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PRMO - 2017
1. How many positive integers less than 1000 have the
property that the sum of the digits of each such
number is divisible by 7 and the number itself is
divisible by 3?
Answer (28)
Sol. Sum of the digits is divisible by 7 and number itself
is divisible by 3
∴ x + y + z = 21, x, y, z are digits 0, 1, 2, ...., 9.
∴ Coefficient x21 is (x0 + x1 + ..... + x9)3
= Coefficient of x21 is
10 3
3
(1 )
(1 )
x
x
−−
= Coefficient of x21 is (1 – 3x
10 + 3x20 – x
30)
(1 + 3C1x + 4C
2x
2 + ....)
= 23C21
– 3.13C11
+ 3.3C1
= 28
2. Suppose a, b are positive real numbers such that
183.a a b b+ = 182.a b b a+ = Find ( )9.
5a b+
Answer (73)
Sol. ∵ 183a a b b+ =
⇒ ( ) ( )3 3
183a b+ = ...(1)
Also 182a b b a+ =
⇒ ( ) ( )2 2
182a b a b+ = ...(2)
Multiply equation (2) by 3 and add to equation (1)
( ) ( ) ( ) ( )( )3 3
3 729a b a b a b+ + + =
∴ ( )3 729a b+ =
∴ 9a b+ =∴ Form (2)
( ) 182ab a b+ =
⇒ 182
9ab =
∴ ( )2 2a b a b ab+ = + −364 365
819 9
= − =
∴ 9( ) 73
5a b+ =
3. A contractor has two teams of workers : team A and
team B. Team A can complete a job in 12 days and
team B can do the same job in 36 days. Team A
starts working on the job and team B joins team A
after four days. The team A withdraws after two more
days. For how many more days should team B work
to complete the job?
Answer (16)
Sol. Team A worked for 6 days ⇒ work completed 1
2=
Team B worked for 2 days ⇒ work complete
2 1
36 18= =
∴ Remaining work = 1 1 4
12 18 9
⎛ ⎞− + =⎜ ⎟⎝ ⎠
∴ Number of days required = 4
36 16 days9
× =
∴ B needs 16 more days
Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005
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Answers & Solutions
forforforforfor
PRMO - 2017
2
PRMO - 2017
4. Let a, b be integers such that all the roots of the
equation (x2 + ax + 20)(a2 + 17x + b) = 0 are
negative integers. What is the smallest possible value
of a + b?
Answer (25)
Sol. ∵ Roots of (x2 + ax + 20)(a2 + 17x + b) = 0 are
negative integers and we required (a + b)min
∴ a, b each should be minimum
a2 – 80 is perfect square and a is a positive
integer
⇒ amin
= 9
289 – 4b is a perfect square and b is a positive
integer
⇒ bmin
= 16 (using 8, 15, 17)
∴ (a + b)min
= 25
5. Let u, v, w be real numbers in geometric progression
such that u > v > w. Suppose u40 = vn = w60. Find
the value of n.
Answer (48)
Sol. Let u40 = vn = w60 = K
∴1 11
40 60, ,
nu K v K w K= = =
∵ v2 = uw
∴1 12
40 60nK K K= ⋅
∴2 1
24nK K=
∴ 2 148
24n
n
= ⇒ =
6. Let the sum ( )( )
9
1
1
1 2nn n n
=
+ +∑ written in its
lowest terms be p
q. Find the value of q – p.
Answer (83)
Sol. ( )( )
9
1
1
1 2n
p
n n n q=
=
+ +∑
⇒ ( )( )
9
1
1 2
2 1 2n
n n p
n n n q=
+ −=
+ +∑
⇒ ( ) ( )
9
1
1 1 2–
1 ( 1) 2n
p
n n n n q=
=
+ + +∑
⇒9
1
1 1 1 1 2– – –
1 1 2n
p
n n n n q=
⎧ ⎫⎛ ⎞ ⎛ ⎞ =⎨ ⎬⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠+ + +⎩ ⎭∑
⇒1 1 1 1 1
1 – ... –2 2 3 9 10
⎛ ⎞− + + +⎜ ⎟⎝ ⎠ –
1 1 1 1 1 1 2– – ... –
2 3 3 4 10 11
p
q
⎛ ⎞+ + + =⎜ ⎟⎝ ⎠
∴ q – p = 110 – 27 = 83
7. Find the number of positive integers n, such that
1 11n n+ + < .
Answer (29)
Sol. 1 11n n+ + <
1 11–n n+ <
n + 1 < 121 + n – 22 n [∵both sides are positive]
⇒ 11 n < 60 ⇒ n <3600
29.7121
�
∴ n = 29
8. A pen costs `11 and a notebook costs `13. Find the
number of ways in which a person can spend
exactly `1000 to buy pens and notebooks.
Answer (07)
Sol. Let pens be x and notebooks be y.
∴ 11x + 13y = 1000
11x = 1000 – 13y
when y = 5 then x = 85
Now every common number will occur at the interval
of 11 × 13 = 143 is RHS starting with y = 5, RHS
is
935, 792, ..., 66
∴ We obtain total 7 such ordered pairs.
9. There are five cities A, B, C, D, E on a certain
island. Each city is connected to every other city by
road. In how many ways can a person starting from
city A come back to A after visiting some cities
without visiting a city more than once and without
taking the same road more than once? (The order in
which he visits the cities also matters : e.g., the
routes A → B → C → A and A → C → B → A are
different)
Answer (60)
Sol. He can visit all 4 cities or 3 cities or 2 cities from
B, C, D, E.
If he visits all 4 cities = 4! = 24
If he visits 3 cities = 4C3.3! = 24
If he visits 2 cities = 4C2.2! = 12
∴ Total ways = 60
3
PRMO - 2017
10. There are eight rooms on the first floor of a hotel,
with four rooms on each side of the corridor,
symmetrically situated (that is each room is exactly
opposite to one other room). Four guests have to be
accommodated in four of the eight rooms (that is,
one in each) such that no two guests are in
adjacent rooms or in opposite rooms. In how many
ways can the guests be accommodated?
Answer (48)
Sol. Clearly guests will stay either in ‘ ’ or in ‘X’
× ×
××
∴ Required way = 2 × 4! = 48
11. Let f(x) = 3
sin cos3 10
x x+ for all real x. Find the least
natural number n such that f(nπ + x) = f(x) for all real x.
Answer (60)
Sol. f(x) = 3
sin cos3 10
x x+ . We need to find period of f(x).
Period of sin 63
x = π
Period of 3 20
cos10 3
x π=
(∵ sinaθ, cosaθ are periodic with period 2
| |a
π)
LCM of 20
6 ,3
π⎛ ⎞π⎜ ⎟⎝ ⎠ = 60π
∴ Period of f(x) = 60π
∴ n = 60
12. In a class, the total numbers of boys and girls are in
the ratio 4 : 3. On one day it was found that 8 boys
and 14 girls were absent from the class and that the
number of boys was the square of the number of girls.
What is the total number of students in the class?
Answer (42)
Sol. Let boys be 4x and girls be 3x.
∵ 4x – 8 = (3x – 14)2
∴ 9x2 – 88x + 204 = 0
9x2 – 54x – 34x + 204 = 0
(9x – 34) (x – 6) = 0
x = 6, 34
9
∴ Total number of students = 7x = 42
13. In a rectangle ABCD, E is the midpoint of AB : F is a
point on AC such that BF is perpendicular to AC : and
FE perpendicular to BD. Suppose BC = 8 3 . Find
AB.
Answer (24)
Sol. Let ∠CAB = ∠ADB = α
∴ ∠NMB = 2α (From figure)
M
F
α 90 + α 90 – α α
N
90 – 2α 2α
D C
A BE
Let AE = EB = x
BN = xcosα (from ΔENB) ...(1)
= BF sin2α (from ΔBFN) ...(2)
∵ BF = 2xsinα (from ΔAFB)
∴ xcosα = (2xsinα) (2sinα cosα) (from (1) & (2))
⇒ sinα = 1
2
⇒ α = 30°
∵ BC = 8 3
∴ In ΔABC, AB
BC = cot30°
∴ AB = 8 3 3× = 24
14. Suppose x is a positive real number such that {x}, [x]
and x are in a geometric progression. Find the least
positive integer n such that xn > 100 . (Here [x] denotes
the integer part of x and {x} = x – [x])
Answer (10)
Sol.∵ {x}, [x], x are in GP let these be a, ar, ar2
Also as {x} = x – [x]
∴ a = ar2 – ar
∴ r2 – r – 1 = 0 ⇒ r = 1 5
2
±
∴ r = 5 1
2
+
∵ ar = [x] = integer = k (say)
∴ a = ( )5 –12
25 1
kk =+
4
PRMO - 2017
∵ a = {x} which lies in [0,1)
∴ ( )5 –1
0 12
k
< < (as a ≠ 0)
⇒ 0 < k < 5 1
2
+
⇒ k = 1
∴ r = 5 1
2
+
∴ xn = (ar
2)n = 5 1
1002
n
⎛ ⎞+ >⎜ ⎟⎝ ⎠
∵
5 1 2.23 11.6
2 2
+ +� �
∴ We require (1.6)n > 100
10
16log
10n
⎛ ⎞⎜ ⎟⎝ ⎠ > 2
⇒ n(4log10
2 – 1) > 2
nx(.204) > 2
nmin
= 10
15. Integers 1, 2, 3........n, where n > 2. are written on a
board. Two numbers m.k such that l < m < n, l < k < n
are removed and the average of the remaining numbers
is found to be 17. What is the maximum sum of the
two removed numbers?
Answer (51)
Sol. Let two numbers moved be 1, 2.
Then average of remaining (n – 2) numbers
=
n n
n nx
n n
2
1
( 1)– 3
– 62
– 2 2( – 2)
++= =
n n
n
( 3)( – 2)
2( – 2)
+=
n 3
2
+=
If two numbers be n, n – 1 then
n nn
x
n2
( 1)– (2 –1)
2
– 2
+
=
n n
n
2 – 3 2
2( – 2)
+=
n –1
2=
Now, the removed numbers are m, k, then new
average is 17 i.e.,
n nm k
n
( 1)– ( )
217– 2
+ +=
Clearly, n n–1 3
172 2
+< <
⇒ n < 35 & n > 31
∴ n = 32, 33, 34
∵
n nm k
n
( 1)– ( )
217– 2
+ +=
∴n n n
m k( 1) – 34( – 2)
( )2
++ =
When n = 32 ⇒ m + k = 18
n = 33 ⇒ m + k = 34
n = 34 ⇒ m + k = 51
∴ (m + k)max
= 51
16. Five distinct 2-digit numbers are in a geometric
progression. Find the middle term.
Answer (36)
Sol. Since all five numbers are 2 digit
∴ Number should have 24 or 34 included no other
possibility and common ratio should be 3
2 or
2
3.
∴ Numbers are 16, 24, 36, 54, 81.
17. Suppose the altitudes of a trianlge are 10, 12 and
15. What is its semi-perimeter?
Answer (Non-integer)
Sol. Altitudes are 10, 12, 15
a ha1
2Δ = ×
∴ aha
2Δ=
∴ Sides
a b c
a b ch h h
1 1 1: : : :=
1 1 1: :
10 12 15=
= 6 : 5 : 4 ⇒ 6λ, 5λ, 4λ
∴ s15
2
λ=
∴ s s a s b s c( – )( – )( – )Δ =
5
PRMO - 2017
215 3 5 7 15
72 2 2 2 4
λ λ λ λ λ= ⋅ ⋅ ⋅ =
Also a
a h1 1
6 10 302 2
Δ = × = λ × = λ
∴2
1530 7
4
λλ =
⇒8
7λ =
∴ s60
7= (non-integer)
18. If the real numbers x, y, z are such that x2 + 4y2 +
16z2 = 48 and xy + 4yz + 2zx = 24, what is the
value of x2 + y2 + z2?
Answer (21)
Sol. x2 + 4y
2 + 16z2 = 48 ...(i)
xy + 4yz + 2zx = 24 ...(ii)
Multiply (ii) by 2 and subtract from (i)
∴ x2 + 4y
2 + 16z2 – 2xy – 8yz – 2zx = 0
(x – 2y)2 + (2y – 4z)2 + (4z – x)2 = 0
(∵ a2 + b2 + c2 – ab – bc – ca
= (a – b)2 + (b – c)2 + (c – a)2)
⇒ x = 2y = 4z ⇒ x y z
k4 2 1= = =
Put x, y, z in (i)
16k2 + 16k
2 + 14k2 = 48 ⇒ k = 1
∴ x2 + y2 + z2 = 16 + 4 + 1 = 21
19. Suppose 1, 2, 3 are the roots of the equations x4 +
ax2 + bx = c. Find the value of c.
Answer (36)
Sol. ∵ Sum of the roots = 0
∴ 1 + 2 + 3 + α = 0 ⇒ α = –6
∴ Product of roots = –c = 1 × 2 × 3 × –6
⇒ c = 36
20. What is the number of triples (a, b, c) of positive
integers such that (i) a < b < c < 10 and (ii) a, b,
c, 10 form the sides of a quadrilateral?
Answer (73)
Sol. a + b + c > 10 , a < b < c < 10
⇒ c > 10 – (a + b)
a + b No. of c (>10 – Values Total
values a + b) of c possibilities
for (a, b)
3 1 c > 7 8, 9 2
4 1 c > 6 7, 8, 9 3
5 2 c > 5 6, 7, 8, 9 2 × 4 = 8
6 2 c > 4 5, 6, 7, 8, 9 4 + 5 = 9
7 3 c > 3 5, 6, 7, 8, 9 3 + 4 +5 = 12
8 3 c > 2 7, 8, 9 2 + 3 + 4 = 9
9 4 c > 1 6, 7, 8, 9 1 + 2 + 3 + 4 = 10
10 3 c > 0 7, 8, 9 1 + 2 + 3 = 6
11 3 Any value 7, 8, 9 1 + 2 + 3 = 6
12 2 Any value 8, 9 1 + 2 = 3
13 2 Any value 8, 9 1 + 2 = 3
14 1 Any value 9 1
15 1 Any value 9 1
∴ Total ways = 73
21. Find the number of ordered triples (a, b, c) of positive
integers such taht abc = 108.
Answer (60)
Sol. abc = 108
= 22 . 33
Two 2’s in 3 boxes ⇒ x1 + y
1 + z
1 = 2
⇒ 3 2 1
3 16c
+ −− =
Three 3s in 3 boxes = x2 + y
2 + z
2 = 3
⇒ 3 3 1
3 110c
+ −− =
∴ Required ways = 6 × 10 = 60
22. Suppose in the plane 10 pairwise non-parallel lines
intersect one another. What is the maximum
possible number of polygons (with finite areas) that
can be formed?
Answer (36)
Sol. If we consider the case.
In given figure there are 3 polygon in all. But if we
count non-overlapping polygons then it is 2.
If we consider non-overlapping, then the solution is
as follows :
Total number of regions in which 10 lines divide to
plane = ( 1)
12
n n + +
= 10 11
1 562
× + =
of which exactly 20 are open (unbounded)
∴ Polygons = 56 – 20 = 36
6
PRMO - 2017
23. Suppose an integer x a natural number n and a
prime number p satisfy the equation 7x2 – 44x + 12
= pn. Find the largest value of p.
Answer (47)
Sol. 7x2 – 44x + 12 = pn
⇒ 7x2 – 42x – 2x + 12 = pn
⇒ (7x – 2)(x – 6) = pn
Let 7x – 2 be the prime number
⇒ x – 6 must be the same prime number or 1
Now, x – 6 = 1
⇒ x = 7
Then, 7x – 2 = 7 × 7 – 2
= 47, which is a prime number
Further, 7x – 2 and x – 6 must be the same prime
numbers
∴ 7x – 2 = x – 6
⇒ 6x = – 4
⇒ x = –2
,3
Hence, x ∉ Z
∴ pn = 47
p = 47 and n = 1
24. Let P be an interior point of a triangle ABC whose
sidelengths are 26, 65, 78. The line through P
parallel to BC meets AB in K and AC in L. The line
through P parallel to CA meets BC in M and BA in
N. The line through P parallel to AB meets CA in S
and CB in T. If KL, MN, ST are of equal lengths, find
this common length.
Sol. No such configuration possible.
25. Let ABCD be a rectangle and let E and F be points
on CD and BC respectively such that area(ADE) =
16, area(CEF) = 9 and area(ABF) = 25. What is the
area of triangle AEF?
Answer (30)
Sol.
A B
CD E
F
a
b
x
y16
9
25
x – b
y – a
ar(ΔAFB) = 25 = x y a1
( – )2
⇒ xy – ax = 50 ...(i)
ar(ΔCEF) = 9 = a b1
2× ×
⇒ ab = 18 ...(ii)
ar(ΔADE) = 16 = x b y1( – )
2
⇒ xy – by = 32 ...(iii)
From (i) ⇒ xy
a
x
– 50 = ...(iv)
From (iii) ⇒ xy
by
– 32 = ...(v)
Using the values of a and b in (ii), we get
xy xy
x y
– 50 – 3218
⎛ ⎞⎛ ⎞× =⎜ ⎟⎜ ⎟⎝ ⎠ ⎝ ⎠
(xy – 50) (xy – 32) = 18xy
(xy)2 – 50xy – 32xy + 1600 = 18xy
Let xy = P
⇒ P2 – 100P + 1600 = 0
⇒ P2 – 80P – 20P + 1600 = 0 [By splitting
middle term]
⇒ P(P – 80) – 20(P – 80) = 0
⇒ P = 20 or 80
Now, xy cannot be 20
⇒ xy = 80
Area of ΔAEF = 80 – (16 + 9 + 25)
= 30 sq. units
26. Let AB and CD be two parallel chords in a circle
with radius 5 such that the centre O lies between
these chords. Suppose AB = 6, CD = 8. Suppose
further that the area of the part of the circle lying
between the chords AB and CD is (mπ + n)/k,
where m, n, k are positive integers with
GCD(m, n, k) = 1. What is the value of m + n + k?
Answer (75)
Sol.
A B
C D
α
β
M
N
5
8
6
α
β5
5
5
O
Area of the part of the circle lying between AB and
CD is = 2[(ar ΔAOM) + area of sector AOC + (ar
ΔCON)]
7
PRMO - 2017
1 180 – ( ) 12 3 4 5 5 4 3
2 360 2
° α + β⎡ ⎤= × × + × π× × + × ×⎢ ⎥°⎣ ⎦
180 – ( )2 12 25
360
° α + β⎡ ⎤= + × π⎢ ⎥°⎣ ⎦
Now, 3
tan4
α =
and 4
tan3
β =
( ) tan tantan
1– tan tan
α + βα + β =α ⋅ β
3 4 25
4 3 12
3 4 01–
4 3
+= =
×
tan(α + β) = Not defined
⇒ α + β = 90°
⇒ Required area = 180 – 90
2 12 25360
° °⎡ ⎤+ × π⎢ ⎥°⎣ ⎦
48 25
2
+ π=
= (m + nπ)/k
⇒ m = 48, n = 25, k = 2
⇒ m + n + k = 75
27. Let Ω1 be a circle with centre O and let AB be a
diameter of Ω1. Let P be a point on the segment OB
different from O. Suppose another circle Ω2 with
centre P lies in the interior of Ω1. Tangents are
drawn from A and B to the circle Ω2 intersecting Ω
1
again at A1 and B
1 respectively such that A
1 and B
1
are on the opposite sides of AB. Given that A1B = 5,
AB1 = 15 and OP = 10, find the radius of Ω
1.
Answer (20)
Sol.
A B
A1
B1
R
Q
Ox 10 P
y x – 10
90–θ y
15
5
θ
In ΔAPQ
y
xsin
10θ =
+ ...(i)
In ΔAA'B
x
5sin
2θ = ...(ii)
From (i) & (ii)
y
x x
5
10 2=
+ ...(iii)
Similarly
y
x x
15
–10 2= ...(iv)
From (iii) & (iv)
5x + 50 = 15x – 150
⇒ 5x – 15x = –150 – 50
⇒ –10x = –200
⇒ x = 20
∴ Radius of Ω1 = 20
28. Let p, q be prime numbers such that n3pq – n is a
multiple of 3pq for all positive integers n. Find the
least possible value of p + q.
Answer (28)
Sol. ∵ n3pq – n = 0 (mod 3)
n3pq
– n = 0 (mod p)
n3pq
– n = 0 (mod q)
∴ We need to fulfill following conditions :
(i) (3 – 1) | (pq – 1) ⇒ pq is odd
(ii) (p – 1) | (3q – 1)
Now 3 must not divide (p – 1) as it doesn’t divide
(3q – 1)
∴ p – 1 = 3k + 1 or 3k + 2, for some integer k.
∴ ⇒ p = 3k + 2 or 3k + 3
But p ≠ 3k + 3 (as its prime)
∴ p = 3k + 2 clearly p > 3 and k = odd = 2λ +
1 (say)
⇒ p = 6λ + 5
(ii) (q – 1) | (3p – 1) so q will also be 5 (mod 6)
by trial least values for p & q are 17 and 11.
∴ p + q = 28
29. For each positive integer n, consider the highest
common factor hn of the two numbers n! + 1 and
(n + 1)!. For n < 100, find the largest value of hn.
Answer (97)
Sol. HCF of ((n! + 1), (n + 1)!) = h , n < 100
If n + 1 is prime then n! + 1 is divisible by (n + 1).
Also (n + 1)! is divisible by it.
∴ For n + 1 = 97
HCF of 96! + 1 & 97! = 97
8
PRMO - 2017
30. Consider the areas of the four triangles obtained by
drawing the diagonals AC and BD of a trapezium
ABCD. The product of these areas, taken two at
time, are computed. If among the six products so
obtained, two products are 1296 and 576, determine
the square root of the maximum possible area of the
trapezium of the nearest integer.
Answer (13)
Sol. Case-I
A B
CD
a
ak
ak2
ak
a2k = 576
a2k
2 = 1296
1296 9
576 4k = =
⇒ a = 16
Case-II
a2k
3 = 1296
a2k = 576
2 1296 9
576 4k = =
3
2k =
⇒ a2 = 384
⇒ 8 6a =
Case-III
a2k
3 = 1296
a2k
2 = 576
1296 9
576 4k = =
⇒ 2 1024
9a =
⇒32
3a =
Area of trapezium = a + ak + ak + ak2
= a(k + 1)2
Case I, when a = 16, 9
4k =
29
Area 16 1 169 134
⎛ ⎞= + = =⎜ ⎟⎝ ⎠
Case 2, when a =8 6 , 3
2k =
23
Area 8 2.4 1 120 10.95 approx.2
⎛ ⎞= × + = =⎜ ⎟⎝ ⎠
Case 3, when 32
3a = ,
9
4k =
232 9 2
Area 1 133 4 3
⎛ ⎞= + =⎜ ⎟⎝ ⎠
∴ Square root of maximum possible area is 13
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