The document contains 6 problems related to algebra and numbers along with their solutions. Problem 1 involves a number guessing game between two players and determining the minimum number of rounds needed. Problem 2 examines properties of a polynomial where the polynomial equals certain values for distinct integer inputs. Problem 3 finds all integer solutions to a system of equations involving cubes of variables. Problem 4 determines the value of a polynomial of degree 8 at a particular input, given its values at other integers. Problems 5 and 6 involve finding the smallest integer greater than an expression and the minimum possible value of a product of variables, respectively, given an equation relating the variables.

1. Problem solving in algebra and numbers
Compiled by Gary Tsang
January 31, 2010
Problem 1. Mary and David play the following number-guessing game. Mary
writes down a list of positive integers x1 , x2 , . . . , xn . She does not reveal them
to David but only tell him the value n. David chooses a list of positive integers
a1 , a2 , . . . , xn and ask Mary to tell him the value of a1 x1 + a2 x2 + · · · + an xn . Then
David chooses another list of positive integers b1 , b2 , . . . , bn and asks Mary for the
value of b1 x1 + b2 x2 + · · · + bn xn . Play continues in this way until David is able to
determine Mary’s numbers. Find the least number of rounds that David needs in
order to determine Mary’s number.
Solution : Answer: 2 rounds. First of all, David choose a1 = a2 = · · · = an = 1
so that he obtains the value m = x1 + x2 + · · · + xn , then he choose b1 = 1, b2 =
m + 1, b3 = (m + 1)2 , . . . , bn = (m + 1)n−1 and obtain the value m = b1 x1 + b2 x2 +
· · · + bn xn = x1 + (m + 1)x2 + (m + 1)2 x3 + · · · + (m + 1)n−1 xn which is the base-
(m + 1) representation of m . By the uniqueness of base-n representation of an
integer, David can determine x1 , x2 , . . . , xn .
Alternative: Once Davaid obtains the value m = x1 + x2 + · · · + xn and let 10k−1 ≤
m < 10k , he can choose b1 = 1, b2 = 10k , b3 = 102k , . . . bn = 10(n−1)k and he can
still determine the value x1 , x2 , . . . , xn .
Problem 2. Suppose P(x) is a polynomial with integer coefficients and P(a) =
P(b) = P(c) = P(d) = 2 for distinct integers a, b, c and d. Which of the following
is true?
a. There is no integer k such that P(k) = 1, 3, 5, 7 or 9.
b. There is an integer k such that P(k) = 1, 3, 5, 7 or 9.
c. Neither a. nor b. is true in general and there are examples for both.
Solution : Answer: a. Proof: Since P(a) = P(b) = P(c) = P(d) = 2, then we can
let P(x) = (x − a)(x − b)(x − c)(x − d)Q(x) + 2.Since Q(x) is an integer and a, b, c
and d are distinct, therefore P(x) − 2 must have at least 4 distinct integral factors:
(x − a), (x − b), (x − c) and (x − d). Which is impossible if P(k) = 1, 3, 5, 7, 9.
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2. Problem 3. Find all integers solutions to the system of equations:
x3 + y3 + z3 = x + y + z = 3
Solution : (From Kin Y. Li, Math Problem) Consider
(x + y + z)3 − x3 − y3 − z3 = 3(x + y)(y + z)(z + x)
8 = (x + y)(y + z)(z + x)
23 = (3 − x)(3 − y)(3 − z)
By checking the factorization of 8, and note that (3 − x) + (3 − y) + (3 − z) = 6,
we have {x, y, z} = {1, 1, 1} and {4, 4, −5}.
Problem 4. Suppose P(x) is a polynomial of degree 8 with real coefficients and
P(k) = 1 for k = 1, 2, 3, . . . , 9. Determine the number P(10).
k
Solution : Since kP(k) − 1 = 0 for k = 1, 2, 3, . . . , 9. Thus xP(x) − 1 = a0 (x −
1)(x − 2) · · · (x − 9) for some constant a0 .Because P(x) is a polynomial, we need
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the constant term of a0 (x − 1)(x − 2) · · · (x − 9) + 1 equal 0. Thus a0 = 9! and so
(x−1)(x−2)···(x−9)+9!
P(x) = 9!x and so P(10) = 1 .
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√ √
Problem 5. Find the smallest integer B such that B > ( 2 + 3)6 .
√ √ √ √ √ √
Solution : Note that ( 3 + 2)( 3 − √ =√ and 0 < ( 3√ 2)n < 1 for all
√ √ 2) 1 − √
n ∈ N. Now we let sn = ( 3 + 2)2n + ( 3 − 2)2n = (5 + 2 6)n + (5 − 2 6)n ,
One can show that sn satisfies the recurrent relation sn = 10sn−1 − sn−2 . By direct
calculation we can get√1 = 10 and s2 = 98. Thus s3 = 10(98) − 10 = 970. This
√ √ s √ √ √
yields ( 3 + 2)6 + ( 3 − 2)6 = 970 and B = ( 3 + 2)6 = 970.
Problem 6. Problem: Given that a, b and c are positive integers and
abc + ab + bc + ca + a + b + c = 2003
Find the least possible value of abc.
Solution : Note that
abc + ab + bc + ca + a + b + c = 2003
(a + 1)(b + 1)(c + 1) = 2004
Since a, b and c are positive integers, we factorize 2004 and only consider the
factors greater than 1:
(a + 1)(b + 1)(c + 1) = 22 × 3 × 167
To minimize abc, we need as many 1 as possible, so choose (a+1)(b+1)(c+1) =
2 × 2 × 501, then {a, b, c} = {1, 1, 500} and so minimum value of abc = 500.
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