) so 1035 335 and we can focus on the latter. Look at
the successive powers of 3, each number being 3 times the preceding
number: 3, 2, 6, 4, 5, 1. So 36
1. That means 335 = 36 5+5 =
5
36 35 15 5 = 5.
(c) 335 mod 7.
Hmm. Didn’t I just prove that in part b?
15.11 Let gcd (m1 ; m2 ) = 1. Prove that
a
b (mod m1 ) and a
b (mod m2 )
if and only if
a
b (mod m1 m2 )
We know that if gcd (m1 ; m2 ) = 1, then lcm (m1 ; m2 ) = m1 m2 . The …rst
condition says that a b is a common multiple of m1 and m2 . The second
condition says that a b is a multiple of lcm (m1 ; m2 ). But we know
that the common multiples of m1 and m2 are exactly the multiples of
lcm (m1 ; m2 ).
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