AMC Marathon

[stupidityismygam]

Welcome to the AMC Marathon!!

Ill post a problem, and then someone (or multiple people) will post a solution. Once the solution is satisfactory (by my standards :) :)) ill post another problem and the marathon will continue.

Ill keep score of who gets the solution correct first and the winner...gets nothing!!

So here we go! (ill try my hardest without latex)

If a,b, and c are positive integers and a and b are odd, then 3^a+c*(b-1)^2 is

a) odd for all choices of c
b) even for all choices of c
c) odd, if c is even; even, if c is odd
d) odd, if c is odd; even, if c is even
e) odd, if c is not a multiple of 3; even, if c is a multiple of 3

Source: 1992 AHSME #5
(guess and check will not receive a point)

[mathslug]

3^a + c*(b-1)^2

Let a and b be odd.

3^(anypower) will be an odd number, b/c odd*odd = odd.

(b-1)^2 will be even, b/c one less than an odd is an even, raised to anypower is still even.

Even times anything = even, so the second value will be even.

even + odd = odd

The answer is always odd.

The answer is A

[Quelloquialism]

Yeah...my solution was to let a = b = 1, so the expression simplifies to 3, which is odd independent of c, so the answer is (A). Not "formal" or "rigorous" or any of that, but it got the answer crazy-fast, anyway. =)

[stupidityismygam]

Point Count:

mathslug: 1

Problem Number 2:

If f(x)=ax^4-bx^2+x+5 and f(-3)=2, then f(3)=

a) -5 b) -2 c) 1 d) 3 e) 8

Source: ASHME 1995

Note: Calculators are not allowed (I'm not sure if I mentioned that before, the rules changed this year)

[Quelloquialism]

f(x)=ax^4-bx^2+x+5
f(-3) = a(-3)^4 - b(-3)^2-3+5 = 2
81a - 9b + 2 = 2
81a - 9b = 0
f(3) = a(3)^4 - b(3)^2+3+5 = ?
81a - 9b + 8 = ?
(0) + 8 = ?
e) 8

[stupidityismygam]

Point Count:

mathslug: 1
Quelloquialism: 1

Problem #3:

What is the product of all positive odd integers less than 10000?

a) 10000!/(5000!)^2
b) 10000!/(2^5000)
c) 9999!/(2^5000)
d) 10000!/(2^5000*5000!)
e) 5000!/(2^5000)

Source: 2001 AMC 12

[mathslug]

Think about it like starting with 10,000! and eliminating out all the evens.

To do so, take 5000! and multiply each entry by 2. This causes you to go from (1*2*3...*5,000) to having (2*4*6..*99998). How do we express this? You multiply each value from 1, to 5,000 by 2, which can be expressed as 2^5,000. Therefore, the expression should be...

10,000! / (2^(5,000)*5,000!).

Which is D.

Hard for #3 I do say.

[stupidityismygam]

[Quelloquialism]

f(500) = 3 = f(100)/5 so f(100) = 15
f(600) = f(100)/6 = 15/6 = 5/2 (c). Does that work? Or am I misunderstanding the f(x) definition?

[stupidityismygam]

[Quelloquialism]

[mathslug]

[stupidityismygam]

Point Count:

mathslug: 2
Quelloquialism: 3

Problem #6:

If the sum of the first 10 terms and the sum of the first 100 terms of a given arithmetic progression are 100 and 10, respectively, then the sum of the first 110 terms is

a) 90 b) -90 c) 110 d) -110 e) -100

Source: 1980 AHSME

[Quelloquialism]

General arithmetic sequence is a, a+d, a+2d, a+3d, a+4d, ...
The sum of the first n terms will be n*a + T(n-1)*d, where T(n) is the n-th triangular number (n*(n+1)/2); this is n*a+(n-1)*n/2*d.

So the sum of the first 10 terms 10*a+9*10/2*d = 100.
(cleanup: 10*a + 45*d = 100)
The sum of the first 100 terms 100*a+99*100/2*d = 10.
(cleanup: 100*a + 4950*d = 10)

The sum of the first 110 terms should be 110*a + 109*110/2*d = x, cleanup to 110*a + 5995*d = x. (x is the answer here)

The sum of the first 10 and 100 terms together will be 110*a + 4995*d = 110. Subtract this equation from the equation from the 110 terms one, and you have that 1000*d = x - 110.

So we need to figure out what 1000*d is. Let's try getting d alone in an equation, then; we do this by subtracting 10 times the first-10-terms equation from the first-100-terms equation, getting (100*a + 4950*d) - (100*a + 450*d) = 10 - 1000, cleanup to 4500*d = -990; divide both sides by 4.5, and you'll have 1000*d = -990/4.5, whatever that is.

So -990/4.5 = x - 110...um...x = 110 - 990/4.5. Lesse. 990/4.5 = 9900/45 = 1100/5 = 220. So x = 110 - 220 = -110 (d). Maybe.

Err...does brutish inelegance win a prize?

[stupidityismygam]

[Quelloquialism]

[stupidityismygam]

[Quelloquialism]

I don't want to reread it, but I'd guess my error would be that I defined the bounds backwards with the guesses where the actuals should be and the actuals where the guesses should be, so I'd think the answer would instead be 60,000/.9 - 50,000 * .9 = 66667 - 45000 = 21667 = 22000 (e). But to be honest, I'm no more confident in that than I was in the other, probably because I have n brains, where n is a number on the interval 0 < n < 1, likely far closer to the lower bound than the upper.

[stupidityismygam]

[stupidityismygam]

Point Count:

mathslug: 2
Quelloquialism: 5

Problem #8:

Let x be a real number such that sec(x)-tan(x)=2. Then sec(x)+tan(x)=

a) .1 b) .2 c) .3 d) .4 e) .5
Source: 1999 AHSME

[collegebookworm]

[stupidityismygam]

Point Count:

mathslug: 2
Quelloquialism: 5
collegebookworm: 1

Problem #9:

Consider the non-decreasing sequence of positive integers

1,2,2,3,3,3,4,4,4,4,5,5,5,5,5,...

in which the nth positive integer appear n times. The remainder when the 1993rd term is divided by 5 is

a) 0 b) 1 c) 2 d) 3 e) 4

Source: 1993 AHSME

[collegebookworm]

[Quelloquialism]

[stupidityismygam]

Point Count:

mathslug: 2
Quelloquialism: 5
collegebookworm: 2

Problem #10:
How many ordered pairs (m,n) of positive integers are solutions to 4/m+2/n=1?

a) 1 b) 2 c) 3 d) 4 e) More than 4

Source: 1993 AHSME