MathDB
Problems
Contests
National and Regional Contests
India Contests
India LIMIT
2019 LIMIT
2019 LIMIT Category B
2019 LIMIT Category B
Part of
2019 LIMIT
Subcontests
(12)
Problem 12
2
Hide problems
system of inequalities
The system of inequalities
a
−
b
2
≥
1
4
a-b^2\ge\frac14
a
−
b
2
≥
4
1
b
−
c
2
≥
1
4
b-c^2\ge\frac14
b
−
c
2
≥
4
1
c
−
d
2
≥
1
4
c-d^2\ge\frac14
c
−
d
2
≥
4
1
d
−
a
2
≥
1
4
d-a^2\ge\frac14
d
−
a
2
≥
4
1
where
a
,
b
,
c
,
d
a,b,c,d
a
,
b
,
c
,
d
are real numbers has
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
A
)
<
/
s
p
a
n
>
no solutions
<span class='latex-bold'>(A)</span>~\text{no solutions}
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
A
)
<
/
s
p
an
>
no solutions
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
B
)
<
/
s
p
a
n
>
exactly one solution
<span class='latex-bold'>(B)</span>~\text{exactly one solution}
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
B
)
<
/
s
p
an
>
exactly one solution
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
C
)
<
/
s
p
a
n
>
exactly two solutions
<span class='latex-bold'>(C)</span>~\text{exactly two solutions}
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
C
)
<
/
s
p
an
>
exactly two solutions
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
D
)
<
/
s
p
a
n
>
infinitely many solutions
<span class='latex-bold'>(D)</span>~\text{infinitely many solutions}
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
D
)
<
/
s
p
an
>
infinitely many solutions
rational points on circles
Find the number of rational solutions of the following equations (i.e., rational
x
x
x
and
y
y
y
satisfy the equations)
x
2
+
y
2
=
2
x^2+y^2=2
x
2
+
y
2
=
2
x
2
+
y
2
=
3
x^2+y^2=3
x
2
+
y
2
=
3
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
A
)
<
/
s
p
a
n
>
2
and
2
<span class='latex-bold'>(A)</span>~2\text{ and }2
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
A
)
<
/
s
p
an
>
2
and
2
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
B
)
<
/
s
p
a
n
>
2
and
0
<span class='latex-bold'>(B)</span>~2\text{ and }0
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
B
)
<
/
s
p
an
>
2
and
0
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
C
)
<
/
s
p
a
n
>
2
and infinitely many
<span class='latex-bold'>(C)</span>~2\text{ and infinitely many}
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
C
)
<
/
s
p
an
>
2
and infinitely many
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
D
)
<
/
s
p
a
n
>
Infinitely many and
0
<span class='latex-bold'>(D)</span>~\text{Infinitely many and }0
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
D
)
<
/
s
p
an
>
Infinitely many and
0
Problem 11
2
Hide problems
counting with triangle inequality
Let
S
=
{
1
,
2
,
…
,
10
}
S=\{1,2,\ldots,10\}
S
=
{
1
,
2
,
…
,
10
}
. Three numbers are chosen with replacement from
S
S
S
. If the chosen numbers denote the lengths of sides of a triangle, then the probability that they will form a triangle is:
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
A
)
<
/
s
p
a
n
>
101
200
<span class='latex-bold'>(A)</span>~\frac{101}{200}
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
A
)
<
/
s
p
an
>
200
101
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
B
)
<
/
s
p
a
n
>
99
200
<span class='latex-bold'>(B)</span>~\frac{99}{200}
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
B
)
<
/
s
p
an
>
200
99
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
C
)
<
/
s
p
a
n
>
1
2
<span class='latex-bold'>(C)</span>~\frac12
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
C
)
<
/
s
p
an
>
2
1
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
D
)
<
/
s
p
a
n
>
110
200
<span class='latex-bold'>(D)</span>~\frac{110}{200}
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
D
)
<
/
s
p
an
>
200
110
floor function, size comparison
⌊
(
1
⋅
2
+
2
⋅
2
2
+
…
+
100
⋅
2
100
)
⋅
9
−
901
⌋
=
?
\left\lfloor\left(1\cdot2+2\cdot2^2+\ldots+100\cdot2^{100}\right)\cdot9^{-901}\right\rfloor=?
⌊
(
1
⋅
2
+
2
⋅
2
2
+
…
+
100
⋅
2
100
)
⋅
9
−
901
⌋
=
?
Problem 10
2
Hide problems
counting six-digit numbers with 239 and 6|n
Using only the digits
2
,
3
2,3
2
,
3
and
9
9
9
, how many six-digit numbers can be formed which are divisible by
6
6
6
?
sum of 1/(√n+√(n+2))
1
1
+
3
+
1
3
+
5
+
1
5
+
7
+
…
+
1
2017
+
2019
=
?
\frac1{1+\sqrt3}+\frac1{\sqrt3+\sqrt5}+\frac1{\sqrt5+\sqrt7}+\ldots+\frac1{\sqrt{2017}+\sqrt{2019}}=?
1
+
3
1
+
3
+
5
1
+
5
+
7
1
+
…
+
2017
+
2019
1
=
?
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
A
)
<
/
s
p
a
n
>
2019
−
1
2
<span class='latex-bold'>(A)</span>~\frac{\sqrt{2019}-1}2
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
A
)
<
/
s
p
an
>
2
2019
−
1
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
B
)
<
/
s
p
a
n
>
2019
+
1
2
<span class='latex-bold'>(B)</span>~\frac{\sqrt{2019}+1}2
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
B
)
<
/
s
p
an
>
2
2019
+
1
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
C
)
<
/
s
p
a
n
>
2019
−
1
4
<span class='latex-bold'>(C)</span>~\frac{\sqrt{2019}-1}4
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
C
)
<
/
s
p
an
>
4
2019
−
1
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
D
)
<
/
s
p
a
n
>
None of the above
<span class='latex-bold'>(D)</span>~\text{None of the above}
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
D
)
<
/
s
p
an
>
None of the above
Problem 9
2
Hide problems
# of sols to trig equation
The number of solutions of the equation
tan
x
+
sec
x
=
2
cos
x
\tan x+\sec x=2\cos x
tan
x
+
sec
x
=
2
cos
x
, where
0
≤
x
≤
π
0\le x\le\pi
0
≤
x
≤
π
, is
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
A
)
<
/
s
p
a
n
>
0
<span class='latex-bold'>(A)</span>~0
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
A
)
<
/
s
p
an
>
0
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
B
)
<
/
s
p
a
n
>
1
<span class='latex-bold'>(B)</span>~1
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
B
)
<
/
s
p
an
>
1
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
C
)
<
/
s
p
a
n
>
2
<span class='latex-bold'>(C)</span>~2
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
C
)
<
/
s
p
an
>
2
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
D
)
<
/
s
p
a
n
>
3
<span class='latex-bold'>(D)</span>~3
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
D
)
<
/
s
p
an
>
3
extrema of |x^2-1|
Let
f
:
R
→
R
f:\mathbb R\to\mathbb R
f
:
R
→
R
be given by
f
(
x
)
=
∣
x
2
−
1
∣
,
x
∈
R
f(x)=\left|x^2-1\right|,x\in\mathbb R
f
(
x
)
=
x
2
−
1
,
x
∈
R
Then
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
A
)
<
/
s
p
a
n
>
f
has local minima at
x
=
±
1
but no local maxima
<span class='latex-bold'>(A)</span>~f\text{ has local minima at }x=\pm1\text{ but no local maxima}
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
A
)
<
/
s
p
an
>
f
has local minima at
x
=
±
1
but no local maxima
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
B
)
<
/
s
p
a
n
>
f
has a local maximum at
x
=
0
but no local minima
<span class='latex-bold'>(B)</span>~f\text{ has a local maximum at }x=0\text{ but no local minima}
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
B
)
<
/
s
p
an
>
f
has a local maximum at
x
=
0
but no local minima
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
C
)
<
/
s
p
a
n
>
f
has local minima at
x
=
±
1
and a local maximum at
x
=
0
<span class='latex-bold'>(C)</span>~f\text{ has local minima at }x=\pm1\text{ and a local maximum at }x=0
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
C
)
<
/
s
p
an
>
f
has local minima at
x
=
±
1
and a local maximum at
x
=
0
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
D
)
<
/
s
p
a
n
>
None of the above
<span class='latex-bold'>(D)</span>~\text{None of the above}
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
D
)
<
/
s
p
an
>
None of the above
Problem 8
2
Hide problems
increasing/decreasing function
If
f
(
x
)
=
cos
(
x
)
−
1
+
x
2
2
f(x)=\cos(x)-1+\frac{x^2}2
f
(
x
)
=
cos
(
x
)
−
1
+
2
x
2
, then
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
A
)
<
/
s
p
a
n
>
f
(
x
)
is an increasing function on the real line
<span class='latex-bold'>(A)</span>~f(x)\text{ is an increasing function on the real line}
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
A
)
<
/
s
p
an
>
f
(
x
)
is an increasing function on the real line
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
B
)
<
/
s
p
a
n
>
f
(
x
)
is a decreasing function on the real line
<span class='latex-bold'>(B)</span>~f(x)\text{ is a decreasing function on the real line}
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
B
)
<
/
s
p
an
>
f
(
x
)
is a decreasing function on the real line
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
C
)
<
/
s
p
a
n
>
f
(
x
)
is increasing on
−
∞
<
x
≤
0
and decreasing on
0
≤
x
<
∞
<span class='latex-bold'>(C)</span>~f(x)\text{ is increasing on }-\infty<x\le0\text{ and decreasing on }0\le x<\infty
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
C
)
<
/
s
p
an
>
f
(
x
)
is increasing on
−
∞
<
x
≤
0
and decreasing on
0
≤
x
<
∞
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
D
)
<
/
s
p
a
n
>
f
(
x
)
is decreasing on
−
∞
<
x
≤
0
and increasing on
0
≤
x
<
∞
<span class='latex-bold'>(D)</span>~f(x)\text{ is decreasing on }-\infty<x\le0\text{ and increasing on }0\le x<\infty
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
D
)
<
/
s
p
an
>
f
(
x
)
is decreasing on
−
∞
<
x
≤
0
and increasing on
0
≤
x
<
∞
polygon with prime # of sides
Given a regular polygon with
p
p
p
sides, where
p
p
p
is a prime number. After rotating this polygon about its center by an integer number of degrees it coincides with itself. What is the maximal possible number for
p
p
p
?
Problem 7
2
Hide problems
segments intersecting outside circle
A
B
‾
\overline{AB}
A
B
and
C
D
‾
\overline{CD}
C
D
are segments of a circle that intersect at a point
P
P
P
outside the circle. Calculate the value of
x
x
x
. https://services.artofproblemsolving.com/download.php?id=YXR0YWNobWVudHMvNy9lL2RkZGQwNDViNTA1MzM5MDI0NDQ5MDEyOTZhZGUyNTEyYjgyZTNkLnBuZw==&rn=U2NyZWVuIFNob3QgMjAyMS0wNC0yOCBhdCAxMC4wMy4zMSBBTS5wbmc=
1/a+1/b=4/2019 over N
Find the number of ordered pairs of positive integers for which
1
a
+
1
b
=
4
2019
\frac1a+\frac1b=\frac4{2019}
a
1
+
b
1
=
2019
4
Problem 5
2
Hide problems
inequality with parameter
The set of values of
m
m
m
for which
m
x
2
−
6
m
x
+
5
m
+
1
>
0
mx^2-6mx+5m+1>0
m
x
2
−
6
m
x
+
5
m
+
1
>
0
for all real
x
x
x
is
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
A
)
<
/
s
p
a
n
>
m
<
1
4
<span class='latex-bold'>(A)</span>~m<\frac14
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
A
)
<
/
s
p
an
>
m
<
4
1
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
B
)
<
/
s
p
a
n
>
m
≥
0
<span class='latex-bold'>(B)</span>~m\ge0
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
B
)
<
/
s
p
an
>
m
≥
0
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
C
)
<
/
s
p
a
n
>
0
≤
m
≤
1
4
<span class='latex-bold'>(C)</span>~0\le m\le\frac14
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
C
)
<
/
s
p
an
>
0
≤
m
≤
4
1
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
D
)
<
/
s
p
a
n
>
0
≤
m
<
1
4
<span class='latex-bold'>(D)</span>~0\le m<\frac14
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
D
)
<
/
s
p
an
>
0
≤
m
<
4
1
polygon has twice as many diagonals as it has sides
A polygon has twice as many diagonals as it has sides. How many sides does it have?
Problem 4
2
Hide problems
graph of equation
The equation
x
3
y
+
x
y
3
+
x
y
=
0
x^3y+xy^3+xy=0
x
3
y
+
x
y
3
+
x
y
=
0
represents
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
A
)
<
/
s
p
a
n
>
a circle
<span class='latex-bold'>(A)</span>~\text{a circle}
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
A
)
<
/
s
p
an
>
a circle
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
B
)
<
/
s
p
a
n
>
a circle and a pair of straight lines
<span class='latex-bold'>(B)</span>~\text{a circle and a pair of straight lines}
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
B
)
<
/
s
p
an
>
a circle and a pair of straight lines
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
C
)
<
/
s
p
a
n
>
a rectangular hyperbola
<span class='latex-bold'>(C)</span>~\text{a rectangular hyperbola}
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
C
)
<
/
s
p
an
>
a rectangular hyperbola
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
D
)
<
/
s
p
a
n
>
a pair of straight lines
<span class='latex-bold'>(D)</span>~\text{a pair of straight lines}
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
D
)
<
/
s
p
an
>
a pair of straight lines
find locus with angle condition
A particle
P
P
P
moves in the plane in such a way that the angle between the two tangents drawn from
P
P
P
to the curve
y
2
=
4
a
x
y^2=4ax
y
2
=
4
a
x
is always
9
0
∘
90^\circ
9
0
∘
. The locus of
P
P
P
is
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
A
)
<
/
s
p
a
n
>
a parabola
<span class='latex-bold'>(A)</span>~\text{a parabola}
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
A
)
<
/
s
p
an
>
a parabola
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
B
)
<
/
s
p
a
n
>
a circle
<span class='latex-bold'>(B)</span>~\text{a circle}
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
B
)
<
/
s
p
an
>
a circle
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
C
)
<
/
s
p
a
n
>
an ellipse
<span class='latex-bold'>(C)</span>~\text{an ellipse}
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
C
)
<
/
s
p
an
>
an ellipse
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
D
)
<
/
s
p
a
n
>
a straight line
<span class='latex-bold'>(D)</span>~\text{a straight line}
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
D
)
<
/
s
p
an
>
a straight line
Problem 1
2
Hide problems
product of sequence and limit
Let
a
1
=
1
a_1=1
a
1
=
1
and
a
n
=
n
(
a
n
−
1
+
1
)
a_n=n(a_{n-1}+1)
a
n
=
n
(
a
n
−
1
+
1
)
for
n
≥
2
n\ge2
n
≥
2
. Define
p
n
=
∏
i
=
1
n
(
1
+
1
a
i
)
p_n=\prod_{i=1}^n\left(1+\frac1{a_i}\right)
p
n
=
i
=
1
∏
n
(
1
+
a
i
1
)
Then
lim
n
→
∞
p
n
\lim_{n\to\infty}p_n
lim
n
→
∞
p
n
is
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
A
)
<
/
s
p
a
n
>
1
+
e
<span class='latex-bold'>(A)</span>~1+e
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
A
)
<
/
s
p
an
>
1
+
e
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
B
)
<
/
s
p
a
n
>
e
<span class='latex-bold'>(B)</span>~e
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
B
)
<
/
s
p
an
>
e
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
C
)
<
/
s
p
a
n
>
1
<span class='latex-bold'>(C)</span>~1
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
C
)
<
/
s
p
an
>
1
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
D
)
<
/
s
p
a
n
>
∞
<span class='latex-bold'>(D)</span>~\infty
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
D
)
<
/
s
p
an
>
∞
inscribing n circles inside big circle
Let
n
≥
3
n\ge3
n
≥
3
be integer. Assume that inside a big circle, exactly
n
n
n
small circles of radius
r
r
r
can be drawn so that each small circle touches the big circle and also touches both its adjacent small circles. Then, the radius of big circle is
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
A
)
<
/
s
p
a
n
>
r
csc
π
n
<span class='latex-bold'>(A)</span>~r\csc\frac{\pi}n
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
A
)
<
/
s
p
an
>
r
csc
n
π
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
B
)
<
/
s
p
a
n
>
r
csc
(
1
+
2
π
n
)
<span class='latex-bold'>(B)</span>~r\csc\left(1+\frac{2\pi}n\right)
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
B
)
<
/
s
p
an
>
r
csc
(
1
+
n
2
π
)
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
C
)
<
/
s
p
a
n
>
r
csc
(
1
+
π
2
n
)
<span class='latex-bold'>(C)</span>~r\csc\left(1+\frac{\pi}{2n}\right)
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
C
)
<
/
s
p
an
>
r
csc
(
1
+
2
n
π
)
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
D
)
<
/
s
p
a
n
>
r
csc
(
1
+
π
n
)
<span class='latex-bold'>(D)</span>~r\csc\left(1+\frac{\pi}n\right)
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
D
)
<
/
s
p
an
>
r
csc
(
1
+
n
π
)
Problem 6
2
Hide problems
differentiability of absolute value function at 0
Let
f
(
x
)
=
a
0
+
a
1
∣
x
∣
+
a
2
∣
x
∣
2
+
a
3
∣
x
∣
3
f(x)=a_0+a_1|x|+a_2|x|^2+a_3|x|^3
f
(
x
)
=
a
0
+
a
1
∣
x
∣
+
a
2
∣
x
∣
2
+
a
3
∣
x
∣
3
, where
a
0
,
a
1
,
a
2
,
a
3
a_0,a_1,a_2,a_3
a
0
,
a
1
,
a
2
,
a
3
are constant. Then
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
A
)
<
/
s
p
a
n
>
f
(
x
)
is differentiable at
x
=
0
if whatever be
a
0
,
a
1
,
a
2
,
a
3
<span class='latex-bold'>(A)</span>~f(x)\text{ is differentiable at }x=0\text{ if whatever be }a_0,a_1,a_2,a_3
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
A
)
<
/
s
p
an
>
f
(
x
)
is differentiable at
x
=
0
if whatever be
a
0
,
a
1
,
a
2
,
a
3
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
B
)
<
/
s
p
a
n
>
f
(
x
)
is not differentiable at
x
=
0
if whatever be
a
0
,
a
1
,
a
2
,
a
3
<span class='latex-bold'>(B)</span>~f(x)\text{ is not differentiable at }x=0\text{ if whatever be }a_0,a_1,a_2,a_3
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
B
)
<
/
s
p
an
>
f
(
x
)
is not differentiable at
x
=
0
if whatever be
a
0
,
a
1
,
a
2
,
a
3
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
C
)
<
/
s
p
a
n
>
f
(
x
)
is differentiable at
x
=
0
only if
a
1
=
0
<span class='latex-bold'>(C)</span>~f(x)\text{ is differentiable at }x=0\text{ only if }a_1=0
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
C
)
<
/
s
p
an
>
f
(
x
)
is differentiable at
x
=
0
only if
a
1
=
0
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
D
)
<
/
s
p
a
n
>
f
(
x
)
is differentiable at
x
=
0
only if
a
1
=
0
,
a
3
=
0
<span class='latex-bold'>(D)</span>~f(x)\text{ is differentiable at }x=0\text{ only if }a_1=0,a_3=0
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
D
)
<
/
s
p
an
>
f
(
x
)
is differentiable at
x
=
0
only if
a
1
=
0
,
a
3
=
0
8n+1 perfect square, find conditions on n
If
n
n
n
is a positive integer such that
8
n
+
1
8n+1
8
n
+
1
is a perfect square, then
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
A
)
<
/
s
p
a
n
>
n
must be odd
<span class='latex-bold'>(A)</span>~n\text{ must be odd}
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
A
)
<
/
s
p
an
>
n
must be odd
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
B
)
<
/
s
p
a
n
>
n
cannot be a perfect square
<span class='latex-bold'>(B)</span>~n\text{ cannot be a perfect square}
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
B
)
<
/
s
p
an
>
n
cannot be a perfect square
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
C
)
<
/
s
p
a
n
>
n
cannot be a perfect square
<span class='latex-bold'>(C)</span>~n\text{ cannot be a perfect square}
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
C
)
<
/
s
p
an
>
n
cannot be a perfect square
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
D
)
<
/
s
p
a
n
>
None of the above
<span class='latex-bold'>(D)</span>~\text{None of the above}
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
D
)
<
/
s
p
an
>
None of the above
Problem 2
2
Hide problems
units digit of factorial sum
The digit in unit place of
1
!
+
2
!
+
…
+
99
!
1!+2!+\ldots+99!
1
!
+
2
!
+
…
+
99
!
is
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
A
)
<
/
s
p
a
n
>
3
<span class='latex-bold'>(A)</span>~3
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
A
)
<
/
s
p
an
>
3
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
B
)
<
/
s
p
a
n
>
0
<span class='latex-bold'>(B)</span>~0
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
B
)
<
/
s
p
an
>
0
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
C
)
<
/
s
p
a
n
>
1
<span class='latex-bold'>(C)</span>~1
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
C
)
<
/
s
p
an
>
1
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
D
)
<
/
s
p
a
n
>
7
<span class='latex-bold'>(D)</span>~7
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
D
)
<
/
s
p
an
>
7
complex subsets
Let
C
\mathbb C
C
denote the set of all complex numbers. Define
A
=
{
(
z
,
w
)
∣
z
,
w
∈
C
and
∣
z
∣
=
∣
w
∣
}
A=\{(z,w)|z,w\in\mathbb C\text{ and }|z|=|w|\}
A
=
{(
z
,
w
)
∣
z
,
w
∈
C
and
∣
z
∣
=
∣
w
∣
}
B
=
{
(
z
,
w
)
∣
z
,
w
∈
C
and
z
2
=
w
2
}
B=\{(z,w)|z,w\in\mathbb C\text{ and }z^2=w^2\}
B
=
{(
z
,
w
)
∣
z
,
w
∈
C
and
z
2
=
w
2
}
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
A
)
<
/
s
p
a
n
>
A
=
B
<span class='latex-bold'>(A)</span>~A=B
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
A
)
<
/
s
p
an
>
A
=
B
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
B
)
<
/
s
p
a
n
>
A
⊂
B
and
A
≠
B
<span class='latex-bold'>(B)</span>~A\subset B\text{ and }A\ne B
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
B
)
<
/
s
p
an
>
A
⊂
B
and
A
=
B
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
C
)
<
/
s
p
a
n
>
B
⊂
A
and
B
≠
A
<span class='latex-bold'>(C)</span>~B\subset A\text{ and }B\ne A
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
C
)
<
/
s
p
an
>
B
⊂
A
and
B
=
A
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
D
)
<
/
s
p
a
n
>
None of the above
<span class='latex-bold'>(D)</span>~\text{None of the above}
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
D
)
<
/
s
p
an
>
None of the above
Problem 3
2
Hide problems
factors of positive integer, find sum of reciprocals
Let
d
1
,
d
2
,
…
,
d
k
d_1,d_2,\ldots,d_k
d
1
,
d
2
,
…
,
d
k
be all factors of a positive integer
n
n
n
including
1
1
1
and
n
n
n
. If
d
1
+
d
2
+
…
+
d
k
=
72
d_1+d_2+\ldots+d_k=72
d
1
+
d
2
+
…
+
d
k
=
72
then
1
d
1
+
1
d
2
+
…
+
1
d
k
\frac1{d_1}+\frac1{d_2}+\ldots+\frac1{d_k}
d
1
1
+
d
2
1
+
…
+
d
k
1
is
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
A
)
<
/
s
p
a
n
>
k
2
72
<span class='latex-bold'>(A)</span>~\frac{k^2}{72}
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
A
)
<
/
s
p
an
>
72
k
2
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
B
)
<
/
s
p
a
n
>
72
k
<span class='latex-bold'>(B)</span>~\frac{72}k
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
B
)
<
/
s
p
an
>
k
72
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
C
)
<
/
s
p
a
n
>
72
n
<span class='latex-bold'>(C)</span>~\frac{72}n
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
C
)
<
/
s
p
an
>
n
72
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
D
)
<
/
s
p
a
n
>
None of the above
<span class='latex-bold'>(D)</span>~\text{None of the above}
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
D
)
<
/
s
p
an
>
None of the above
ringness of sets
A subset
W
W
W
of the set of real numbers is called a ring if it contains
1
1
1
and if for all
a
,
b
∈
W
a,b\in W
a
,
b
∈
W
, the numbers
a
−
b
a-b
a
−
b
and
a
b
ab
ab
are also in
W
W
W
. Let
S
=
{
m
2
n
∣
m
,
n
∈
Z
}
S=\left\{\frac m{2^n}|m,n\in\mathbb Z\right\}
S
=
{
2
n
m
∣
m
,
n
∈
Z
}
and
T
=
{
p
q
∣
p
,
q
∈
Z
,
q
odd
}
T=\left\{\frac pq|p,q\in\mathbb Z,q\text{ odd}\right\}
T
=
{
q
p
∣
p
,
q
∈
Z
,
q
odd
}
. Then
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
A
)
<
/
s
p
a
n
>
neither
S
nor
T
is a ring
<span class='latex-bold'>(A)</span>~\text{neither }S\text{ nor }T\text{ is a ring}
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
A
)
<
/
s
p
an
>
neither
S
nor
T
is a ring
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
B
)
<
/
s
p
a
n
>
S
is a ring,
T
is not a ring
<span class='latex-bold'>(B)</span>~S\text{ is a ring, }T\text{ is not a ring}
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
B
)
<
/
s
p
an
>
S
is a ring,
T
is not a ring
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
C
)
<
/
s
p
a
n
>
T
is a ring,
S
is not a ring
<span class='latex-bold'>(C)</span>~T\text{ is a ring, }S\text{ is not a ring}
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
C
)
<
/
s
p
an
>
T
is a ring,
S
is not a ring
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
D
)
<
/
s
p
a
n
>
both
S
and
T
are rings
<span class='latex-bold'>(D)</span>~\text{both }S\text{ and }T\text{ are rings}
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
D
)
<
/
s
p
an
>
both
S
and
T
are rings