MO-ARML – 1/25/15
Practice Question Solutions
1. Let S be the sum of the base 10 logarithms of all the proper divisors (all divisors of a number excluding itself) of 1000000. What is the integer nearest to ?
Since 1000000 = 2656, there are 7×7 = 49 divisors. Excluding 1000, we can pair the other divisors into 24 groups of two so each group multiplies to 106. So the product of all proper divisors is equal to . Hence, the sum of the logarithms of all the proper divisors is equal to the logarithm of the product of the all the proper divisors and is equal to .
2. A certain fraction r is represented in base b by 0.1111… while in base 2b it takes the simpler form 0.2b. What is b?
0.1111…b = and 0.2b2b = . Hence,
3. Let m be a positive integer, and let be a sequence of real numbers such that and , and for . Find
For 0 < k < m, we have . Thus, the product decreases by 3 each time k increases by 1. Since , we have m – 1 = , and thus m = 889.
4. A triangle has vertices (2, 3), (6, -2), and (-1, -4). Find the area of the triangle.
The quickest way is to use the formula A = , where (x1, y1), (x2, y2), (x3, y3)
are the vertices of the triangle, and . Then, the area is equal to .
5. What is the shortest distance between the circle x2 + y2 = 25 and the line 3x + 4y = 48 (it doesn’t intersect the circle)?
In general, the distance from a point (a, b) to a line ax + by + c = 0 is , so the distance from the origin (0, 0) to the line 3x + 4y = 48 is , and the shortest distance between the circle and the line would be .
6. What is the closest distance that the line comes to a lattice point on the xy-plane? A lattice point is a point with integer coordinates.
In general, the distance from a point (a, b) to a line ax + by + c = 0 is , so the distance from any point (a, b) to the line (or or 20x – 35y + 7 = 0)
is , and we would like to minimize the numerator with integers a and b. Since 20a – 35b must be a multiple of 5, we could make it -5 (for example, a = -2,
b = -1), so the smallest value for the numerator is 2, and the closest the line comes to a lattice point is.
7. A certain complex number z satisfies z2 = z – 1. What are the possible values of z2015?
z2 - z + 1 = 0 Þ z3 + 1 = (z + 1)(z2 - z + 1) = 0 Þ z3 = -1 Þ z2013 = (-1)671 = -1
Þ z2015 = -z2 = 1 – z = , since z2 - z + 1 = 0 Þ z = .
Note: There are real cubit root of -1: -1, .
OR
z2 - z + 1 = 0 Þ z =
Þ z2015 =
= .
8. Find the sum of the cube of the roots of the equation x10 + x9 + x8 + × × × + x2 + x + 1 = 0.
Since x11 – 1 = (x – 1)(x10 + x9 + x8 + × × × + x2 + x + 1) = 0, these roots of the equation
x10 + x9 + x8 + × × × + x2 + x + 1 = 0 are (1) the 11th roots (¹ 1) of 1. Since, together with x = 1, they are also the zeros of the equation x11 – 1 = 0, their total sum of 0, so the sum of these ten roots of the equation x10 + x9 + x8 + × × × + x2 + x + 1 = 0 is -1.
Being the 11th roots of 1, these roots space out around the unit circle at the angles:
, and their cubes space out around the unit circle at the angles: .
Hence, the sum of the cube of these roots is the same as the sum of these roots = -1.
Note: we are essentially using the fact that
3×{2, 4, 6, 8, 10, 12, 14, 16, 18, 20} º {2, 4, 6, 8, 10, 12, 14, 16, 18, 20} (mod 22).
It is also true that
k×{2, 4, 6, 8, 10, 12, 14, 16, 18, 20} º {2, 4, 6, 8, 10, 12, 14, 16, 18, 20} (mod 22) as long as k and 22 are relatively prime. Therefore, the sum of the fifth (seventh, ninth, …) power of the above roots would also be -1.
9. The equation 2333x-2 + 2111x+2 = 2222x+1 +1 has three real roots. Given that their sum is , where m and n are relatively prime positive integers, find m + n.
Let y = 2111x. Then,, or (*). If r1, r2, and r3 are the three original real roots, then are the three real roots of the equation (*), hence , implying . Therefore, , and the answer is 2 + 111 =113.
10. For how many positive integers n less than or equal to 1000 is for all real t? Note: This is an attempt to see how we can convert the De Mviore’s Theorem which says that for all positive integer n and all real t.
= for all real t.
If , then = 1, or , so n º 1 (mod 4).
Among the first 1000 positive integers, there 250 of them.