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Conference rusure::math

Title:Mathematics at DEC
Moderator:RUSURE::EDP
Created:Mon Feb 03 1986
Last Modified:Fri Jun 06 1997
Last Successful Update:Fri Jun 06 1997
Number of topics:2083
Total number of notes:14613

1628.0. "1992 American Invitational Math Exam" by BEING::EDP (Always mount a scratch monkey.) Mon Jun 15 1992 12:28

    Here are the first four questions from the 1992 American Invitational
    Mathematics Examination, via _Crux Mathematicorum_, April, 1992.
    
    
    				-- edp
    
    
    1. Find the sum of all positive rational numbers that are less than 10
    and that have denominator 30 when written in lowest terms.
    
    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?
    
    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?
    
    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?
T.RTitleUserPersonal
Name
DateLines
1628.1TRACE::GILBERTOwnership ObligatesMon Jun 15 1992 15:0122
1628.2ZFC::deramoDan D'EramoTue Jun 16 1992 17:346
1. 400
2. 502
3. 164
4.  62 (starting counting with the zeroth row)

Dan
1628.3GUESS::DERAMODan D'Eramo, zfc::deramoFri Jun 19 1992 23:5267
>    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
1628.4Formula for general case of #1CRONIC::CRONIC::MCINTYREMon Jun 29 1992 04:1717
>    1. Find the sum of all positive rational numbers that are less than 10
>    and that have denominator 30 when written in lowest terms.
    
    If you change the wording to "Find the sum of all positive rational 
    numbers that are less than N and that have denominator D when written 
    in lowest terms, where N and D are positive integers", then a formula 
    for the answer is:
                           
                     N^2 * rprime(D) /2,
    
    where rprime() means the number of positive integers less than D which 
    are relatively prime to D.
    
    The stated problem then becomes the special case where N = 10 and 
    D = 30.
    
    Jon
1628.5A "batting average" shortcut for #3CRONIC::CRONIC::MCINTYREMon Jun 29 1992 04:5340
>    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
1628.6Slightly different viewpoint on itCRONIC::CRONIC::MCINTYREMon Jun 29 1992 06:2031
>    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