| > 4. In Pascal's triangle, each entry is the sum of the two entries above
> it. In which row of Pascal's triangle do three consecutive entries
> occur that are in the ratio 3 : 4 : 5?
r!/(i-1)!/(r-i+1)! : r!/i!/(r-i)! : r!/(i+1)!/(r-i-1)! :: 3 : 4 : 5
(i-1)!(r-i+1)! : i!(r-i)! : (i+1)!(r-i-1)! :: 1/3 : 1/4 : 1/5
(r-i+1)(r-i) : i(r-i) : (i+1)i :: 1/3 : 1/4 : 1/5
(r-i+1)/i = 4/3, (r-i)/(i+1) = 5/4
r = 7i/3 - 1, r = 9i/4 + 5/4
=> i = 27, r = 62
The values C(62,26) = 209769429934732479, C(62,27) = 279692573246309972,
and C(62,28) = 349615716557887465 are in the ratio 3 : 4 : 5.
In general, consecutive entries in the ratio A : B : C can be found at
C(r,i-1), C(r,i), C(r,i+1)
where r = (AB + 2AC + BC)/(B�-AC) and i = (AB + AC)/(B�-AC), when these
values of r and i are integers.
|
| > 1. Find the sum of all positive rational numbers that are less than 10
> and that have denominator 30 when written in lowest terms.
This is sum n+i/30 for n in {0,...,9} and i in {1,7,11,13,17,19,23,29}.
The sum is 8 times the sum of {0,...,9} plus 10/30 times the sum
of {1,7,11,13,17,19,23,29}, which is 8 * 45 + 1/3 * 120 or 400.
> 2. A positive integers is called "ascending" if, in its decimal
> representation, there are at least two digits and each digit is less
> than any digit to its right. How many ascending positive integers are
> there?
The count for k digits is just the number of ways of choosing
9-k digits out of "123456789" to be deleted (leaving a k-digit
ascending positive integer). So the total number is sum(k=2,...,9)
C(9, 9-k) = 2^9 - C(9,8) - C(9,9) = 502.
> 3. A tennis player computes her "win ratio" by dividing the number of
> matches she has won by the total number of matches she has played. At
> the start of a weekend, her win ratio is exactly .500. During the
> weekend she plays four matches, winning three and losing one. At the
> end of the weekend her win ratio is greater than .503. What is the
> largest number of matches that she could have won before the weekend
> began?
Initially w = l (i.e., w / (w + l) = 1/2), and we are given that
(w + 3)/(w + l + 4) > 503/1000.
(w+3) / (2w+4) > 503/1000
1000(w+3) > 503(2w+4)
1000w + 3000 > 1006w + 2012
988 > 6w
6w < 988
w < 164 2/3
The largest number of matches which she could have won before the
weekend began was 164.
> 4. In Pascal's triangle, each entry is the sum of the two entries above
> it. In which row of Pascal's triangle do three consecutive entries
> occur that are in the ratio 3 : 4 : 5?
The entries of row n are C(n, k) for k in {0,...,n}, where
C(n, k) = n! / (k! (n-k)!). So we set
3a = C(n, k) = n! / (k! (n-k)!)
4a = C(n, k+1) = n! / ((k+1)! (n-(k+1))!) = C(n, k) (n-k)/(k+1)
5a = C(n, k+2) = n! / ((k+2)! (n-(k+2))!)
= C(n, k) (n-k)(n-k-1) / ( (k+1)(k+2) )
So 4/3 = (n-k)/(k+1) and 5/3 = (4/3) (n-k-1)/(k+2)
Just solve for n and k in 4(k+1) = 3(n-k) and 5(k+2) = 4(n-k-1)
4k+4=3n-3k or 3n-7k=4
5k+10=4n-4k-4 or 4n-9k=14
12n-28k = 16 27n-63k = 36
12n-27k = 42 28n-63k = 98
------------ ------------
k = 26 n = 62
In "row 62" assuming you count: row 0 1
row 1 1 1
row 2 1 2 1
etc.
Dan
|
|
> 3. A tennis player computes her "win ratio" by dividing the number of
> matches she has won by the total number of matches she has played. At
> the start of a weekend, her win ratio is exactly .500. During the
> weekend she plays four matches, winning three and losing one. At the
> end of the weekend her win ratio is greater than .503. What is the
> largest number of matches that she could have won before the weekend
> began?
I took a slightly different approach to solve this, by using a short
cut that I use to recalculate baseball player's batting averages in my
head after a game (you have to know their previous batting average and
at-bat total). The equation I put to use is:
a + c d
----- = a/b + ---(c/d - a/b)
b + d b+d
This is easily proved in a couple of simple steps of simplifying the
right side. You can think of it this way (using baseball for
terminology). If a = hits for the season (before today's game),
and b = at bats for the season (before game)
and c = hits today, and d = at bats today,
then the player's batting average is "pulled" d/(b+d) of the way
from his old batting average of a/b toward his batting average for that
one day (c/d).
In the problem above, c=3, d=4, c/d = .750, a/b is .500 (so b=2a),
and you're asked to find the largest integer a for which
(a+c)/(b+d) > .503,
which by the equation above is the same as saying
d/(b+d) * (c/d - a/b) > .003.
That means 4/(2a+4) * .250 > .003
1000/3 > 2a + 4
164 2/3 > a
So the answer, as you already know, is 164.
Jon
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| > 4. In Pascal's triangle, each entry is the sum of the two entries above
> it. In which row of Pascal's triangle do three consecutive entries
> occur that are in the ratio 3 : 4 : 5?
The solutions provided in earlier replies were great, but a bit too
involved for my tastes. I like to look for alternative solutions
that allow simpler calculations, especially when taking a timed exam
which is designed to be difficult to finish, like the AIME.
There is such a short cut for this problem which is based on the
observation that to generate the nth row of Pascal's triangle (counting
the first row as row 1 instead of row 0) you can start with the number
1, then multiply by (n-1)/1, then multiply by (n-2)/2, then by
(n-3)/3, and continue until you get back to 1. Note that the numerator
and denominator of these fractions add up to the number of the row.
So the problem becomes one of finding fractions of the form 4a/3a and
5b/4b (a & b are positive integers) such that
row number is same: 4a+3a = 5b+4b
and they are "consecutive": 4a - 1 = 5b, and 3a + 1 = 4b
Our approach will be (step 1) solve the first equation with the smallest
values of a and b possible, then (step 2) check against the last two
equations. Step 1 is the same as simply finding the Least Common
Multiple of 4+3 and 5+4, which is 63. Now all we have to do is check.
So a is 9 and b is 7, which means our fractions are 36/27 and 35/28.
These satisfy the "consecutivity" equations above, so we have our
answer. The row number is 63 (again, I started counting with row 1,
not row 0, so my answer is one higher).
Jon
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