MATH 433
January 30, 2015
Quiz 2: Solutions
Problem 1. Using the induction principle, prove that
for every positive integer n.
1
1
1
1
+
+· · ·+
= 1−
1·2 2·3
n(n + 1)
n+1
First consider the case n = 1. In this case the formula reduces to
Now assume that the formula holds for n = k, that is,
1
2
= 1 − 12 , which is a true equality.
1
1
1
1
+
+ ··· +
=1−
.
1·2 2·3
k(k + 1)
k+1
Then
1
1
1
1
1
1
+
+ ··· +
+
=1−
+
1·2 2·3
k(k + 1) (k + 1)(k + 2)
k + 1 (k + 1)(k + 2)
1
k+1
1
1
1
=1−
·
=1−
,
1−
=1−
k+1
k+2
k+1 k+2
k+2
which means that the formula holds for n = k + 1 as well.
By induction, the formula holds for every positive integer n.
Problem 2. A positive integer m has the prime decomposition 7p1 p2 p3 p4 , where p1 , p2 , p3 , p4 are
some prime numbers different from 7 (not necessarily distinct). The integer m + 100 has the prime
decomposition 25 q1 q2 , where q1 , q2 are odd prime numbers (not necessarily distinct). The integer
m + 200 has the prime decomposition 52 r1 r2 r3 r4 , where r1 , r2 , r3 , r4 are prime numbers different from
5 (not necessarily distinct). Find m.
Solution: m = 700.
The prime decomposition of 100 is 22 · 52 . Since the number m + 100 is divisible by 25 = 32, it follows
that m = (m + 100) − 100 is divisible by 22 = 4.
The prime decomposition of 200 is 23 · 52 . Since the numbers m + 200 and 200 are divisible by 52 = 25,
so is the number m = (m + 200) − 200.
By the above the prime decomposition of m contains 22 · 52 · 7. As there are only 5 factors in this
decomposition, the number m is exactly 22 · 52 · 7 = 700. Then m + 100 = 800 = 25 · 52 and m + 200 =
900 = 22 · 32 · 52 .