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SOLUTION SET #3 FOR MATH 55
Note.
Any typos or errors in this solution set should be reported to the GSI at
isammis@math.berkeley.edu
2.4.10.
What are the quotient and remainder when
(a) 19 is divided by 7?
(b) 111 is divided by 11?
(c) 789 is divided by 23?
(d) 1001 is divided by 13?
(e) 0 is divided by 19?
(f) 3 is divided by 5?
(g) 1 is divided by 3?
(h) 4 is divided by 1?
Solution.
The key thing for this entire problem is to remember the deﬁnition of the
division algorithm, which requires the remainder
r
of
a
divided by
b
to be in the
range 0
≤
r < b
. So:
(a) Quotient 2, Reminder 5: 19 = 2(7) + 5.
(b) Quotient 11, Remainder 10:

111 =

11(11) + 10
(c) Quotient 34, Remainder 7: 789 = 34(23) + 7
(d) Quotient 77, Remainder 0: 1001 = 77(13) + 0
(e) Quotient 0, Remainder 0: 0 = 0(19) + 0
(f) Quotient 0, Remainder 3: 3 = 0(5) + 3
(g) Quotient 1, Remainder 2:

1 =

1(3) + 2
(h) Quotient 4, Remainder 0: 4 = 4(1) + 0
2.4.14.
How many zeros are there at the end of 100!?
Solution.
A zero at the end of the factorial represents a 10 divisor. Since twos
appear in the prime factorization of 100! far more frequently than do ﬁves, we
can simply count the number of factors of 5 that must appear in 100!’s prime
factorization.
First, every multiple of ﬁve contributes one factor of ﬁve to 100!. Since there
are 100
/
5 = 20 mutliples of ﬁve in 100!, we have twenty factors of ﬁve. Now, each
multiple of 25 contributes an additional factor of ﬁve, so we have four more, for a
grand total of 24 factors of 5.
Again, there are many more than 24 factors of two in 100! (there are 50 just
from multiples of two, before even counting the additional factors from multiples
of 4, 8, 16, and so forth). Thus, ever factor of ﬁve can be matched with a factor of
two, and results in a zero at the end of 100!. 100! thus terminates in a string of 24
zeros.
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SOLUTION SET #3 FOR MATH 55
2.4.16.
Which positive integers less than 12 are relatively prime to 12?
Solution.
Since 12 = 2
2
3
,
the integers that aren’t multiples of two or three are
relatively prime to 12. Those are
{
1
,
5
,
7
,
11
}
.
2.4.20.
We call a positive integer perfect if it equals the sum of its positive divisors
other than itself.
(a) Show that 6 and 28 are perfect
(b) Sow that
2
p

1
(2
p

1)
is a perfect number when
2
p

1
is prime.
Solution.
(a) The positive divisors of 6 (other than 6) are 1, 2, and 3. 1+2+3=6, so 6 is
perfect.
The positive divisors of 28 (other than 28 itself) are 1, 2, 4, 7, and 14. Since
14 + 7 + 4 + 2 + 1 = 28, 28 is also perfect.
(b) Suppose 2
p

1 is prime. Then the factors of 2
p

1
(2
p

1) other than the
number itself are 2
j
, where
j
= 0
,
1
,... ,p

1 and (2
p

1)2
j
for
j
= 0
,
1
,... ,p

2.
Summing these, we see
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 Spring '11
 MichealAlain
 Math, Remainder

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