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Math 1300 Fall 2009
Assignment 3 Solutions
1
2.4 #3
16 ≡ 2 mod 7
24 ≡ 3 mod 7
(24)(16) = 384
≡ 6 mod 7
(16 mod 7)(24 mod 7) = 6
The last two answers are equal.
2
2.4 #13
If the covered digit were D, then the sum would be 55 + 3D. The goal is to
find a value for D that will make the sum divisible by 10, which is easily found
by trial and error. D = 5 is the only digit between 0 and 9 that works. (Note
that we are solving the equation 55 + 3D ≡ 0 mod 10, or equivalently, 3D ≡ 5
mod 10.)
3
2.4 #25
In the first code, the 1st and the 4th digits are changed; instead of 7 × 0 + . . . +
7 × 7 + . . ., we have 7 × 7 + . . . + 7 × 0 + . . . Thus, the sum remains unchanged.
The same type of mistake occurs in the second example where the 6th and 9th
terms are interchanged. Because the 6th and 9th terms are both multiplied by
the same weight, 9, the total sum will remain unchanged.
4
2.4 #29
We have 10061559129884 = 1437365589983 × 7 + 3, so the check digit is 3.
5
2.6 #18
If E = n/m, then 13n = 8m . The prime factorization of the left side is just
a bunch of 13s, whereas the prime factorization of the right side is composed
solely of 2s. Because this represents two different prime factorizations of the
same number, we have derived a contradiction to the uniqueness of prime factorizations. Therefore, our assumption was wrong; E is not rational.
1
6
2.6 #28
If we assume π is rational, this means that π = n/m, where n, m are integers.
So, π 2 = n2 /m2 , which means that we have expressed π 2 as a rational number,
which is a contradiction. We must conclude that our original assumption is
wrong. The only alternative is that π is not rational.
7
2.6 #33
Not always; the counterexample is that π and −π are both irrationals, yet their
sum is 0, which is rational. Of course, the sum is also sometimes irrational:
π + π = 2π.
8
2.6 #34
Sometimes √
the√product
√ is rational, as in
irrational: 2 3 = 6.
9
√ √
2 2 = 2. However, sometimes it is
2.6 #35
Assume that the sum of an irrational and a rational is rational, and try to
derive a contradiction. Let q represent the irrational number, let a/b represent
the rational, and because we are assuming that the sum is rational, let c/d
represent the sum. We then have q + a/b = c/d. Solving for q gives q =
c/d − a/b = (cb − ad)/bd. We’ve just expressed q as the quotient of two integers,
which contradicts the assumption that q represents an irrational number. This
contradiction proves our assumption wrong. The only alternative is that the
original sum is irrational.
10
Extra Credit
Suppose there are 3 colors: white, black, and red. The prisoners agree that
white = 0, black = 1, and red = 2. The first prisoner’s guess encodes the ”sum”
of all the colors he can see, modulo 3. That is, he looks at all the hats, converts
colors to numbers, adds them all up and reduces mod 3, then converts this
number to a color and guesses that color. Every other prisoner can then figure
out what color his hat is using modular arithmetic on the hats he can see and
the fact that he can hear the guess that came before his. A similar strategy
works for an arbitrary number of colors: if there are m colors, do arithmetic
modulo m.
2