George Gavrilopoulos
Problem Set 1,P1
There are 4 possible residues
(mod 4) and we have 7 numbers,thus,
from the pigeonhole principle we get that there are 2 of them (call them
a1 , a2 ) that are congruent
(mod 4) thus their sum is even.
The remaining numbers are 5 thus,again there exist 2 (call them a3 , a4 )
that are congruent
(mod 4)thus their sum is even.
Among the 3 remaining numbers,there are 2 that have the same parity
(call them a5 , a6 ) thus their sum is an even number.
The sums s1 = a1 + a2 , s2 = a3 + a4 , s3 = a5 + a6 are all even.
Hence,each is either of the form 4k , k 2 Z or of the form 4k + 2 , k 2 Z.
From the pigeonhole principle,two of these sums are of the same form i.e.
both are congruent to 0
(mod 4) or both are congruent to 2
(mod 4).
In both cases,the sum of these two sums is a multiple of 4 whence we get
the desired result.
1
George Gavrilopoulos
Problem Set 1,Problem 2
Let M, N, P, Q be the midpoints of AB, BC, CD, DA respectively.
In the triangle 4ABC the segment M N connects the midpoints
of AB, BC thus M N k AC and M N =
AC
2 .
In the triangle 4ADC the segment P Q connects the midpoints of
DA, DC thus P Q k AC and P Q =
AC
2 .
Hence P Q = M N and P Q k M N .
With similar argument we obtain M Q = P N and M Q k P N .
Hence M N P Q is a parallelogram (in fact,the relation P Q =k BC
is on its own sufficient to prove that M N P Q is parallelogram).
1
The opposite angles of a parallelogram are equal,that is 6 M N P = 6 M QP .
Since M, N, P, Q all lie on the same circle we also have
M N P = 180
6
6
M QP .Hence 6 M QP = 180
6
M QP )
) 6 M QP = 90 .
Thus M Q ? QP ,and,since M Q k BD and QP k AC we get
BD ? AC which is the desired result.
2
Problem 5
Take the following sums:
( )
1 1 1
1
1
1
+ + +...+ >10
=
10 11 12
19
20
2
1
1
1
1
1
1
+
+
+...+
>100
=
100 101 102
199
200 2
1
1
1
1
1
1
+
+
+...+
>1000
=
1000 1001 1002
1999
2000
2
...
1
1
1
1
1
1
+ n + n +...+
>10 n
=
n
n
n
2
10 10 +1 10 +2
2( 10) −1
2 (10)
( )
( )
(
)
Let Sn be the sum of all the above sums. We clearly see that S n is the finite sum of the reciprocates of distinct,
positive integers which all contain the digit one, since it is the leading digit in all of them. Thus, if S n > 100, then
the set of denominators of the terms of S n satisfy all the given conditions.. Adding the inequalities above
together, we get that:
1
Sn> n
2
It is clear, then, that S200 is a sum that works. Therefore, the set of denominators of S 200 forms a list that satisfies
the given conditions. That is, the following set is a solution:
{ x : x ∈N , 10n≤x <2(10 n) for some n∈ N ,n≤200 }