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COT3100 Spring 05: Solutions to Problems (3/23/05)
8)
a) The number of divisors, using the formula from class is
(14+1)(9+1)(8+1)(10+1)(3+1)(5+1)(10+1) = 3920400
b) i) All of these numbers have the form 23+a34+b57+c7d112+e13f372+g, where we have the
following restrictions on a, b, c, d, e, f and g:
0  a  11, 0  b  5, 0  c  1, 0  d  10, 0  e  1, 0  f  5, 0  g  8.
Each of these was determined by the fact that each exponent of a divisor must be at least
as big as the corresponding exponent in the number itself, but no exponent could exceed
the corresponding value in the originally given number in part (a).
The total number of 7-tuples (a,b,c,d,e,f,g) that satisfy these conditions using the
multiplication principle is
(11+1)(5+1)(1+1)(10+1)(1+1)(5+1)(8+1) = 171072
ii) 1166400000 = 11664x100000 = 729x16x100000=36242555 = 293655.
Using the same reasoning as in part (i), we find the total number of divisors of the value
above that are also divisors of the original number is:
(14 - 9 + 1)(9 - 6 + 1)(8 - 5 + 1)(10 + 1)(3 + 1)(5 + 1)(10 + 1) = 278784
iii) Here, we must restrict each exponent to be even. The number of even values k that
satisfy the inequality 0  k  n, is (n+2)/2, where we compute an integer division.
(Alternatively, this is floor((n+2)/2).) Calculating the appropriate product, we get
(14+2)/2x(9+2)/2x(8+2)/2x(10+2)/2x(3+2)/2x(5+2)/2x(10+2)/2 = 43200
iv) Same questions as (iii) except we must adjust our ranges for possible values for our
exponents, noting that our squares must be of the form 22+a34+b52+c7d112+e13f37g, here are
the restrictions of a, b, c, d, e, f, and g:
0  a  12, 0  b  5, 0  c  6, 0  d  10, 0  e  1, 0  f  5, 0  g  10.
The total number of perfect squares we can create with these exponents is:
(12+2)/2x(5+2)/2x(6+2)/2x(10+2)/2x(1+2)/2x(5+2)/2x(10+2)/2 = 9072
v) We must count the number of exponents in the given range that are divisible by 3.
We can count these manually and multiply them to get:
5x4x3x4x2x2x4 = 3840
vi) Let these perfect cubes be of the form 2a3b5c7d11e13f37g, then here are the values
we must have for each variable:
a= 12, b= 9, c = 3 or 6, d = 6 or 9, e = 3, f = 3, g = 3, 6 or 9
Thus, there are a total of 2x2x3 = 12 such perfect cubes
vii) These are perfect 6th powers. Using the same set up as in (vi), here are the possible
values for a, b, c, d, e, f and g:
a = 0, 6, 12
b = 0, 6
c = 0, 6
d = 0, 6
Total number of perfect 6th powers = 3x2x2x2x2 = 48.
e=0
f=0
g= 0, 6