Regd. Office Answers & Solutions...1 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

1

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

Ph.: 011-47623456 Fax : 011-47623472

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

� � �