MathDB
Problems
Contests
National and Regional Contests
India Contests
Postal Coaching
2015 Postal Coaching
2015 Postal Coaching
Part of
Postal Coaching
Subcontests
(11)
5
1
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Sum of finite set is not a square
Prove that there exists a set of infinitely many positive integers such that the elements of no finite subset of this set add up to a perfect square.
4
1
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Interesting Recursive Integer Sequence
The sequence
<
a
n
>
<a_n>
<
a
n
>
is defined as follows,
a
1
=
a
2
=
1
a_1=a_2=1
a
1
=
a
2
=
1
,
a
3
=
2
a_3=2
a
3
=
2
,
a
n
+
3
=
a
n
+
2
a
n
+
1
+
n
!
a
n
,
a_{n+3}=\frac{a_{n+2}a_{n+1}+n!}{a_n},
a
n
+
3
=
a
n
a
n
+
2
a
n
+
1
+
n
!
,
n
≥
1
n \ge 1
n
≥
1
. Prove that all the terms in the sequence are integers.
3
1
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Monic Polynomials divisible by powers of (x-1)
Let
n
≥
2
n\ge2
n
≥
2
and let
p
(
x
)
=
x
n
+
a
n
−
1
x
n
−
1
⋯
a
1
x
+
a
0
p(x)=x^n+a_{n-1}x^{n-1} \cdots a_1x+a_0
p
(
x
)
=
x
n
+
a
n
−
1
x
n
−
1
⋯
a
1
x
+
a
0
be a polynomial with real coefficients. Prove that if for some positive integer
k
(
<
n
)
k(<n)
k
(
<
n
)
the polynomial
(
x
−
1
)
k
+
1
(x-1)^{k+1}
(
x
−
1
)
k
+
1
divides
p
(
x
)
p(x)
p
(
x
)
then
∑
i
=
0
n
−
1
∣
a
i
∣
≥
1
+
2
k
2
n
\sum_{i=0}^{n-1}|a_i| \ge 1 +\frac{2k^2}{n}
i
=
0
∑
n
−
1
∣
a
i
∣
≥
1
+
n
2
k
2
2
1
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Natural numbers in Harmonic Progression
Prove that there exists a real number
C
>
1
C > 1
C
>
1
with the following property. Whenever
n
>
1
n > 1
n
>
1
and
a
0
<
a
1
<
a
2
<
⋯
<
a
n
a_0 < a_1 < a_2 <\cdots < a_n
a
0
<
a
1
<
a
2
<
⋯
<
a
n
are positive integers such that
1
a
0
,
1
a
1
⋯
1
a
n
\frac{1}{a_0},\frac{1}{a_1} \cdots \frac{1}{a_n}
a
0
1
,
a
1
1
⋯
a
n
1
form an arithmetic progression, then
a
0
>
C
n
a_0 > C^n
a
0
>
C
n
.
1
1
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Iterative Polynomials
Find all real polynomials
P
(
x
)
P(x)
P
(
x
)
that satisfy
P
(
x
3
−
2
)
=
P
(
x
)
3
−
2
P(x^3-2)=P(x)^3-2
P
(
x
3
−
2
)
=
P
(
x
)
3
−
2
Problem 6
2
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Composition of Number Theoretic Functions
Let
k
∈
N
k \in \mathbb{N}
k
∈
N
, let
x
k
x_k
x
k
denote the nearest integer to
k
\sqrt k
k
. Show that for each
m
∈
N
m \in \mathbb {N}
m
∈
N
,
∑
k
=
1
m
1
x
k
=
f
(
m
)
+
m
f
(
m
)
+
1
\sum_{k=1}^{m} \frac{1}{x_k} = f(m)+ \frac{m}{f(m)+1}
k
=
1
∑
m
x
k
1
=
f
(
m
)
+
f
(
m
)
+
1
m
, where
f
(
m
)
f(m)
f
(
m
)
is the integer part of
4
m
−
3
−
1
2
\frac{\sqrt{4m-3}-1}{2}
2
4
m
−
3
−
1
Numbers that are sums of squares and cubes
Show that there are infinitely many natural numbers which are simultaneously a sum of two squares and a sum of two cubes but which are not a sum of two
6
−
6-
6
−
th powers.
Problem 4
4
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Problem 3
4
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Problem 1
5
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Problem 2
4
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Problem 5
4
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