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Math 335: solutions to assignment 2
1. (2.1–16) There are 100 numbers less than 1000 with a 3 in the hundreds place (300–399).
In every group of 100 consecutive numbers less than 1000 (0–99, 100–199, ..., 900–999),
there are at least 10 numbers with a 3 in the tens place (30–39, 130–139, ..., 930–939). There
are exactly 10 such numbers in the 10 − 1 = 9 groups that haven’t yet been counted (all of
them except 330–339). This gives us 9 × 10 = 90 additional numbers containing a 3.
In every group of 10 consecutive numbers between two multiples of 100, there is at least
1 number with a 3 in the ones place. This gives us 9 × 9 = 81 additional numbers.
In total, we have found
100 + 90 + 81 = 271
(1)
numbers less than 1000 containing a 3. Thus, the proportion of such numbers i
271/1000 = 27.1%.
(2)
2. (2.1–26) The total number of pennies is
1 + 3 + 32 + 33 + 34 + 35 + 36 + 37 + 38 = 9841.
(3)
3. (2.2–6) The first few Fibonacci numbers are given by
1, 1, 1 + 1 = 2, 2 + 1 = 3, 3 + 2 = 5,
5 + 3 = 8, 8 + 5 = 13, 13 + 8 = 21, 21 + 13 = 34.
4. (2.2–16) One set of solutions is given by
52 = 34 + 13 + 5
143 = 89 + 34 + 13 + 5 + 2
13 = 8 + 3 + 2
88 = 55 + 21 + 8 + 3 + 1.
(4)
(5)
(6)
(7)
5. (2.2–19) Writing 100 = 89 + 8 + 3, we see that a good first move would be to take 3 sticks.
6. (2.2–46) One way to proceed is as follows. Multiplying both sides of
x=1+
1
6
x
(8)
by x gives
x2 = x + 6.
(9)
Subtracting x + 6 from both sides then gives
x2 − x − 6 = 0
(10)
which is a quadratic equation. We provide two ways to solve this equation.
(i) One way to solve it is to notice that the left-hand side can be factored as
x2 − x − 6 = (x − 3)(x + 2).
(11)
(x − 3)(x + 2) = 0.
(12)
Thus, we wish to solve
But the left-hand side is 0 when either of the factors is 0, i.e. when x − 3 = 0 or x + 2 = 0.
Thus, the solutions are given by
x = 3, −2.
(13)
(ii) Alternatively, we could use the quadratic formula, which states that
ax2 + bx + c = 0
whenever
x=
−b ±
√
b2 − 4ac
.
2a
With a = 1, b = −1, and c = −6, this yields
√
1±5
1 ± 1 + 24
=
.
x=
2
2
Thus,
x=
(14)
6
−4
= 3,
= −2.
2
2
(15)
(16)
(17)
7. (2.3–8) A prime multiplied by a prime can never be prime. It is always a composite
number with exactly two factors (or one repeated factor if the prime was multiplied by
itself—but this still makes it composite): the two primes that were multiplied together.
A nonprime multiplied by a nonprime never results in a prime, essentially for the same
reason as above.
Note: the number 1 is not prime. The definition of a prime number as “a number with
exactly two factors” explicitly excludes 1, which has only one factor.
8. (2.3–16) Here are two possible solutions:
(i) A well-known fact is that a number is divisible by 3 if and only if its digits sum to a
multiple of three. Thus, 111 . . . 1 will be divisible by 3, and hence non-prime, whenever the
number of digits it contains is divislble by 3.
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(ii) Notice that, if 111 . . . 1 has an even number of digits and at least four digits, then
111 . . . 1 = 1010 . . . 10 + 0101 . . . 01
= 10 × 0101 . . . 01 + 1 × 0101 . . . 01
= 11 × 0101 . . . 01,
where all the numbers above containing an “ellipsis” (three dots) contain the same number
of digits, possibly including a single leading zero. Thus, 111 . . . 1 is non-prime (since it has
at least four digits, it does not equal 11).
Note: Just because 111 . . . 1 has an odd number of digits, doesn’t mean it’s prime. For
instance,
11111 = 41 × 271.
(18)
9. (2.3–25) Let n be the number in question. Then we are told that n is 52 plus an integer
multiple of 91, i.e.
n = 91m + 52
(19)
for some integer m. Thus,
n + 103 = 91m + 155.
(20)
Since 91 = 7 × 13 and 155 = 7 × 22 + 1,
n + 103 = 7 × 13m + 7 × 22 + 1 = 7(13m + 22) + 1.
(21)
In other words, n + 103 is 1 plus an integer multiple of 7. Thus, dividing n + 103 by 7 leaves
a remainder of 1.
10. (2.3–34) Since A and B have remainders a and b (respectively) when divided by n, there
are integers M1 and M2 such that
A = M1 n + a
B = M2 n + b.
(22)
(23)
Multiplying A and B together using the above gives
AB = M1 M2 n2 + M1 nb + M2 na + ab.
(24)
Now let c be the remainder when ab is divided by n. Then there is an integer m such that
ab = mn + c.
(25)
Using this in our expression for AB above, we get
AB = M1 M2 n2 + M1 nb + M2 na + mn + c.
(26)
Factoring out n from all but the last term,
AB = (M1 M2 n + M1 b + M2 a + m)n + c.
(27)
Thus, the remainder when AB is divided by n is c, which is the same as the remainder when
ab is divided by n.
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