Math 126, Number Theory
First Test
22 Feb 2006
Scale. 90–101 A. 80–89 B. 60–79 C. Median 94.
Problem 1. On divisors. [18; 6 points each part]
a. Draw the Hasse diagram of the divisors of 496 = 24·31.
1
2
4
8
16
31
62
124
248
496
b. What are the values of d(496) and σ(496).
d(496) is the number of divisors of 496, and you can see in
your diagram in part a that there are are 10 of them. σ(496)
is the sum of those divisors, which you could compute by
adding all the divisors, but there’s an easier way. Since 24
and 31 are relatively prime, therefore σ(496) = σ(24)σ(31) =
31 ·32 = 992.
c. Is the number 496 a perfect number? Why or why not?
A number nis perfect if σ(n) = 2n. Since σ(496) = 992 =
2·496, therefore 496 is perfect.
Problem 2. On the Euclidean algorithm. [20; 10 points
each part] The Euclidean algorithm shows that the greatest
common divisor of 399 and 703 is 19. Here are the compu-
tations.
703 −399 = 304
399 −304 = 95
304 −3·95 = 19
95 −5·19 = 0
a. Express 19 as a linear combination of 399 and 703.
19 = 304 −3·95
= 304 −3·(399 −304)
= 4 ·304 −3·199
= 4 ·(703 −399) −3·399
= 4 ·703 −7·399
b. Find all the integral solutions of the linear Diophantine
equation 399x+ 703y= 19.
From part a a particular solution to the equation is
x0=−7, y0= 4.
Therefore, the general solution is
x=x0+b
dt, y =y0+a
dt
where a= 399, b= 703, and d= (a, b) = 19, and tis an
arbitrary integer, so we can write the general solution as
x= 4 + 37t, y = 4 −21t.
Problem 3. On divisibility. [15] Recall that we say that
one positive integer adivides another b, written a|b, if there
exists a third integer csuch that ac =b. Carefully prove the
following theorem. (Note that the theorem has two parts.)
Theorem. Let a,b, and dbe positive integers. If a|b, then
ad|bd. Conversely, if ad|bd then a|b.
There are various ways you can prove this theorem. If
you’re careful, you can prove both the statement and its
converse in the same time by using only if and only if state-
ments. Here’s a proof where each half is proved separately.
Proof =⇒:Suppose that a|b. Then there exists an integer
csuch that ac =b. Multiplying both sides of that equation
by dwe find that adc =bd. Therefore, ad|bd.
Proof ⇐=:Suppose that ad|bd. Then there exists an in-
teger csuch that adc =bd. Since dis positive, we can
divide both sides of that equation by dto find that ac =b.
Therefore a|b.q.e.d.
Problem 4. True or false. [15; 3 points each part]
a. A function fdefined for all positive integers is said to
be multiplicative if f(ab) = f(a)f(b) whenever a|b.
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