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RMO
Regional Mathematical Olympiad 2015 (First Round)
Questions with Solutions
1.
Given 100 numbers a1, a2, ………. a100 such that
15
a1 – 3a2 + 2a3  0
a2 – 3a3 + 2a4  0
a3 – 3a4 + 2a5  0
__________________
a99 – 3a100 + 2a1  0
a100 – 3a1 + 2a2  0
Prove that a1 = a2 = ……………… = a100.
Sol. a1 – a2 > 2(a2 – a3)
.
.
.
a100 – a1 > 2(a1 – a2) on multiplying ineqns, we get 1 > 2100  absurd. so, every where equality has to
hold.
 a1 – a2 = 2(a2 – a3) & so on

a1  a 2
a  a3
a
 a1
2 2
 ..........  100
a2  a3
a3  a4
a1  a 2
If each of ai – aj is non-zero, on multiplication we get 1 = 2100  absurd  so at least one of them
has to be zero  those two are equal, which in turn maves every one equal.  Q.E.D
2.
Prove that x12 – x9 + x4 – x + 1 > 0 for all real values of x
10
Sol. f(x)  x12 – x9 + x4 – x + 1, in f(– x) no sign change is there  No negative root.
If x > 1, x12 > x9, x4 – x > 0, f(x) > 0
If 0 < x < 1, 1 – x > 0, x4 – x9 > 0, x12 > 0  f(x) > 0  x.
(None  Decarte’s rule)
(None  f(x) < 0 clearly  x < 0 as well.)
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3.
RMO
96 

A student of 11th class found 11 distinct solutions of the equation f 19x 
  0 . His teacher told
x 

him that being a student of 11th class he could have found one more solution. Justify with proof if
his teacher was right.
15
Sol. Let for some , f() = 0   = 19x 
96
x
 19x2 – x – 96 = 0  D > 0 for  .
 Even no. of roots.  one more root.
4.
In a class of 20 student no two of them have the same ordered pair (written and oral examination)
of scores in mathematics. We say that student A is better than B if his two scores are greater than
or equal to the corresponding scores of B. The scores are integers between 1 and 10. Show that
there exist 3 students A, B, and C such that A is better than B and B is better than C.
Sol.
Let ordered no is (ai , bi)
20
ai {1, 2, ……. , 10}
bi {1, 2, …… , 10}
Clearly for any particular ai, no more than two value of bi are possible or we are done. Similarly for
any particular bj, no more than two value of ai s are possible, or we are done. So, exactly two
students will have one particular. ai & exactly two students will have one particular bj. So clearly
most differentiated case is
(10, 1), (10, 2), (9, 2), (9, 3), (8, 3), (8, 4), (7, 4) (7, 5), (6, 5), (6, 6) (5, 6), (5, 7), (4, 7), (4, 8), (3, 8), (3,
9) (2, 9), (2, 10)
Still two students with exactly 1 mark in written is left to be placed.
 some 3 students will be in same group.  For those three A, B, C ,
A  B  C  Q.E.D
(Note:-) one more seq. with (1, 10), (1, 9), (2, 9), (2, 8) ….. etc. can be made)
5.
Saurabh is walking up a stair that has 10 steps. With each stride (move) he goes up either one step
or two steps. In how many different ways can he go up the stairs?
Sol.
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2x + y = 10, x, y  0 for x = 0, one way
When x = 1  9C8
case
x = 2  8C6
case
x = 3  7C4
case
x = 4 6C2
case
x = 5  5C0
case
total = 88 case
Ans: 88 ways + 1 way = 89 ways
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CHANDIGARH BRANCH: Campus-1: SCO No.-350-351-352, Ground Floor, Sector 34 A, Chandigarh – 160022 
Campus-2: Plot No. 11–12, 1st Floor, Sector 25-D, Chandigarh. Tel. 0172-4612029, 4622029
PANCHKULA BRANCH: SCO 264, 2nd Floor, Sector 14, Panchkula, Tel. 0172-4004028, 4005028
PATIALA BRANCH: SCF 99–102, Chotti Barandari, Patiala. Tel. 0175-5012029, 5012030
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6.
RMO
Determine all primes p for which the system of equations
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p + 1 = 2x2
p2 + 1 = 2y2
Has a solution in integers x, y.
Sol.
Let x, y > 0 I
p + 1 = 2x2, p2 + 1 = 2y2
 p2 – p = 2 (y2 –x2) = p(p – 1)
 p(p –1) = 2 (y – x) (y + x)
p & p – 1 are rel. prime factor.
should be split to rel. prime factor with diff = 1
 case (a): 2y + 2x – (y – x) = 1  3x + 4 = 1
 absurd
Case (b): (y + x) – 2(y – x) = 1  3x – 1 = y
 p2 + 1 = 2y2 = 2(3x – 1)2 = 18x2 – 12x + 2
= 9 (p + 1) –12x + 2
 12x = – (p2 – 10 – 9p) <
121
4
 12x = 12 or 24  x = 1 or 2
x = 1 gives absurd answers  x = 2, y = 5, p = 7
 only p = 7 satisfy the given condition
__________________________________________________________________________________________________________________________________
Visit us at:
CHANDIGARH BRANCH: Campus-1: SCO No.-350-351-352, Ground Floor, Sector 34 A, Chandigarh – 160022 
Campus-2: Plot No. 11–12, 1st Floor, Sector 25-D, Chandigarh. Tel. 0172-4612029, 4622029
PANCHKULA BRANCH: SCO 264, 2nd Floor, Sector 14, Panchkula, Tel. 0172-4004028, 4005028
PATIALA BRANCH: SCF 99–102, Chotti Barandari, Patiala. Tel. 0175-5012029, 5012030
SANGRUR BRANCH: Rajbaha Road, Guru Nanak Colony, Near Dr. D.P. Singla Clinic, Sangrur. Tel. 01672-500136
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7.
RMO
In an acute angled triangle ABC let a, b, c be its sides, m a, mb, mc the lengths of altitudes and from
A, B, C respectively and da, db, dc the distances from the vertices A, B, C respectively to the
orthocentre. Prove that
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ma da  mbdb  mc dc 
Sol.
da = 2R cosA, ma =
  ma da  4R
(
abc = 4R)
a b c
2
2
2
2
2
a
cos A
b2  c2  a 2
 4R
a
2abc
=
a 2  b2  c2
2
Q.E.D
Note
AE = c cos A
AH cos(900 – c) = AE = c cos A
 AH = da =
c cos A
 2R cos A
sin c
Dream on !!

__________________________________________________________________________________________________________________________________
Visit us at:
CHANDIGARH BRANCH: Campus-1: SCO No.-350-351-352, Ground Floor, Sector 34 A, Chandigarh – 160022 
Campus-2: Plot No. 11–12, 1st Floor, Sector 25-D, Chandigarh. Tel. 0172-4612029, 4622029
PANCHKULA BRANCH: SCO 264, 2nd Floor, Sector 14, Panchkula, Tel. 0172-4004028, 4005028
PATIALA BRANCH: SCF 99–102, Chotti Barandari, Patiala. Tel. 0175-5012029, 5012030
SANGRUR BRANCH: Rajbaha Road, Guru Nanak Colony, Near Dr. D.P. Singla Clinic, Sangrur. Tel. 01672-500136