Algebra A
1. [3] On the number line, consider the point x that corresponds to the value 10. Consider 24 distinct
integer points y1 , y2 , ..., y24 on the number line such that for all k such that 1 ≤ k ≤ 12, we have
that y2k−1 is the reflection of y2k across x. Find the minimum possible value of
P24
n=1 (|yn − 1| + |yn + 1|)
2. [3] Alice, Bob, and Charlie are visiting Princeton and decide to go to the Princeton U-Store to
buy some tiger plushies. They each buy at least one plushie at price p. A day later, the U-Store
decides to give a discount on plushies and sell them at p0 with 0 < p0 < p. Alice, Bob, and Charlie
go back to the U-Store and buy some more plushies with each buying at least one again. At the
end of that day, Alice has 12 plushies, Bob has 40, and Charlie has 52 but they all spent the same
amount of money:$42. How many plushies did Alice buy on the first day?
3. [4]A function f has its domain equal to the set of integers 0, 1, ..., 11, and f (n) ≥ 0 for all such
n, and f satisfies
f (0) = 0
f (6) = 1
f (x)+f (y)
If x ≥ 0, y ≥ 0, andx + y ≤ 11, then f (x + y) = 1−f (x)f (y)
Find f (2)2 + f (10)2 .
j n k l n m
4. [4]There is a sequence with a(2) = 0, a(3) = 1 and a(n) = a +a for n ≥ 4. Find
jnk lnm 2 2
n
a(2014). [Note that and denote the floor function (largest integer ≤ 2 ) and the ceiling
2 n
2
function (smallest integer ≥ 2 ), respectively.]
5. [5] Real numbers x, y, z satisfy the following equality:
4(x + y + z) = x2 + y 2 + z 2
Let M be the maximum of xy + yz + zx, and let m be the minimum of xy + yz + zx. Find
M + 10m.
∞
20xn+1 X x3n p
6. [6] Given that xn+2 = , x0 = 25, x1 = 11, it follows that n
= for some positive
14xn n=0
2 q
integers p, q with GCD(p, q) = 1. Find p + q.
7. [7] x, y, z are positive real numbers that satisfy x3 +2y 3 +6z 3 = 1. Let k be the maximum possible
value of 2x + y + 3z. Let n be the smallest positive integer such that k n is an integer. Find the
value of k n + n.
8. [8] For nonnegative integer n, the following are true:
f (0) = 0
f (1) = 1
f (n) = f (n− m(m−1)
2 )−f ( m(m+1)
2 −n) for integer m satisfying m ≥ 2 and m(m−1)
2 <n≤ m(m+1)
2 .
Find the smallest n such that f (n) = 4.