Ding! AMC MATH Q #2





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Discus: SAT/ACT Tests and Test Preparation: June 2004 Archive: Ding! AMC MATH Q #2
By Loop123 (Loop123) on Friday, June 18, 2004 - 02:40 pm: Edit

This problem deals with a sequence, so let a_(n) stand for a with the subset of n.

Let a_(n),n being greater than or equal to one, be a sequence such that a_1 = 1 and
3a_(n+1) - 3a_(n)=1. Find a_(2002).

This problem isn't as difficult as my last one (I actually answered it correctly lol), so speed matters. Be the first to answer correctly for 10 points! Currently the standings in CC AMC are:
Jbaby: 10
Everyone Else: 0

By Optimizerdad (Optimizerdad) on Friday, June 18, 2004 - 03:17 pm: Edit

a_(n+1) = a_(n) + 1/3, so a_(2002) =a_(1) + (2002-1)/3 = 1 + 667 = 668

By Loop123 (Loop123) on Friday, June 18, 2004 - 03:32 pm: Edit

BING! Optimizerdad got it. Good job.
In my method, I just started writing out the sequence, which is as follows: 1, 4/3, 5/3, 6/3...

The numerator is always 2 more than the # of the term. So, 2004/3 is 668.

That's 10 points for Optimizerdad!

By Benzinspeicher (Benzinspeicher) on Friday, June 18, 2004 - 05:37 pm: Edit

Give us some more AMC problems. I loved that problem! Math problems are just like fun puzzles!

By Loop123 (Loop123) on Friday, June 18, 2004 - 06:05 pm: Edit

OK here's another:

The perimeter of a rectangle is 100 and its diagonal has length x. What is the area of this rectangle?

A)625 - x^2
B)625 - (x^2)/2
C)1250 - x^2
D)1250 - (x^2)/2
E)2500 - (x^2)/2

By Optimizerdad (Optimizerdad) on Friday, June 18, 2004 - 06:12 pm: Edit

If the rectangle has length=a and width=b, we have

1. a^2 + b^2 = x^2
2. Perimeter = 2(a+b)= 100, so (a+b)=50
3. Area = ab = 2ab/2 = ( (a+b)^2 - (a^2+b^2))/2
= ( 50^2 - x^2)/2
= 1250 - (x^2)/2

Answer is D

By Loop123 (Loop123) on Friday, June 18, 2004 - 06:20 pm: Edit

Wow you're just too good Optimizerdad.
Here's another:

In how many zeros does (2002!) / (1001!)^2 end?

By Fairyofwind (Fairyofwind) on Friday, June 18, 2004 - 06:41 pm: Edit

that would be
499 - (249*2) = 1

By Link12 (Link12) on Friday, June 18, 2004 - 10:28 pm: Edit

Could you explain?

By Thermodude (Thermodude) on Friday, June 18, 2004 - 11:15 pm: Edit

No...sorry...i'm sure Fairyofwind is incorrect on this one....

By Optimizerdad (Optimizerdad) on Saturday, June 19, 2004 - 08:54 am: Edit

Umm.. let's see if this explanation holds:
(2002!) / (1001!)^2 = (2002 * 2001 * ... *1002) / (1001 * 1000 * ... * 1)
= 2002 * (2001/1001) * (2000/1000) * ... * (1002/2)

Any of the () fractions above that has a numerator that is a multiple of 5,25,125,2,4 and 8 also has a denominator that is such a multiple; every denominator = numerator - 1000, and 1000 is a multiple of 5,25,125,2,4,8. I think that by reducing each of the fractions above, we will be left with no numerator that ends in 5 or 0. I believe the answer to the original question is zero, unless I missed something along the way (which is a distinct possibility :-)

By Texas137 (Texas137) on Saturday, June 19, 2004 - 10:47 am: Edit

people who like problems from AMC and other contests should check out www.artofproblemsolving.com

By Amitoman (Amitoman) on Saturday, June 19, 2004 - 11:18 am: Edit

Fairyofwind is correct.

If you want hard evidence, the value of that expression is:

820489951402314977460608965726619108950216689410325248027852048727829857742191
993001821723256152485126348025680744472290786380491376972232587388252047076142
224658678994288178111823146197682905321743421515004645381221029254760624739633
110561227998143405977012593687244914023499594726925463277080205499106731654279
656806449004729582304785995980953044052000570334573516829950123742353942786051
986265597342151413287601719948908658938912139405106815638163288762112638367134
104798211952009169273328300576127288835445032983018793035057441329429855497482
2101299658045689142804604871387166214383014213216471357018240


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