| > 8. For any sequence of real numbers A = (a[1], a[2], a[3], . . .),
> define dA to be the sequence (a[2]-a[1], a[3]-a[2], a[4]-a[3], . . .),
> whose n-th term is a[n+1]-a[n]. Suppose that all of the terms of the
> sequence dA are 1, and that a[19] = a[92] = 0. Find a[1].
>
> (8 must have a mistake, yes? Can anybody figure it out?)
Perhaps it originally said "...sequence dA are +/-1..." and the
non-ascii "+/-" character got stripped out along the way?
Dan
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| Re. >5. Let S be the set of all rational numbers, 0 < r < 1, that have a
>repeating decimal expansion of the form 0.abcabcabc..., where the
>digits a, b, and c are not necessarily distinct. To write the elements
>of S as fractions in lowest terms, how many different numbers are
>required?
1000*S=abc.abcabc..., so
1000*S - S = abc
abc
S = --- then if a+b+c divisible by 9 one can reduce this fraction.
999
|
| > 5. Let S be the set of all rational numbers, 0 < r < 1, that have a
> repeating decimal expansion of the form 0.abcabcabc..., where the
> digits a, b, and c are not necessarily distinct. To write the elements
> of S as fractions in lowest terms, how many different numbers are
> required?
If r = 0.abcabcabc... then r = abc/999, not necessarily
in reduced terms. 999 = 27 * 37 = 3^3 * 37, so its
divisors are {1,3,9,27}*{1,37} = {1,3,9,27,37,111,333,999}.
For denominator 37, the numerators will be {1,...,36}.
For 999, the numerators will be the (37-1)*(27-9) = 648
smaller positive integers which are relatively prime to
it. 24 of these values will be in {1,...,36}. This
yields 648 - 24 + 36 = 660 numerators. The numbers
needed for denominators that aren't used as numerators
are {37,111,333,999}.
Therefore 664 different numbers are needed to write the
elements of S as fractions in lowest terms.
Dan
|
| > 6. For how many pairs of consecutive integers in { 1000, 1001, 1002, .
> . ., 2000 } is no carrying required when the two integers are added?
There are 4 types of number pairs here. Their four digits go as
follows:
thousands hundreds tens ones #pairs of this type
2,1 0,9 0,9 0,9 1
1,1 n+1,n 0,9 0,9 5
1,1 same n+1,n 0,9 25
1,1 same same n+1,n 125
---
So the answer is 156
> 7. Faces ABC and BCD of tetrahedron ABCD meet at an angle of 30
> degrees. The area of face ABC is 120, the area of face BCD is 80, and
> BC = 10. Find the volume of the tetrahedron.
Because of the 30 degree angle, we know that the height from face ABC
to D is exactly half the height drawn from the line BC up to D (within
the BCD triangle). The latter height is 16, so the height of the
tetrahedron is 8, assuming a base of ABC.
The volume of a tetrahedron is 1/3 * base * height, so
volume of ABCD = 1/3 * 120 * 8 = 320
Jon
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