Download Math 107A Name: Sec # HW #13 December 8, 2010 Score: 1. (a

Math 107A
HW #13
December 8, 2010
Name:
Sec #
Score:
1. (a) Find the prime factorization of 10.
10 = 2 · 5
(b) Find the prime factorization of 100.
100 = 102 = (2 · 5)2 = 22 · 52
(c) Find the prime factorization of 1000.
1000 = 103 = (2 · 5)3 = 23 · 53
(d) Looking at the work you’ve done above, what do you think the prime factorization of
100,000,000 is? Explain how you came up with your answer without doing any computations.
From looking at the previous problems we see that the prime factorization of any power
of ten will have equal amounts of 2’s and 5’s, and we can get that amount by looking at
the number of zeros. Therefore the prime factorization of 100,000,000 is 28 · 58.
(e) A googol is the number 1 with a hundred zeros after it. What is the prime factorization
of a googol?
Using the same idea as above the prime factorization of a googol is 2100 · 5100.
2. Use the prime factorization method to find the greatest common divisor of the following
numbers. Please show your work.
(a) 132 and 504
132 = 22 · 3 · 11
504 = 23 · 32 · 7
GCD = 22 · 31 = 12
(b) 900 and 96 and 630
900 = 22 · 32 · 52
96 = 25 · 3
630 = 2 · 32 · 5 · 7
GCD = 21 · 31 = 6
3. For each of the following problems you should not just be guessing and checking. You should
have some sort of strategy in mind. After giving the answer to the problem, please explain
the strategy you used.
(a) Find two whole numbers A and B such that AB = 1, 000, 000 and neither A nor B
contain any zeros as digits.
From the above problem we know the prime factorization of 1,000,000 is
1, 000, 000 = 26 · 56
Now 26 = 64 and 56 = 15625. Neither of these have any zeros and 64·16525 = 1, 000, 000.
(b) Find a number that has exactly three factors.
Let’s use the partner factors idea here. We know 1 and the number are factors. So we
automically have two factors there. We only want one more factor, so that factor must
be its own partner. Moreover, we don’t want it to have any more factors so it must
be a prime number. Therefore we want a number such that it equals a prime number
squared.
There are lots of possible answers here. For example 9 = 32. The factors of 9 are 1,3
and 9. So 9 has exactly three factors.
(c) Find a three digit number that has exactly three factors.
Using the same strategy we want a number that is equal to a prime squared. We want
to get a three digit number, so 112 = 121 will work. 121 is a three digit number and its
only divisors are 1, 11, and 121. There are lots of other possible answers here as well.
(d) Find the largest four digit number that has exactly three factors.
Again using the same idea as above we want a prime number squared that is four digits
and there are no bigger four digit numbers of that form.
Using my Sieve of Eratosthenes I am going to keep squaring primes until I find the
biggest four digit of that form.
972 = 9409, so we have found a four digit number of the correct form. Moreover, the
next prime on the list is 101 and 1012 = 10201 which is not a four digit number. Thus
9409 is the largest four digit number that has exactly three factors.
4. Use the prime factorization method to find the least common multiple of the following numbers. Please show your work.
(a) 132 and 504
132 = 22 · 3 · 11
504 = 23 · 32 · 7
LCM = 23 · 32 · 7 · 11 = 5544
(b) 900 and 96 and 630
900 = 22 · 32 · 52
96 = 25 · 3
630 = 2 · 32 · 5 · 7
LCM = 25 · 32 · 52 · 7 = 50400
5. Find the smallest positive number that is divisible by 2, 3, 4, 5, 6, 7, 8, 9, 10, and 11.
All I am asking for is the least common multiple of these numbers. Let’s look at the prime
factorization of each of the numbers and compute the LCM.
2
3
4 = 22
5
6 = 2·3
7
8 = 23
9 = 32
10 = 2 · 5
11
LCM = 23 · 32 · 5 · 7 · 11 = 27720
6. You computed the GCD and LCM of 132 and 504 in previous problems. You will be using
those answers in the following questions.
(a) What is the product of 132 and 504?
(b) What is the product of their GCD and their LCM?
(c) You should have found that the two products above are equal. That is in fact true for
any two whole numbers. In other words, the product of two numbers is equal to the
product of their GCD and LCM. Check this fact with the numbers 90 and 24.
(d) Why do you think these two products will always be equal for any two numbers?