|
2 0 , 0 0 0 |
P r o b l e m s |
U n d e r |
t h e |
S e a |
This page contains copyrighted material. Comment by Kevin Costello, costello@its.caltech.edu, entered on 14
Aug 2001: The sum is
Comment by Lekraj Beedassy, beedassylekraj@hotmail.com, entered on 16
Jul 2002: Such problems are classics that pop up quite often in Contests
in some form or other. The crux of the solution requires the
simple insight: for n>5,2n>9 or
3n-9>n,i.e.,3(n-3)>n. Comment by Tom Engelsman, Mathzeus@peoplepc.com, entered on 27 Dec
2001: The centroid of V occurs at (0,
2.60) View First
Total found:
20000
This page lists 14851 through
14900.
Click on problem numbers for additional
information.
To enter comments or corrections, click "enter
comments".14851. PUTNAM 1978/B.2
14852. PUTNAM 1978/B.3
Sum(n,1,inf)Sum(m,1,inf)1/((mn)(m+n+2)
=Sum(n,1,inf)1/nSum(m,1,inf)1/(m)(m+n+2)
=Sum(n,1,inf)(1/n)Sum(m,1,inf)(1/n+2)(1/m-1/m+n+2)
=Sum(n,1,inf)(1/n)(1/n+2)Sum(m,1,n+2)(1/m)
=1/2(Sum(n,1,inf)(1/n-1/(n+2))Sum(m,1,n+2)(1/m))
=1/2((1(1+1/2+1/3))+1/2(1+1/2+1/3+1/4)-Sum(n,1,inf)(1/n+2)(1/(n+3)+1/(n+4)))
=23/16+1/2(Sum(n,1,inf)(1/n+2)(1/n+4)+Sum(n,1,inf)(1/n+2)(1/n+3))
=23/16+1/2(1/2(Sum(n,1,inf)(1/n+2-1/n+4))+Sum(n,1,inf)(1/n+2-1/n+3))
=23/16+1/4(1/3+1/4)+1/2(1/3)=7/4
14853. PUTNAM 1978/B.4 
14854. PUTNAM 1978/B.5 
14855. PUTNAM 1978/B.6 
14856. PUTNAM 1979/A.1 
14857. PUTNAM 1979/A.2
We proceed stepwise.
First,it's
plain that partitioning into 1's is counter to increasing the
product.
Following the above inequality,if we wish to maximize
the product of the parts of the partition of 1979,we derive no
benefit in using summands exceeding 3,for then we can always split
them further to give a greater product.
Also,we observe that
between 2+2+2 and 3+3,the latter partition of 6 gives a greater
product,so we are settled with the fact that the number of 2's may
not exceed two.
In conclusion,the product is maximized when
there are as many 3's as possible.
For this we simply need to
express 1979 (mod 3).One simple way to evaluate this is to compute
its digital root,which we can do prior to casting out
9's.Hence,digital root is 1+7=8 which is congruent to 2 (mod
3),whence we know that number of 3's is (1979-2)/3=659.
From
all the above,its immediate that n=659+1=660 and a1=2 and all a2
up to a659 are 3's.
For information,the sequence relevant
to this problem on the Online Encyclopedia of Integer
sequences(www.research.att.com/~njas/sequences) is
A000792.
14858. PUTNAM 1979/A.3 
14859. PUTNAM 1979/A.4 
14860. PUTNAM 1979/A.5 
14861. PUTNAM 1979/A.6 
14862. PUTNAM 1979/B.1 
14863. PUTNAM 1979/B.2 
14864. PUTNAM 1979/B.3 
14865. PUTNAM 1979/B.4 
14866. PUTNAM 1979/B.5 
14867. PUTNAM 1979/B.6 
14868. PUTNAM 1980/A-1 
14869. PUTNAM 1980/A-2 
14870. PUTNAM 1980/A-3 
14871. PUTNAM 1980/A-4 
14872. PUTNAM 1980/A-5 
14873. PUTNAM 1980/A-6 
14874. PUTNAM 1980/B-1 
14875. PUTNAM 1980/B-2 
14876. PUTNAM 1980/B-3 
14877. PUTNAM 1980/B-4 
14878. PUTNAM 1980/B-5 
14879. PUTNAM 1980/B-6 
14880. PUTNAM 1981/A-1 
14881. PUTNAM 1981/A-2 
14882. PUTNAM 1981/A-3 
14883. PUTNAM 1981/A-4 
14884. PUTNAM 1981/A-5 
14885. PUTNAM 1981/A-6 
14886. PUTNAM 1981/B-1 
14887. PUTNAM 1981/B-2 
14888. PUTNAM 1981/B-3 
14889. PUTNAM 1981/B-4 
14890. PUTNAM 1981/B-5 
14891. PUTNAM 1981/B-6 
14892. PUTNAM 1982/A-1 
14893.
PUTNAM 1982/A-2
14894. PUTNAM 1982/A-3 
14895. PUTNAM 1982/A-4 
14896. PUTNAM 1982/A-5 
14897. PUTNAM 1982/A-6 
14898. PUTNAM 1982/B-1 
14899. PUTNAM 1982/B-2 
14900. PUTNAM 1982/B-3 
View Previous
View Next
View Last