Hello !
I am 15 and am studying in India. This is the the set of problems that was given in AMTI Finals. Could you sove it ?

1. Calculate the sum 1/(2!)+2/(3!)+3/(4!)+¼+99/(100!).
2. Let a, b, c be Real numbers such that 0 < a < 1, 0 < b < 1, 0 < c < 1. Prove that
   
   (a)
   

   __ Öa b

   +

   ________ Ö(1-a)(1-b)

   £ 1

   
   .
   (b)
   

   ____ Öa b c

   +

   ____________ Ö(1-a)(1-b)(1-c)

   £ 1

   
   
3. Find all pairs (x,y) positive Integers such that x³-y³=x y+61.
4. A function f is defined for all Real Numbers and for all Real x, satisfies the equations

f(2+x)=f(2-x) and f(7+x)=f(7-x)

If 0 satisfies f(x)=0, what is the smallest number of solutions that f(x)=0 could have in the interval -2002 £ x £ 2002?

5. In a triangle A B C, let A D be a median. Let L and N be points on the segments A B and A C. Suppose L N cuts A D at M. Prove that A L.L B/A B+A N.N C/A C=2.A M.M D/A D
6. In a rhombus A B C D, angle A=60^°. Let K be a point on the diagonal A C and choose points L, M on A B, B C respectively such that K L B M is a parallelogram. Show that the triangle L M D is equilateral.
7. Let m, n be positive Integers such that 3m+n=3(least common multiple [m,n])+ greatest common divisor [m,n].

Prove that n divides m.
8. Find the max. no. of distinct 4-digit positive Integers consisting only of digits 1, 2 and 3 such that any two of these have equal digits in at most one position.

Good questions. Hints for those I've solved:

1) Use the identity [1 - 1/(N-1)!] + (N-1)/N! = [1 - 1/N!], and induction.

A: 1 - 1/100!

2a) make a substitution along the lines of a = cos² (alpha ), and so on...
2b) use a similar trick, and then use factor formulas for sine.

6) This can be done fairly easily using analytic geometry, but I'd suspect there's an easier solution yet.

7) Assume that n does not divide m. Then there exists integers k, m', and n' such that m = k^. m', n = k^. n', and GCD(n',m') = 1. Then we have

k(3m' + n') = k(3m'n' + 1),

upon subsitution, and we can go on to derive a contradiction from this when n' is not 1.

Brad

for 3,
x³ - y³ = xy + 61
put y=x+c .. where (c is an integer)
substituting and rearranging we get,
(1+3c)x² + (c+3c² )x + (c³ +61)=0 .. (x)

(Note:for a quadratic equation,ax² +bx+c = 0
sum of roots = -b/a
product of roots = c/a)

let x₁ and x₂ be the roots of the above equation.
x₁ +x₂ =-c .... (1)
x₁ xx₂ = [c³ +61]/[1+3c] ... (2)

now we are looking for integer values of x therefore assume one of the roots (Say x₁ ) to be an integer.Then equation (1) forces x₂ also to be an integer.

since x₁ and x₂ are integers,this forces the RHS of equation (2) to be an integer.

so now we have to find the possible values of c for which [c³ +61]/[1+3c] is an integer.

intuitionally i feel (-1,0) are the only possible values of c for which the above is an integer..(i am not able to prove that these are the only solution .. so someone please give this a try)

substituting c=0 in (x) we see that we don't get real roots.

substituting c=-1 in (x)we get,
-2x² + 2x + 60 = 0
i.e
x² - x - 30 = 0
therefore,x=-5 or x=6
we need positive integers therefore x=-5 is discarded.....

therefore we get x=6 as the answer,
y=x+c
now x=6 and c=-1
therefore y=5

hence ... (x,y) = (6,5)
This is the only possible solution

love arun

Hint on Q4:

Step 1: Show that f(4-x)=f(x)=f(14-x) (x)
Step 2: Use (x) to show that f(x)=f(x+10) (xx)
Step 3: f(0)=f(4)=0 (xxx)
Step 4: Use (xx) and (xxx) to show that f(10n)=f(10n+4)=0 for any integer n. (both positive or negative)
Step 5: Find a function that has only those zeros in the required interval. (Don't try to think of anything complicated. Just a function which will make your life easier to check that indeed it satisfies the required conditions)
[To check that your fuction satisfies the required conditions it might help to use (x)]
Step 6: Count the number of zeros

Demetres