1

REGIONAL MATHEMATICAL OLYMPIAD (RMO) - 2009

Time : 3 Hours
Instructions :
• Calculators (in any form) and protractors are forbidden.
• Rulers and compasses are allowed.
• Answer all the questions. Maximum marks : 100
• Answer to each question should start on a new page. Clearly
indicate the question number.

1. Let ABC be a triangle in which AB = AC and let I be its in-centre.

Suppose BC = AB + AI. Find  BAC. [16]

2. Show that there is no integer a such that a2 – 3a – 19 is divisible by

289. [15]

3. Show that 32008 + 42009 can be written as product of two positive

integers each of which is larger than 2009182. [16]

4. Find the sum of all 3-digit natural numbers which contain at least
one odd digit and at least one even digit. [15]

5. A convex polygon  is such that the distance between any two

vertices of  does not exceed 1.
(i) Prove that the distance between any two points on the
boundary of  does not exceed 1.
(ii) If X and Y are two distinct points inside  , prove that there
exists a point Z on the boundary of  such that XZ + YZ  1.
[19]

6. In a book with page numbers from 1 to 100, some pages are torn off.
The sum of the numbers on the remaining pages is 4949. How many
pages are torn off ? [19]
 2
REGIONAL MATHEMATICAL OLYMPIAD - 2009

HINTS & SOLUTIONS

1. We observe that  AIB = 90° + (C/2). Extend CA to D such that
AD = AI. Then CD = CB by the hypothesis.
Hence  CDB =  CBD = 90° – (C/2). Thus
 AIB +  ADB = 90° + (C/2) + 90° – (C/2) = 180°
Hence ADBI is a cyclic quadrilateral. C
This implies that
B
ADI =  ABI =
2 I
But ADI is isosceles,
since AD = AI this gives A B
 DAI = 180° – 2 ( ADI) = 180° – B
Thus  CAI = B and this gives A = 2B.
D
Since C = B,
we obtain 4B = 180° and hence B = 45°. We thus get A = 2B = 90°.

2. We write a2 – 3a – 19 = a2 – 3a – 70 + 51 = (a – 10) (a + 7) + 51
Suppose 289 divides a2 – 3a – 19 for some integer a. Then 17 divides it
and hence 17 divides (a – 10) (a + 7). Since 17 is a prime, it must divide
(a – 10) or (a + 7). But (a + 7) – (a – 10) = 17. Hence whenever 17 divides
one of (a – 10) and (a + 7), it must divide the other also. Thus 172 = 289
divides (a – 10)(a + 7). It follows that 289 divides 51, which is impossible.
Thus, there is no integer a for which 289 divides a2 – 3a – 19.

3. We use the standard factorisation :

x4 + 4y4 = (x2 + 2xy + 2y2) (x2 – 2xy + 2y2)
we observe that for any integers x, y
x2 + 2xy + 2y2 = (x + y)2 + y2  y2
and x2 – 2xy + 2y2 = (x – y)2 + y2  y2 .
We write 32008 + 42009 = 32008 + 4(42008) = (3502)4 + 4(4502)4
Taking x = 3502 and y = 4502, we se that 32008 + 42009 = ab, where
a  (4502)2 , b  (4502)2
But we have (4502)2 = 22008 > 22002 = (211)182 > (2009)182,
since 211 = 2048 > 2009

4. Let X denote the set of all 3-digit natural numbers; let O be those
numbers in X having only odd digits; and E be those numbers in X
X
having only even digits. Then
O  E is the set of all 3-digit natural
numbers having at least one odd digit and at least one even digit. The
desired sum is therefore  x –  y –  z
x X y O zE

It is easy to compute the first sum ;

 3

999 99
999  1000 99  100
 x   j – k = – = 50 × 9891 = 494550
x X j1 k 1 2 2
Consider the set O. Each number in O has its digits from the set
{1, 3, 5, 7, 9}. Suppose the digit in unit’s place is 1. we can fill the digit in
ten’s place in 5 ways and the digit in hundred’s place in
5 ways. Thus there are 25 numbers in the set O each of which has 1 in
its unit’s place. Similarly, there are 25 numbers whose digit in unit’s
place is 3 ; 25 having its digit in unit’s place as 5; 25 with
7 amd 25 with 9. Thus the sum of the digits in unit’s place of all the
numbers in O is 25(1 + 3 + 5 + 7 + 9) = 25 × 25 = 625
A similar argument shows that the sum of digits in ten’s place of all the
numbers in O is 625 and that in hundred’s place is also 625. Thus the
sum of all the numbers in O is
625(102 + 10 + 1) = 625 × 111 = 69375
Consider the set E. The digits of numbers in E are from the set
{0, 2, 4, 6, 8}, but the digit in hundred’s place is never 0. Suppose the
digit in unit’s place is 0. There are 4 × 5 = 20 such numbers. Similarly, 20
numbers each having digits 2, 4, 6, 8 in their unit’s place. Thus the sum
of the digits in unit’s place of all the numbers in E is
20(0 + 2 + 4 + 6 + 8) = 20 × 20 = 400
A similar reasoning shows that the sum of the digits in ten’s place of all
the numbers in E is 400, but the sum of the digits in hundred’s place of
all the numbers in E is 25 × 20 = 500. Thus the sum of all the numbers
in E is
500 × 102 + 400 × 10 + 400 = 54400
The required sum is
494550 – 69375 – 54400 = 370775

5. Let S and T be two points on the Q

boundary of  , with S lying on the T
side AB and T lying on the side PQ P
of  . (See Fig. 1.) Join TA, TB, TS.
Now ST lies between TA and TB in
triangle TAB. One of
AST and BST is at least 90°, say
AST  90°. Hence AT  TS. But AT
lies inside triangle APQ and one of
ATP and ATQ is at least 90°, say A S B
ATP  90°. Then AP  AT. Thus we
get TS  AT  AP  1.
 4
F
Z2
E
Y

C Z1 D
(ii) Let X and Y be points in the interior  . Join XY and produce them on
either side to meet the sides CD and EF of  at Z1 and Z2 respectively..
We have
(XZ1 + YZ1) + (XZ2 + YZ2) = (XZ1 + XZ2) + (YZ1 + yZ2) = 2Z1Z2  2
by the first part. Therefore one of the sums XZ 1 + YZ 1 and
XZ2 + YZ2 is at most 1. We may choose Z accordingly as Z1 or Z2.

6. Suppose r pages of the book are torn off. Note that the page
numbers on both the sides of a page are of the form 2k – 1 and 2k, and
their sum is 4k –1. The sum of the numbers on the torn pages must be
of the from
4k1 – 1 + 4k2 – 1 + . . . + 4kr – 1 = 4(k1 + k2 + . . . + kr) – r
The sum of the numbers of all the pages in the untorn book is
1 + 2 + 3 + . . . + 100 = 5050
Hence the sum of the numbers on the torn pages is
5050 – 4949 = 101
We therefore have
4(k1 + k2 + . . . + kr) – r = 101
This shows that r  3 (mod 4). Thus r = 4 + 3 for some   0.
Suppose r  7, and suppose k1 < k2 < k3 < . . . < kr. Then we see that
4(k1 + k2 + . . . + kr) – r  4(k1 + k2 + . . . + k7) – 7
 4(1 + 2 + . . . + 7) – 7
= 4 × 28 – 7 = 105 > 101
Hence r = 3. This leads to k1 + k2 + k3 = 26 and one can choose distinct
positive integers k1, k2, k3 in several ways.
 5

REGIONAL MATHEMATICAL OLYMPIAD (RMO) - 2010

Time : 3 Hours
Instructions :
• Calculators (in any form) and protractors are not allowed.
• Rulers and compasses are allowed.
• Answer all the questions. Maximum marks : 100
• Answer to each question should start on a new page. Clearly
indicate the question number.

1. Let ABCDEF be a convex hexagon in which the diagonals

AD, BE, CF are concurrent at O. Suppose the area of triangle OAF
is the geometric mean of those of OAB and OEF; and the area of
triangle OBC is the geometric mean of those of OAB and OCD.
Prove that the area of triangle OED is the geometric mean of those
of OCD and OEF. [16]
2. Let P1(x) = ax2 – bx – c, P2(x) = bx2 – cx – a, P3(x) = cx2 – ax – b be
three quadratic polynomials where a, b, c are non-zero real numbers.
Suppose there exists a real number  such that
P1() = P2() = P3(). Prove that a = b = c. [17]
3. Find the number of 4-digit numbers (in base 10) hav ing
non-zero digits and which are divisible by 4 but not by 8. [16]
4. Find three distinct positive integers with the least possible sum
such that the sum of the reciprocals of any two integers among
them is an integral multiple of the reciprocal of the third integer.
[17]
5. Let ABC be a triangle in which A = 60°. Let BE and CF be the
bisectors of the angle B and C with E on AC and F on AB. Let M
be the reflection of A in the line EF. Prove that M lies on BC.
[17]
 n 
6. For each integer n  1, define an =   , where [x] denotes the
 [ n ] 
largest integer not exceeding x, for any real number x. Find the
number of all n is the set {1, 2, 3,....,2010} f or which
an > an+1. [17]
 6

REGIONAL MATHEMATICAL OLYMPIAD (RMO) - 2010

HINTS & SOLUTIONS

1. Let OA = a, OB = b, OC = c, OD = d, OE = e, OF = f,
[OAB] = x, [OCD] = y, [OEF] = z, [ODE] = u, [OFA] = v and [OBC]
= w. We are given that v 2 = zx, w2 = xy and we have to prove that u2
= yz.
Since AOB = DOE, we have

1
de sin DOE
u 2 de
= 1 = .
x ab sin AOB ab
2

v fa w bc
Similarly, we obtain = , =
y cd z ef
Multiplying, these three equalities, we get uvw = xyz. Hence
x2y2z2 = u2v 2w2 = u2(zx) (xy).
This gives u2 = yz, as desired.

2. We have three relations :

a2 – b – c = , b2 – c – a = , c2 – a – b = ,
Where  is the common value. Eliminating 2 from these, taking
these equations pair -wise, we get three relations :
(ca – b2) – (bc – a2) = (b – a), (ab – c2) – (ca – b2) = (c – b), (bc
– a2) – (ab – c2) = (a – c).
Adding these three, we get
(ab + bc + ca – a2 – b2 – c2) ( – 1) = 0.
(Alternatively, multiplying above relations respectively by
b – c, c – a and a – b, and adding also leads to this.) Thus either ab
+ bc + ca – a2 – b2 – c2 = 0 or  = 1. In the first case
1
0 = ab + bc + ca – a2 – b2 – c2 = ((a – b)2 + (b – c)2 + (c – a)2)
2
shows that a = b = c. If  = 1, then we obtain
a – b – c = b – c – a = c – a – b,
and one again we obtain a = b = c.
 7
3. We divide the even 4-digit numbers having non-zero digits into
4 classes : those ending in 2, 4, 6, 8.

(A) Suppose a 4-digit number ends in 2. Then the second right digit
must be odd in order to be divisible by 4. Thus the last 2 digits must
be of the form 12, 32, 52,72 or 92. If a number ends in
12, 52 or 92, then the previous digit must be even in order not to be
divisible by 8 and we have 4 admissible even digits. Now the left
most digit of such a 4-digit number can be any non-zero digit and
there are 9 such ways, and we get 9 × 4 × 3 = 108 such number. If
a number ends in 32 or 72, then the previous digit must be odd in
order not to be divisible by 8 and we have 5 admissible odd digits.
Here again the left most digit of such a 4-digit number can be any
non-zero digit and there are 9 such ways, and we get 9 × 5 × 2 = 90
such number. Thus the number of 4-digit number having non-zero
digits, ending in 2, divisible by 4 not by 8 is 108 + 90 = 198.

(B) If the number ends in 4, then the previous digit must be even for
divisibility by 4. Thus the last two digits must be of the form
24, 44, 54, 84. If we take numbers ending with 24 and 64, then the
previous digit must be odd for non-divisibility by 8 and the left most
digit can be any non-zero digit. Here we get 9 × 5 × 2 = 90 such
numbers. If the last two digits are of the form 44 and 84, then previous
digit must be even for non-divisibility by 8. And the left most digit
can take 9 possible values. We thus get 9 × 4 × 2 = 72 numbers.
Thus the admissible numbers ending in 4 is 90 + 72 = 162.

(C) If a number ends with 6, then the last two digits must be of the form
16, 36, 56, 76, 96. For numbers ending with 16, 56, 76, the previous
digit must be odd. For numbers ending with 36, 76, the previous
digit must be even. Thus we get here (9 × 5 × 3) + (9 × 4 × 2)
= 135 + 72 = 207 numbers.

(D) If a number ends with 8, then the last two digits must be of the form
28, 48, 68, 88. For numbers ending with 28, 68, the
previous digit must be even. For numbers ending with 48, 88 the
previous digit must be odd. Thus we get (9 × 4 × 2) + (9 × 5 × 2)
= 72 + 90 = 162 numbers.
 8
Thus the number of 4-digit numbers, having non-zero digits, and
divisible by 4 but not by 8 is 198 + 168 + 207 + 162 = 729.

Alternative Solution : If we take any four consecutive even numbers

and divide them by 8, we get remainders 0, 2, 4, 6 in some order.
Thus there is only one number of the from 8k + 4 among them
which is divisible by 4 but not by 8. Hence if we take four even
consecutive numbers.

1000 a + 100 b + 10c + 2, 1000a + 100b + 10c + 4,

1000 a + 100b + 10c + 6, 1000a + 100b + 10 c + 8,
there is exactly one among these four which is divisible by 4 but
not by 8. Now we can divide the set of all 4-digit even numbers with
non-zero digits into groups of 4 such consecutive even numbers
with a, b, c nonzero. And in each group, there is exactly one number
which is divisible by 4 but not by 8. The number of such groups is
precisely equal to 9 × 9 × 9 = 729, since we can vary a, b.c in the
set {1, 2, 3, 4, 5, 6, 7, 8, 9}.

4. Let x, y, z be three distinct positive integers satisfying the given

conditions. We may assume that x < y < z. Thus we have three
1 1 b 1 1
relations :  = a, 1 1 = ,  = c,
y z x z x y x y z
For some positive integers a, b, c. Thus
1 1 1 b 1
  = a1 = =
c 1
= r,,
x y z x y z
say, Since x < y < z, we observe that a < b < c. We also get
1 r 1 r 1 r
= , = , = ,
x a1 y b 1 z c 1
1 1 1 r r r
Adding these, we obtain r =   =   ,
x y z a1 b1 c 1
1 1 1
or   = 1. (1)
a1 b1 c 1

1  1 3
Using a < b < c, we get 1 =   <
a1 b1 c 1 a1
Thus a < 2. We conclude that a = 1. Putting this in the relation (1),
 9

1 1 1 1
we get  =1– = .
b 1 c 1 2 2
1 2
Hence b < c gives <
2 b 1
Thus b + 1 < 4 or b < 3. Since b > a = 1, we must have b = 2. This
1 1 1 1
gives = – = ,
c 1 2 3 6
or c = 5. Thus x : y : z = a + 1 : b + 1 : c + 1 = 2 : 3 : 6. Thus the
required numbers with the least sum are 2, 3, 6.
Alternative Solution : We first observe that (1, a, b) is not a
solution whenever 1 < a < b. Otherwise we should have
1 1  ab
 = =  for some integer . Hence we obtain =
a b 1 ab
showing that a|b and b|a. Thus a = b contradicting a  b. Thus the
least number should be 2. It is easy to verify that (2, 3, 4,) and (2,
3, 5) are not solutions and (2, 3, 6) sati sf ies all the
conditions. (We may observe) (2, 4, 5) is also not a solution.) Since
3 + 4 + 5 = 12 > 11 = 2 + 3 + 6, it follows that (2, 3, 6) has the
required minimality.
5. Draw AL  EF and extend it to meet AB in M. We show that
AL = LM. First we show that A, F, I, E are con-cyclic. We have
A
BIC = 90° + = 90° + 30° = 120°.
2
Hence FIE = BIC = 120°.
Since A = 60°,
it follows that A, F, , E are concyclic.
Hence BEF = IEF = IAF = A/2.
This gives
B A
AFE = ABE + BEF =  .
2 2
Since ALF = 90°, we see that
B A C
FAM = 90° – AFE = 90° – – = = FCM.
2 2 2
This implies that F, M, C, A are concyclic. It follows that
C
FMA = FCA = = FAM.
2
Hence FMA is an isosceles triangle. But FL  AM. Hence L is the
mid-point of AM or AL = LM.
 10

6. Let us examine the first few natural number : 1, 2, 3, 4, 5, 6, 7, 8, 9.

Here we see that an = 1, 2, 3, 2, 2, 3, 3, 4, 3. We observe that
an  an+1 for all n except when n + 1 is a square in which case
an > an+1. We prove that this observation is valid in general. Consider
the range
m2, m2 + 1, m2 + 2,....., m 2 + m, m2 + m + 1,....., m 2 + 2m.
Let n take v alues in this range so that n = m 2 + r,

where 0  r 2m. Then we see that [ n ] = m and hence

 n   m2  r  r 
  =   =m+  
 [ n ]   m  m 
m, m, m,...., m, m  1, m  1,..., m  1
Thus a takes the v alues 
n m times
  ,
m times
2
m + 2, in this range, But when n = (m + 1) , we see that
an = m + 1. This shown that an–1 > an whenever n = (m + 1)2. When
we take n in the set {1, 2, 3,..., 2010}, we see that the only squares
are 1 2, 2 2 , . .... .... .... ., 44 2 (since 44 2 = 1936 and
452 = 2025) and n = (m + 1)2 is possible for 43 values of m. Thus
an > an+1 for 43 values of n.
(These are 22 – 1, 32 – 1,......, 442 – 1).
 11

REGIONAL MATHEMATICAL OLYMPIAD (RMO) -2011

Time : 3 hours
Instruction :

• Calculators (in any form) and protractors are not allowed.

• Rulers and compasses are allowed.
• Answer all the questions. Maximum marks : 100
• Answer to each question should start on a new page.
Clearly indicate the questions number.

1. Let ABC be a triangle. Let D, E, F be points respectively on the

segments BC, CA, AB such that AD, BE, CE concur at the point K.
Suppose BD / DC = BF / FA and  ADB =  AFC. Prove that
 ABE =  CAD. [12]
2. Let (a1, a2, a3 . . . , a2011) be a permutation (that is a rearrangement)
of the numbers 1, 2, 3 . . . , 2011. Show that there exist two
numbers j, k. such that 1  j < k  2011 and |aj – j| = |ak –k|.
[19]
3. A natural number n is chosen strictly between two consecutive
perfect square. The smaller of these two squares is obtained by
subtracting k from n and the larger one is obtained by adding  to n.
Prove that n – k is a perfect square. [12]
4. Consider a 20-sided conv ex polygon K, with v ertices
A1, A2, . . . . , A20 in that order. Find the number of ways in which
three sides of K can be chosen so that every pair among them has
at least two sides of K between them. (For example (A1A2, A4A5,
A11A12) is an admissible triple while (A1A2, A4A5, A19A20) is not).
[19]
5. Let ABC be a triangle and let BB1, CC1 be respectively the bisectors
of B, C with B1 on AC and C1 on AB. Let E, F be the feet of
perpendiculars drawn from A onto BB1, CC1 respectively. Suppose
D is the point at which the incircle of ABC touches AB. Prove that
AD = EF. [19]
2y 2
6. Find all pairs (x, y) of real numbers such that 16 x  16 x  y  1.
[19]
 12
REGIONAL MATHEMATICAL OLYMPIAD (RMO) - 2011

HINTS & SOLUTIONS

1.  =  ADB =  AFC  BFC =  – 
 BDKF is cyclic quadrilateral.
FK in chord of circle through B, D, K, F
  FBK = FDK ...(1)
BD BF
=   CBA similar to DBF
DC FA

 FD parallel to AC
  FDK =  DAC
  ABE =  CAD Hence proved.

2. Total numbers  2011

Total number of a’s  2011
difference (ai – i) may be from 0 to 2010.
Case-1 If difference is not zero
means ai – i  0
ai  i
total differences = 2010
Numbers = 2011
Hence two numbers will have same difference
 |aj – j| = | ak – k |
Case-2 If difference is zero
 13
2 2
3. Let Squares are p and (p + 1)
given p2 = n – k
(p + 1)2 = n + 
Now n – k
= n – (n – p2) ((p + 1)2 – n)
= n – n(p + 1)2 + n2 + p2(p + 1)2 – np2
= n – np2 – 2np – n + n2 – np2 + p2(p + 1)2
= n2 – 2np2 – 2np + p2(p + 1)2
= n2 – 2np(p + 1) + p2(p + 1)2
= [n – (p + 1)p]2
Which is a perfect square.

20
4. Any side can be selected in C1 ways
Let x, y, z are gapes between two sides and
x  2 , y  2, z  2
also x + y + z = 17
Let x = t1 + 2 , y = t2 + 2, z = t3 + 2

so t1 + t2 + t3 = 11 where t1, t2, t3  W

11 3 1 13
so total ways C3 1 = C2
20
C1  13 C 2
Now total required ways = = 520
3

5. Let radius of incircle = r

A
 AI = r cosec
2

B C  A  A
BIC=  –  2  2  =  = C1IB1  FAE = –
  2 2 2 2
 14
If AI is diameter of circle then this circle passes through F & E and
center of this circle is O
FOE =  – A
Now In the  FOE
FE2 = OF2 + OE2 – 2OF . OF cos( – A)

A 2 A 2 A 2 
 OF  OE 
A 

=   cos A
4 4 2  2 

A 2
= (1 + cos A)
2
A A A
= AI2 cos2 = r2 cosec2 cos2
2 2 2

A
= r2 cot2
2

A
FE = r cot ..(1)
2
ID = r
A A
In  ADIDAI = AD = r cot  AD = FE
2 2

6. Let two numbers are 16 x2  y , 16 x  y2 use A.M.  G.M.

2 2
16 x y
 16 x  y 2 2
 (16 x y
16 x  y )1/ 2
2

1 2 2
 (16 x  x  y  y )1 / 2
2
1
 16 x 2  x y2  y   4 2( x 2  x  y  y 2 )  4–1
4
2(x2 + x + y2 + y)  –1
(x + 1/2)2 + (y + 1/2)2  0
which is possible only if
x = –1/2, y = –1/2
 15

REGIONAL MATHEMATICAL OLYMPAID (RMO) - 2012

Time : 3 hours
Instruction :
• Calculators (in any form) and protractors are not allowed.
• Rulers and compasses are allowed.
• Answer all the questions. Maximum marks : 100
• Answer to each question should start on a new page.
Clearly indicate the questions number.

1. Let ABCD be a unit square. Draw a quadrant of circle with A as

centre and B, D as end points of the arc. Similarly, draw a quadrant
of a circle with B as centre A, C as end points of the arc. Inscibe
a circle  touching the arcs AC and BD both externally and also
touching the side CD. Find the radius of the circle  .

2. Let a, b, c be positive integers such that a divides b5, b divides c5

and c divides a5. Prove that abc divides (a + b + c)31.

3. Let a and b be positive real numbers such that a + b = 1. Prove

that
aabb + abba  1.

4. Let X = {1, 2, 3,.....,10}. Find the the number of pairs {A, B} such
A  X. B  X. A  B and
A  B = {5, 7, 8}.

5. Let ABC be a triangle. Let D, E be points on the segment BC

such that BD = DE = EC. Let F be the mid-point of AC. Let BF
intersect AD in P and AE in Q respectively. Determine the ratio of
the area of the triangle APQ to that of the quadrilateral PDEQ.

6. Find all positive integers n such that 32n + 3n2 + 7 is a perfect

sqaure.
 16

REGIONAL MATHEMATICAL OLYMPIAD (RMO) - 2012

HINTS & SOLUTIONS

1
1. Ans.
16
1
By pythagores (r + R)2 = (R – r)2 +
4

1
(r + R)2 = (1 – r)2 + Here R = 1
4
1 1
(r + 1)2 = (1 – r)2 + ; r= .
4 16

2. Method 1 :
Consider the expansion of (a + b + c)31. All terms in it are of the
form of rkm ak b cm, where rkm is a constant (in the form of
binomial coefficient) and k, , m are non-negative integers such
that
k +  + m = 31.
If k  1,   1, m  1 then abc always divides ak b cm. hence we
have to consider terms in which one or two of k, , m are zero.
Now let k =  = 0, then a31 = a. a5. a25
= a. (ck1) (ck1)5
= a (ck1) (k2b k15)
= abc k16 k2
Clearly, abc divides a and similarly we can say about b31 and
31

c31.
Now let us consider two out of k, , m are not zero. Let k = 0,
, m  0 then term would be
rkm bcm, where  + m = 31.
 17
b c = bc b c ,  ,m  1
 m –1 m–1

Now atleast one of –1 and m–1 must be greater than 25 and
5 respectively, which clearly shows the divisiblity of abc.
Method 2 :
a | b5  b5 = k1 a ; b | c5  c5 = k2b ; c | a5  a5 = k3c
31 31
 b5 a5   b5 b25 
(a + b + c)  
31 b  =  b 5 
 k1 k3   k1 k1 k 3 
31 31
 b4 b24   b4 b24 
=b 
31  1   = b. b5 . b25   1  
 k1 k15k3   k1 k15 .k3 


125 4 126 125 80
= abc k1 b  k1  k1 k 3 .b
24

Which is clearly divisible by abc.

3. aabb + abba  1
Given a + b = 1
To prove that
aabb + abba  aa+b + ba+b
aa(bb – ab) + ba(ab – bb)  0
(aa – ba)(bb – ab)  0 .............(1)
If a = b then above is obviously true.
If a  b then assume a > b
a > b  aa > ba
ab > bb  bb – ab < 0 this prove (1)

4. Ans. 2186
X = {1, 2, 3, ......., 0}
So X – (A  B) has 7 elements.
A will has 5, 7, 8. Rest elements can be assigned in 2 ways
'1' can either go to A of B or none.
So total pairs = 37 – (1).

(When no elements has
been assigned to A or B.)
 18

9
5. Ans.
11


ADE =
3

x+y= ...(1) where x = area of  APQ & y = area
3
of PQED
Consider ADC Consider AEC
AP BD CF AQ BE CF
  1   1
PD CB AF QE BC AF
AP 1 1 AQ 2
  1  1  1
PD 3 1 QE 3

AP AQ 3
3 
PD QE 2
3
So AP = AD
4
3
AQ = AE
5
3 3
APQ = AD  AF sin A
4 5
9 9  3
APQ =   x= × ; x=
20 ADE 20 3 20
 19

9     9 
Thus (1)  +y=  y= 1 – 
20 3 3 3  20 

11
y=
3  20

3
x 20 9
= =
y 11 11
3  20

6. 32n + 3n2 + 7
(a) If n is odd n = 2k + 1
n2 is of form of 4 + 1 here 
34k+2 + 3(4 + 1) + 7  9.81k + 12 + 10
k
(81)  1 mod(4)
9(81)k  1 mod(4)
32n + 3n2 + 7 3 mod(4) and a perfect square can’t be
of form of 4 + 3

(b) If n is even n = 2k
92k + 12k2 + 7
Now 92k < 92k + 12k2 + 7  (9k + 1)2
where equality will hold only at k = 1
Rest it will be in between perfect square of 9k and 9k +1
i.e. two consecutive no. Hence n = 2 is only solution.