AMC Marathon

[mathslug]

4/m + 2/n = 1
4n + 2m = mn
4n - mn + 2m = 0

Solve for m and n -->

m = 4n / (n-2)
n = 2m / (m-4)

To have both m and n be positive numbers, we assume m>4 and n>2. This means that there can be a limited amount of integers that fit the description.

Lets start on the n's at 3.
n = 3, m = 12
n = 4, m = 8
n = 5, m = 20/3 (doesn't count, not an integer)
n = 6, m = 6
n = 7, m = 28/5 (doesn't count, not an integer)
n = 8, m = 16/3 (doesn't count, not an integer)
n = 9, m = 36/7 (doesn't count, not an integer)
n = 10, m = 5

No need to go any further, since the spread of possible integer values has been covered.

The answer is 4.

D!

Zack

[Roel]

with some algebra we get:
m=4n/(n-2)
By plugging in the first positive integers for "n" we get the ordered pairs(m,n): (12,3) (8,4) (6,6) (5,10)
We can see that this are the only positive pairs because m<4 according to our equation.

answer d)4

[stupidityismygam]

Another approach:

We have mn=4n+2m so mn-2m-4n=0

By SFFT (simons favorite factoring trick)

(m-4)(n-2)=8

So, m-4=8,n-2=1 OR m-4=4,n-2=2 OR m-4=2,n-2=4 OR m-4=1,n-2=8

Which is 4 solutions!

Point Count:

mathslug: 3
Quelloquialism: 5
collegebookworm: 2
Roel: 1/2

Very long-winded
Problem #11:

Three cards, each with a positive integer written on it, are lying face-down on a table. Casey, Stacy, and Tracy are told that

i) the numbers are all different
ii) they sum to 13, and
iii) they are in increasing order, left to right

First, Casey looks at the number on the leftmost card and says, "I don't have enough information to determine the other two numbers." Then
Tracy looks at the number of the rightmost card and says, "I don't have enough information to determine the other two numbers." Finally, Stacy looks at the number on the middle card and says, "I don't have enough information to determine the other two numbers." Assume that each person knows that the other two reason perfectly and hears their comments. What number is on the middle card?

a) 2 b) 3 c) 4 d) 5 e) There is not enough information to determine the number.

Source: 1998 AHSME

[Quelloquialism]

[stupidityismygam]

Point Count:

mathslug: 3
Quelloquialism: 6
collegebookworm: 2
Roel: 1/2

Problem #12:

Four positive integers a,b,c, and d have a product of 8! and satisfy

ab+a+b=524,
bc+b+c=146, and
cd+c+d=104.

What is a-d?

a) 4 b) 6 c) 8 d) 10 e)12

Source: 2001 AMC 12

[Roel]

By using SFFT:
(a+1)(b+1)=524+1=525=25x21
(b+1)(c+1)=146+1=147=21x7
(c+1)(d+1)=104+1=105=7x15

On the last two equations we see that they have (c+1) and 7 in common so we assume:
c+1=7
c=6
By plugging this value back in to the equations we find d=14 and b=20. Finally we go back to the first equation (a+1)21=525
a=24

So...
a-d=24-14=(d) 10

[stupidityismygam]

Point Count:

mathslug: 3
Quelloquialism: 6
collegebookworm: 2
Roel: 1

Problem #13:

Let P(x) be a polynomial such that when P(x) is divided by x-19, the remainder is 99, and when P(x) is divided by x-99, the remainder is 19. What is the remainder when P(x) is divided by (x-19)(x-99)?

a) -x+80 b) x+80 c) -x+118 d) x+118 e) 0

Source: 1999 AHSME

After this the problems get a little harder

[Michael T]

[stupidityismygam]

Point Count:

mathslug: 3
Quelloquialism: 6
collegebookworm: 2
Roel: 1
Michael T: 1

Im posting 2 problems
Problem #14:

In triangle ABC, angle C is a right angle and CB>CA. Point D is located on BC so that angle CAD is twice angle DAB. If AC/AD=2/3, then CD/BD=m/n, where m and n are relatively prime positive integers. Find m+n.

a) 10 b) 14 c) 18 d) 22 e) 26

Source: 1998 AHSME

Problem #15:

There are unique integers, a_2,a_3,a_4,a_5,a_6,a_7 such that

5/7=a_2/2!+a_3/3!+a_4/4!+a_5/5!+a_6/6!+a_7/7!,

where 0<=a_i
a) 8 b) 9 c) 10 d) 11 e) 12

Source: 1999 AHSME

[stupidityismygam]

oh by the way, now that we are getting into the more difficult problems I am willing to give hints (if I know how to work it)

So, if/when wants some hints just ask and ill deliver

[collegebookworm]

[stupidityismygam]

Nice, that was my solution too (i love trig bash)

Hint to the other: Try getting rid of those nasty fractions

Point Count:

mathslug: 3
Quelloquialism: 6
collegebookworm: 3
Roel: 1
Michael T: 1

Problem #15:

There are unique integers, a_2,a_3,a_4,a_5,a_6,a_7 such that

5/7=a_2/2!+a_3/3!+a_4/4!+a_5/5!+a_6/6!+a_7/7!,

where 0<=a_i
a) 8 b) 9 c) 10 d) 11 e) 12

Source: 1999 AHSME

Problem #16:

The total area of all the faces of a rectangular solid is 22 cm^2, and the total length of all its edges is 24 cm. Then the length in cm of any one of its internal diagonals is

a) 11^(1/2) b) 12^(1/2) c) 13^(1/2) d) 14^(1/2) e) not uniquely determined

[collegebookworm]

[stupidityismygam]

[collegebookworm]

[Quelloquialism]

[Quelloquialism]

[stupidityismygam]

Point Count:

mathslug: 3
Quelloquialism: 8
collegebookworm: 3
Roel: 1
Michael T: 1

Problem #17:

How many ordered triples (x,y,z) of integers satisfy the system of equations below?

x^2-3xy+2y^2 -z^2= 31
-x^2 +6yz+2z^2= 44
x^2+ xy +8z^2=100

a) 0 b) 1 c) 2 d) a finite number greater than two e) infinitely many

Source: AHSME 1980

Problem #18:

The polynomial x^(2n)+1+(x+1)^(2n) is not divisible by x^2+x+1 if n equals

a) 17 b) 20 c) 21 d) 64 e) 65

Source: AHSME 1980

Problem #19:

The sum (5+2\sqrt(13))^(1/3)+(5-2\sqrt(13))^(1/3) equals

a) 3/2 b) (65)^(1/3)/4 c) (1+(13)^(1/6))/2 d) (2)^(1/3) e) none of these (i realize that without latex this is difficult to read)

Source: AHSME 1980

[Roel]

(5+2\sqrt(13))^(1/3)+(5-2\sqrt(13))^(1/3)
Let a=5 and b=2sqrt(13)

(a+b)^(1/3)+(a-b)^(1/3)=x
Cubing both sides we get:
(a+b)+3(a+b)^(2/3)*(a-b)^(1/3)+3(a+b)^(1/3)*(a-b)^(2/3)+(a-b)=x^3
After rewriting the left side and factoring out:
2a+3(a+b)^(1/3)(a-b)^(1/3)[(a+b)^(1/3)+(a-b)^(1/3)]=x^3
Inside of the brackets is what we originally we had as "x":
2a+3(a+b)^(1/3)(a-b)^(1/3)(x)=x^3
2a+3(a^2-b^2)^(1/3)(x)=x^3
Plugging in a=5 and b=2sqrt(13):
2*5+3*(5^2-(2sqrt(13))^2)^(1/3)(x)=x^3
10+3*(-27)^(1/3)(x)=x^3
Sending everything to one side:
x^3+9x-10=0
(x-1)(x^2+x+10)=0
So our only real solution is 1
(a+b)^(1/3)+(a-b)^(1/3)=x =1
Answer e) none of these

[stupidityismygam]