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MATH 145: Homework Solutions #6

1. Brualdi 6.2

Find the number of integers between 1 and 10,000 inclusive which are not divisible by
4,6,7, or 10.

Answer:

Let be the set of integers between 1 and 10,000 that are divisible by 4, be the
set of integers between 1 and 10,000 that are divisible by 6, be the set of integers
between 1 and 10,000 that are divisible by 7 and let be the set of integers between
1 and 10,000 that are divisible by 10. Then by the inclusion-exclusion principle the
number of integers between 1 and 10,000 inclusive which are not divisible by 4,6,7, or
10 is

because:

2. Brualdi 6.3

Find the number of integers between 1 and 10,000 which are neither perfect squares
nor perfect cubes .

Answer:

Let S = {1, 2, ..., 10000} be the set of all integers between 1 and 10,000. Then |S| =
10000. Let be the set of all perfect squares in S . Then since
all integers less than 100 and their perfect squares are in S. Let be the set of all
perfect cubes in S . Then since 9261 is the largest number that
is a perfect cube in S . Now are integers in S that are both perfect squares
and perfect cubes. Therefore if n is in , and the prime factorization of n is
then each exponent is divisible by 6. That is are integers
in S that are 6 th powers of an integer . The largest integer which is 6th power in S is
4096. so .

Therefore by the inclusion-exclusion principle we have the number of integers between
1 and 10,000 which are neither perfect squares nor perfect cubes is

3. Brualdi 6.7

Determine the number of solutions of the equation in non-negative
integers and not exceeding 8.

Answer:

Let S be the set of all non- negative integral solutions of the equation .
Then, the number of non-negative integral solutions of the given equation is ,


Let the set consist of solutions in S for which We make a change of variable ,
to get|| which is the same as the number of non-negative solutions of
the equation
. Therefore

Similarly, if we let be the set of solutions in S for which , be the set of
solutions in S for which and be the set of solutions in S for which ,
we get

The set consists of solutions in S which have and . Let
. Then, || is the same as the non-negative integral
solutions of the equation

Thus, || = 0. Similarly, we can easily verify that

By inclusion-exclusion principle we get that the number of solutions of the equation
in non-negative integers and not exceeding 8 is

(It is quite obvious that the sets , . . . , are disjoint because if two numbers are
greater than 8, then the sum of them together with other non-negative integers exceeds
14. Thus 680 − 4 × 56 = 456.)
 

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