MAD 4203 HOMEWORK 1
SOLUTION
Ch1.
15. The sum of five positive numbers is 100. Prove that there are two numbers
among them whose difference is at most 10.
Proof. By contradiction. Suppose the five numbers are a1 ≤ a2 ≤ a3 ≤ a4 ≤ a5 ,
and the difference between every pair of them is larger than 10. Then we have
a1 > 0, a2 > a1 + 10 > 10, a3 > a2 + 10 > 20, a4 > a3 + 10 > 30, a5 > a4 + 10 > 40.
Their sum a1 + a2 + a3 + a4 + a5 > 0 + 10 + 20 + 30 + 40 = 100, contradicts to the
condition that their sum is 100.
Hence there are two numbers among them whose difference is at most 10.
21. We select n + 1 different integers from the set {1, 2, . . . , 2n}. Prove that
there will always be two among the selected integers whose largest common divisor
is 1.
Proof. Partition the set {1, 2, . . . , 2n} into n sets, namely
{1, 2}, {3, 4}, {5, 6}, . . . , {2n − 1, 2n}.
Let the n + 1 different integers from the set {1, 2, . . . , 2n} be balls and the n sets
be boxes.
Then the number of balls = n + 1 > n = the number of boxes, by pigeon-hole
principle, there is a box that has at least two balls in it. In our terms, there are
two numbers chosen from the same set, say {2k − 1, 2k} for some 1 ≤ k ≤ n. Then
these two integers 2k − 1, 2k are relatively prime to each other.
Thus, there will always be two among the selected integers whose largest common
divisor is 1.
22. Let n ≥ 2. We select n + 1 different integers from the set {1, 2, . . . , 2n}. Is it
true that there will always be two among the selected integers so that one of them
is equal to twice the other.
Solution:
It is not true. Consider the following counter example:
Let n = 2, select n + 1 = 3 integers from the set {1, 2, 3, 4 = 2n}. If we select
1, 3, 4, then there does not exist two integers so that one of them is equal to twice
the other.
1
2
SOLUTION
Ch2.
17. Prove that for all positive integers n, we have
13 + 23 + . . . + n3 = (1 + 2 + . . . + n)2 .
Proof. By mathematical induction.
Initial case: n = 1, 13 = 1 = 12 .
Inductin step: Assume the equation holds for n, then 13 + 23 + . . . + n3 =
(1 + 2 + . . . + n)2 . For n + 1, we have
13 +23 +. . .+n3 +(n+1)3 = (13 +23 +. . .+n3 )+(n+1)3 = (1+2+. . .+n)2 +(n3 +3n2 +3n+1).
On the other hand,
(1+2+. . .+n+(n+1))2 = (1+2+. . .+n)2 +2(1+2+. . .+n)(n+1)+(n+1)2 = (1+2+. . .+n)2 +(n3 +3n2 +3n+1).
Hence the right hand side and the left hand side of the equation are equal, the
equation holds for n + 1. By mathematical induction, the equation holds for any
positive integer n.
23. Prove that for any positive integer n, it is possible to partition any triangle
T into 3n + 1 similar triangles.
Proof. By mathematical induction.
Initial case: n = 1, we will show that any triangle T can be partitioned into 4
similar triangles.
Simply connect the middle points of the 3 edges and we have 4 similar triangles.
Induction step: Suppose any triangle T can be partitioned into 3n + 1 similar
triangles. Choose one of these 3n + 1 similar triangles, we already proved that it
can be partitioned into 4 similar triangles. Thus we have 3 more similar triangles,
which means any triangle T can be partitioned into 3n + 4 = 3(n + 1) + 1 similar
triangles.
Hence, by mathematical induction, for any positive integer n, it is possible to
partition any triangle T into 3n = 1 similar triangles.
MAD 4203 HOMEWORK 1
3
Ch3.
26. How many four-digit positive integers are there that contain the digit 1?
Solution:
All together there are 9 · 10 · 10 · 10 four-digit positive integers. We only need to
find out the number of four-digit positive integers that do not contain the digit 1.
There are 8 · 9 · 9 · 9 of them.
Hence, there are 9 · 10 · 10 · 10 − 8 · 9 · 9 · 9 four-digit positive integers that contain
the digit 1.
30. Let P be a convex n-gon in which no three diagonals intersect in one point.
How many intersection points do the diagonals of P have?
Outline of solution (for a complete solution, you need to fill in the proof of the
bijection):
For a convex n-gon with no three diagonals intersect in one point, there is a
bijection between the set of intersection points and the set of four vertices sets.
Hence the number of intersection points of diagonals is the same as the number of
ways to choose four vertices out of n, which is equal to n4 .
31. Andy and Brenda play with dice. They throw four dice at the same time.
If at least one of the four dice shows a six, then Andy wins, if not, then Brenda.
Who has a greater chance of winning?
Solution:
The number of ways to have four dice without any dice showing six is 54 . Since
there are 64 possible results of rolling four dice, the number of ways to have four
dice with at least a die showing six is 64 − 54 .
Compare these two numbers, we have 64 − 54 = 1296 − 625 = 671 > 625 = 54 .
Thus, Andy has a greater chance of winning.