Points through the end of the first week of the summer problem solving marathon:
Matthew: 28
Amelia: 16
Sophia: 13
Michael: 5
Paul: 3
Lily: 2
Ben: 2
Monday, June 6, 2011
Summer Problem Solving Marathon Question #7
[Value = 6 points]
X, Y, and Z are pairwise disjoint sets of people. The average age of people in X is 37. The average age of people in Y is 23. The average age of people in Z is 41. The average age of people in X ∪ Y is 29. The average age of people in X ∪ Z is 39.5. The average age of people in Y ∪ Z is 33.
What is the average age of people in X ∪ Y ∪ Z?
X, Y, and Z are pairwise disjoint sets of people. The average age of people in X is 37. The average age of people in Y is 23. The average age of people in Z is 41. The average age of people in X ∪ Y is 29. The average age of people in X ∪ Z is 39.5. The average age of people in Y ∪ Z is 33.
What is the average age of people in X ∪ Y ∪ Z?
Sunday, June 5, 2011
Summer Problem Solving Marathon Solution #6
Since n + 125 and n + 199 are both perfect squares, we have n + 125 = a2 and n + 199 = b2 for some a and b. Subtracting the first equation from the second, we have:
b2 - a2 = 76
or:
(b + a)(b - a) = 76
This yields three possibilities:
#1: b + a = 76, b - a = 1
#2: b + a = 38, b - a = 2
#3: b + a = 19, b - a = 4
But only the second yields an integer value for b. So b = 20, and a = 18. So n + 201 = 202 = 400, and n = 199.
b2 - a2 = 76
or:
(b + a)(b - a) = 76
This yields three possibilities:
#1: b + a = 76, b - a = 1
#2: b + a = 38, b - a = 2
#3: b + a = 19, b - a = 4
But only the second yields an integer value for b. So b = 20, and a = 18. So n + 201 = 202 = 400, and n = 199.
Saturday, June 4, 2011
Summer Problem Solving Marathon Solution #5
We have d = 4⁄c. c = πd, so by substitution d = 4⁄πd. Cross-multiplying, πd2=4, and d2=4⁄π.
Since r = d⁄2, r2 = d2⁄4. So the area of the circle is πr2 = π(d2⁄4) = π(1⁄π) = 1.
Since r = d⁄2, r2 = d2⁄4. So the area of the circle is πr2 = π(d2⁄4) = π(1⁄π) = 1.
Friday, June 3, 2011
Summer Problem Solving Marathon Question #6
[Value = 2 points]
What is the smallest positive integer n such that n + 125 and n + 201 are both perfect squares?
What is the smallest positive integer n such that n + 125 and n + 201 are both perfect squares?
Summer Problem Solving Marathon Solution #4
Consider first the general form of the binomial theorem. If we have (x + y)n, the full product will have terms of the form xny0, xn-1y1, xn-2y2, ..., x1yn-1, x0yn.
The coefficients of these terms will then be C(n,0), C(n,1), C(n,2), ..., C(n,n-1), C(n,n), where C is the "choose" function.
In our case, y is -(x-1⁄2). If we number the terms in the expansion from 0 to n=7, the kth term will be C(7,k) x7-k(-(x-1⁄2))k. This simplifies to (-1)kC(7,k) x(7-k) - k⁄2 = (-1)kC(7,k) x7- 3k⁄2.
We want the term in which the exponent of x is -1&frasl2. So we want 7 - 3k⁄2 = -1&frasl2, or 14 - 3k = -1, or 3k = 15, or k = 5.
The coefficient is then (-1)5C(7,5) = -7!⁄(5!2!) = -7*6/2 = -21.
The coefficients of these terms will then be C(n,0), C(n,1), C(n,2), ..., C(n,n-1), C(n,n), where C is the "choose" function.
In our case, y is -(x-1⁄2). If we number the terms in the expansion from 0 to n=7, the kth term will be C(7,k) x7-k(-(x-1⁄2))k. This simplifies to (-1)kC(7,k) x(7-k) - k⁄2 = (-1)kC(7,k) x7- 3k⁄2.
We want the term in which the exponent of x is -1&frasl2. So we want 7 - 3k⁄2 = -1&frasl2, or 14 - 3k = -1, or 3k = 15, or k = 5.
The coefficient is then (-1)5C(7,5) = -7!⁄(5!2!) = -7*6/2 = -21.
Thursday, June 2, 2011
Summer Problem Solving Marathon Question #5
[Value = 1 point]
In a certain circle, the diameter is four times the reciprocal of the circumference. What is the area of the circle?
In a certain circle, the diameter is four times the reciprocal of the circumference. What is the area of the circle?
Subscribe to:
Posts (Atom)