Download 1 - Homework Tutoring

1. Randomly cut a segment into three parts. What is the probability that the parts
can form a triangle?
Let’s take a segment of length 1, for instance. To cut it into three parts is to take such
numbers x, y, z that:
x>0
y>0
z>0
x+y+z=1
If x and y are picked, z is determined as z = 1 – x – y.
So, the division is determined uniquely by two numbers x, y such that:
x>0
y>0
x+y<1
Let’s graph this area on xy-plane:
The measure of this area is 1/2.
Let’s determine which points of this area correspond to such segment divisions after
which the parts x, y, z can form a triangle. This is so when the triangle inequalities are
fulfilled:
x+y>z
x+z>y
y+z>x
They can be rewritten as
x + y > 1 – x – y; 2x + 2y > 1; x + y > 1/2
x + (1 – x – y) > y; 1 > 2y; y < 1/2
y + (1 – x – y) > x; 1 > 2x; x < 1/2
Let’s graph this area:
It is easy from the geometry that its measure is 1/8.
So, the probability of getting onto it is 1/8 : 1/2 = 1/4.
2. Show that if N is a positive integer, so is (N5/5) + (N3/3) + 7N/15
(N5/5) + (N3/3) + 7N/15 = N * (3N4 + 5N2 + 7) / 15
It is an integer if N * (3N4 + 5N2 + 7) is divisible by 15.
Let’s look through all possible reminders after division of N by 15, and calculate the
reminder after division of N * (3N4 + 5N2 + 7) by 15. The equalities in the table are
modulo 15.
N mod 15
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
N2 mod 15
0
1
4
9
1
10
6
4
4
6
10
1
9
4
1
N4 mod 15
0
1
1
6
1
10
6
1
1
6
10
1
6
1
1
N*(3N4 + 5N2 + 7) mod 15
0
1*(3*1 + 5*1 + 7) = 15 = 0
2*(3*1+5*4 + 7) = 2*30 = 0
3*(3*6+5*9+7) = 3*70 = 0
4*(3*1 + 5*1 + 7) = 4*15 = 0
5*(3*10 + 5*10 + 7) = 5*87 = 0
6*(3*6 + 5*6 + 7) = 6*55 = 0
7*(3*1 + 5*4 + 7) = 7*30 = 0
7*(3*1 + 5*4 + 7) = 7*30 = 0
9*(3*6 + 5*6 + 7) = 9*55 = 0
10*(3*10 + 5*10 + 7) = 5*87 = 0
11*(3*1 + 5*1 + 7) = 11*15 = 0
12*(3*6+5*9+7) = 12*70 = 0
13*(3*1+5*4 + 7) = 13*30 = 0
14*(3*1 + 5*1 + 7) = 14*15 = 0
In any case N * (3N4 + 5N2 + 7) is divisible by 15.
3. Find the smallest positive integer all of whose substrings are prime numbers?
I guess it is 2. The only substring is 2. It is a prime number.
4. What is the one’s digit of 311,141,953
30 = 1; 31 = 3; 32 = 9; 33 = 27; 34 = 81; 35 = 243; and so on
The last digit pattern is 1, 3, 9, 7, repeating over and over.
So, the last digit of 3k depends on the value k modulo 4.
We have to find 11,141,953 modulo 4; it is equal to 53 modulo 4, that is, 1.
Hence, the last digit of 311,141,953 is 3.
5. A bowl contains 16 poker chips: 6 are red, 7 are white, and 3 are blue. A
blindfolded person draws 4 chips from the bowl at random. What is the
probability that all three colors are represented.
The total number of ways to draw 4 chips from 16 is C(16,4) = 16! / (4! 12!) = 1820.
Let’s calculate the number of ways to draw such 4, that all three colors are represented.
This can be achieved in 3 different cases:
1) 2 red, 1 white and 1 blue
2) 1 red, 2 white and 1 blue
3) 1 red, 1 white and 2 blue
For the first case, the number of possibilities is C(6,2)*7*3 = 15*7*3 = 315;
For the second case, it is 6*C(7,2)*3 = 6*21*3 = 378;
For the third case, 6*7*C(3,2) = 6*7*3 = 126.
The total is 315 + 378 + 126 = 819.
The required probability is 819 / 1820 = 0.45.