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Q.1
RMO 1990
There are two boxes. So one of the boxes must contain at least 33
balls (i.e. > 33 balls)
(PHP Principle)
This box have more than 33 balls and four colours, white, black, red
and yellow.
So there are 9 balls of same color (PHP principle)
There are at most 4 different sized available for these 9 balls of same
color.
So among these 9 balls of same color and in the same box, There
must be at least 3 balls of same size
Q.2
(PHP principle).
RMO 1990
For positive numbers b + c, c + a, a + b
(b  c)  (c  a)  (a  b)
>
3
b  a c  a a  b
1
3
……(1)
for positive numbers
1
1
1
,
,
bc ca ab
AM > GM
1 1
1
1   1  1  1 



  



3  b  c c  a a  b   b  c  c  a  a  n 
Multiplying (1) and (2) corresponding sides
2
a  b  c  1  1  1  1   1
3
3 b  c c  a a b 
1
3
….(2)
1
1  9
 1
(a  b  c)



bc ca a b 2

abc abc abc
9



bc
ca
ab
2

a
b
c
9


 3
bc ca ab 2

a
b
c
3



bc ca ab 2
Q.3
RMO 1990
When, in figure below, the folding taker place and B touches M a
new crease PQ is formed.
M
a/2
D
C
x
Q
a-x
P
PQ is perpendicular bisector A MB. So, MQ = BQ
A
If QC = x units, let side of square = a units
B
Then
MQ = BQ = a – x,
By Pythagoras,
2
So
a
(a  x)2     ( x) 2
2

3
x a
8
as a ≠ o
So
3
a
CQ
8

3
QB
a a
8
=
3
5
Q. 4 RMO 1990
Fermat’s Little theorem:
If p is prime and p x a
Then ap-1 = 1 (mod p)
Corollary:
If p is a prime Then
ap = a (mod. p) for any integer a.
Here 1990 = 199 x 10, 199 is prime number
On application
2199 = 2 (mod 199)

199  2199 – 2

199n = 2199 – 2

2199 = 199n + 2

21990
=
(199n +2)10
=
(199n)10 + 10c1 (199n)9 . 2 + …. + 1010
 21990 - 1010 =
(199n)10 + 10c1 (199n)9.2 + …. + 10C9199n92
Here, we know the last – digit on power of 2
21, 25, 29, … = 2
22, 26, 210, … = 4
23, 27, 211, … = 8
24, 28, 212, … = 6
i.e last digit for 210, = 4
and last digit for 21990 = 4
22+m.4
1990 = 2 + 497 x 4
22+497 x 4
So, in 21990 – 210 both terms have last digit 4. Hence it,s difference
has last digit 0 i.e. dividable by 10.
So 21990 – 210 is divisible by 10
Further 21990 – 210 is divisible by 199
So 21990 – 210 is divisible by 1990
So when 21990 is divided by 1990
The remainder will be 210
i.e. 1024
Q. 5 RMO 1990
In the figure
PA + PB > c
PB + PC > a
PC + PA > b
2(PA + PB + PC) > a + b + c

PA + PB + PC >
As P is inside
Further,
AB
>
PB  PA
2
BC
>
PB  PC
2
CA
>
PC  PA
2
a + b + c > PA + PB + PC

2s
>
PA + PB + PC
Proof let there is reflex for point P i.e.p.
A
The
C
b

b + c > 2AP

bc
 AP
2
Similarly
ca
 BP
2
P
B
C
P’
b
C
A’
b + c > AA’
[AP = A’P’]
ab
 CP
2
a + b + c > AP + BP + CP
2S > AP + BP + CP
QED
Q. 6 1990
Method (1)
A number is divisible by 13 if and only if the following operation
gives us a number derisible by 13. Starting at right of the number group the
number, group the digits in threes and alternatively add and subtract the
number. Here the 26th digit may be 0 to 9
e.g.
10345678
678 – 345 + 10
= 343
26th digit
II
III
III
III
III
III
III
III
III
IIx
III
III
III
III
III
III
Operation starting at right group of three numbers alternatively subtracting
and adding gives.
III – III + III - III + III - III + III - III + II x - III + III - III + III – III + III – III + II
=0


+
0 +


0+ 0

+ III x – III +

0
+
0


+ 0
+II
= (IIIx - III) + II
expression
for
x = 0 
=
10
for
x=1
=
11
x=2
=
12
x=3
=
13
x=4
=
14
x=5
=
15
x=6
=
16
x=7
=
17
x=8
=
18
x=9
=
19

So the 26th digit is 3
for divisible by 13
III
Q. 7 RMO 1990
As total combination = 6
So
n( n  1)
6
2
So
n=4
let their ages are a b c and d. such that a < b < c < d
such that
a + b = 30
So, c + d = max
a + d = 41
= 69
a + c = 33
a>d>c
b + c = 58
so b + c = 58
b + d = 66
b + d = 69
c + d = 69
a+c>a+b
Gives
so a + c = 33
a = 2.5
so a + b = 30
b = 27.5
a+d>a+c
c = 30.5
so a + d = 41
d = 38.5
so a + e = 33
Q. 8 1990
As I in circum centre of triangle ABC
A
So
Q
R
I
BPI = IPC = 900
B
C
P
Further I is centriod also
So
BP = PC and PI passes through vertex A.
This gives in Δ ABP & Δ APC
BP = PC
AP is common side
BPI = IPC = 900
Hence both triangles are congruent. Similarly all six triangles are
congruent.
Hence BAC = BCA = CAB
and BP = PC = QC = QA = AR = R3
so Δ ABC is equilateral.
QED