a b b a 182. Possible values of (a, b, c) Permutation 3 2 = = = 6 = 3 = = 6 = 1

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1 FIITJEE Solutions to PRMO How many positive integers less than 1000 have the property that the sum of the digits of each Sol. such number is divisible by 7 and the number itself is divisible by? Let the number be abc a + b + c = 7k (divisible by 7) number is divisible by i.e. a + b + c = m a + b + c is divisible by 1. 0 a 9, 0 b 9, 0 c 9 0 a + b + c 7 a + b + c = 1 Possible values of (a, b, c) 9 9 Permutation = = = = = = Total Permutations = 8 lternatively, = 8 = 1. Suppose a, b are positive real numbers such that a a b b 18. a b b a 18. Find 9 (a + b). Sol. a / + b / = 18 (i) a 1/ b 1/ (a 1/ + b 1/ ) = 18 (ii) (a 1/ + b 1/ ) = a / + b / + a 1/ b 1/ (a 1/ + b 1/ ) = = 79 = 9 a 1/ + b 1/ = 9 a 1/ b 1/ = 18/9 add (i) and (ii) (a+b) a b 6 6 a b 9 9 a b 7 FIITJEE Ltd., FIITJEE House, 9-, Kalu Sarai, Sarvapriya Vihar, New elhi , Ph , 66949, Fax 6194

2 PRMO contractor has two teams of workers: team an team. Team can complete a job in 1 days and team can do the same job in 6 days. Team starts working on the job and team joins team after four days. The team withdraws after two more days. For how many more days should team work to complete the job? Sol. Let total work is W units and team does a unit work per day and team does b unit work per day. 1a = W a = W 1 6 b = W b = W 6 (i) (ii) let n days more are needed to complete remaining work 6a + (n +) b = W W W 6 n W n = Let a, b be integers such that all the roots of the equation (x + ax + 0) (x + 17x + b) = 0 are negative integers. What is the smallest possible value of a + b? Sol. Roots are ve hence a is +ve and b is +ve integer x + ax + 0 = 0 a = sum of factors of 0 0 = 1 0 = 10 = 4 possible values of a are 1, 1 and 9. x + 17x + b = 0 if 17 = p + q then b = pq 17 = = + 1 = + 14 = = + 1 = = = possible vales of p are 16, 0, 4,, 60, 66 70, 7. minimum value of a + b = =. Let u, v, w be real numbers in geometric progression such that u > v > w. Suppose u 40 = v u = w 100. Find the value of n. Sol. v = wu v n = u 40 v n = u 10 v n = w 60 v n = w 10 v n v n = u 10 w 10 v n = (v ) 10 n = 40 n = Let the sum Sol. q p. S 9 9 n1 1 nn 1n 9 1 n1n n 1 n written in its lowest terms be 1 Let Tn nn 1n Tn nn 1n 1n S9 T1 T T...T = p 7 S q 110 qp p. q Find the value of FIITJEE Ltd., FIITJEE House, 9-, Kalu Sarai, Sarvapriya Vihar, New elhi , Ph , 66949, Fax 6194

3 PRMO Find the number of positive integers n, such that n n Sol. n n 1 11 n 1 11 n n 1 11 n n n n n 11 n 9.7 n = 9 8. pen costs Rs. 11 and a notebook costs Rs. 1. Find the number of ways in which a person can spend exactly Rs to buy pens and notebooks. Sol. let x pens and y notebooks 11x + 1y = 1000 y =, 16, 7, 8, 49, 60, 71 Hence maximum 7 ways 9. There are five cities,,,, 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 come back to 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 and are different.) Sol. Let cities be,, and E Number of ways/routes of the type ( ) = 4! 1 ways i.e. selecting cities of out of,,, E and permuting them. Number of ways /routes of the type ( ) = 4! 4 ways Number of ways /routes of the type ( ) = 4 4 4! 4 ways Total = 60 ways 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? Sol. Total number of ways = 4 4 = = 48 FIITJEE Ltd., FIITJEE House, 9-, Kalu Sarai, Sarvapriya Vihar, New elhi , Ph , 66949, Fax 6194

4 PRMO Let f(x) = sin x x + cos for all real x. Find the least natural number n such that f(n + x) = f(x) 10 for all real x. x x Sol. f(x) = sin cos 10 n x n x f(n + x) = sin cos = n and n should be multiple of 10 n n and should be even 10 n should be multiple 6 & 0 both Hence L..M of 6 & 0 will be In a class, the total numbers of boys and girls are in the ratio 4 :. On one day it was fond 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? Sol. Let number of boys = 4x and number of girls = x Given 4x 8 = (x 14) 4x 8 = 9x 84x x 88x + 04 = 0 (x 6)(9x 4) = 0 4 x = 6, x (not possible) 9 Total number of student = 7x = 7 6 = 4 1. In a rectangle, E is the midpoint of ; F is a point on such that F is perpendicular to ; and FE perpendicular to. Suppose = 8. Find. Sol. Let = a F 8 cos F a P = F sin = 8 cos sin cos 8 a 16 sin... (1) P 8 tan... () a E tan a 16 1tan 8 16 a 19 1 a a = 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 x n > 100. [Here [x] denotes the integer part of x and {x} = x [x].) Sol. Given {x}, [x], x are in G.P. [x] = x{x} = x(x [x]) x x[x] [x] = 0 x x x4x FIITJEE Ltd., FIITJEE House, 9-, Kalu Sarai, Sarvapriya Vihar, New elhi , Ph , 66949, Fax 6194

5 PRMO-017- x x 1 1 x x x x 1 x 1 0 x 1 0 x x as {x}, [x], x are in G.P. [x] cannot be zero [x] = x = [x] + {x} = 1 n 1 n = Integers 1,,,.., n, where n >, are written on a board. Two numbers m, k such that 1 < m < n, 1 < 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? nn 1 n n n 6 Sol. Min..M. n n n n 1 n n 10 Max..M. n n Now n n 6 n n n n 1 n < nn 1 mk Now 17 n nn 1 m + k = 17n (m + k)max occurs at n = 4 (m + k)max = Five distinct -digit numbers are in a geometric progression. Find the middle term. Sol. Only possible five distinct numbers for a = 16, r Numbers are 16, 4, 6, 4, 81 Middle term = 6 FIITJEE Ltd., FIITJEE House, 9-, Kalu Sarai, Sarvapriya Vihar, New elhi , Ph , 66949, Fax 6194

6 PRMO Suppose the altitudes of a triangle are 10, 1 and 1. What is the its semi-perimeter? Sol.. rea of = 1 a10 1 b1 1 c 1 10a = 1b = 1c a b c k (say) 6 4 a = 6k, b = k, c = 4k lso area of ss as bs c = 1k 1 7 6k 10 0k k k Semi-perimeter = 7 7 1k k k 7k 1k 7 c a b Note : Not possible to write the answer in integer format. Hence bonus marks be awarded. 18. If the real numbers x, y, z are such that x + 4y + 16z = 48 and xy + 4yz + zx = 4. What is the value of x + y + z? Sol. x + 4y + 16z xy 8yz 8yz 4zx = = 0 (x y) + (y 4z) + (4z x) = 0 x = y = 4z = k (say) x + 4y + 16z = 48 k + k + k = 48 k = 16 Hence, x + y + z k k 1k = k x + y + z = Suppose 1,, are the roots of the equation x 4 + ax + bx = c. Find the value of c. Sol. Fourth root of equation is 6, since sum of roots is zero c = 1 ( 6) = 6 0. 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 from the sides of a quadrilateral? Sol. 0 < a < b < c < 10 lso a + b + c > 10 Total possible case = 9 = 84 ut a + b + c = 6, 7, 8, 9, 10 not possible a + b + c = 6 one way (1,, ) a + b + c = 7 one way (1,, 4) a + b + c = 8 two ways (1,, ) and (1,, 4) a + b + c = 9 three ways (1,, 6) (1,, ) (,, 4) a + b + c = 10 four ways (1,, 7) (1,, 6) (1, 4, ) and (,, ) Hence total ways = = 7 1. Find the number of ordered triples (a, b, c) of positive integers such that abc = 108. Sol. abc = 8 = This is similar to distribution of identical coins among beggars. Here, we have to distribute powers of among terms x, y & z. FIITJEE Ltd., FIITJEE House, 9-, Kalu Sarai, Sarvapriya Vihar, New elhi , Ph , 66949, Fax 6194

7 PRMO x + y + z = Number of ways = + = = 10 Similarly, we distributed powers of among three terms x,y & z x + y + z = Number of ways = + 1 = 4 = 6 Hence, total ways / total number of triplets = 10 6 = 60 possible triplets.. Suppose in the plane 10 pairwise nonparallel lines intersect one another. What is the maximum possible number of polygons (with finite areas) that can be formed? Sol. Maximum number of polygon = = 10 ( ) = 10 ( ) = = 968 polygons Note : The answer must be digit integer, hence bonus marks.. Suppose an integer x, a natural number n and a prime number p satisfy the equation 7x 44x +1 = p. Find the largest value of p. Sol. 7x 44x + 1 = p n (7x )(x 6) = p n ase-i: Either x 6 = 1 and 7x = p n x = 7 Then 7x = 7 7 = 47 = p n p = 47 ase-ii: x 6 = p n and 7x = 1 ut x 7 x 6 p n Hence not possible p = Let P be an interior point of a triangle whose side length s are 6, 6, 78. The line through P parallel to meets in K and in L. The line through P parallel to meets in M and in N. The line through P parallel to meets in S and in T. If KL, MN, ST are of equal lengths, find this common length. Sol. Method-I Let MN = KL = ST = LK ~ 6 So, L, K ST ~ So, S =, T NM ~ So, N, M 6 6 N K P M S L 78 T 6 Now, SL = L + S 78 = 1 78 S = L SL = 78 L = S SL = 78 6 S = NP and L = PM So S + L = NP + PM FIITJEE Ltd., FIITJEE House, 9-, Kalu Sarai, Sarvapriya Vihar, New elhi , Ph , 66949, Fax 6194

8 PRMO = 0 Which is not possible as point P lies inside the triangle, and possible only when point P lies outside the circle Method-II Let PM = 1, PL = and PS = SL = 6 and TM = 78 TPM = NPS = Similarly SPL = also PLS = and we can find other angles as shown in figure Now apply Sine Rule in PTM 78 sin sin and apply Sine Rule in PSL sin 6 sin N 6 K S 6 P 6 L T M 1 Now again by using Sine Rule sin sin sin and eliminating Which is not possible as point P lies inside the triangle, and possible only when point P lies outside the circle. Let be a rectangle and let E and F be point son and respectively such that are (E) = 16, area (EF) = 9 and area (F) =. What is the area of triangle EF? Sol. ar (E) = 16 ar (EF) = 9 ar (F) = Let be the area of rectangle Let E and F k E 1 F 1 l E k F l FIITJEE Ltd., FIITJEE House, 9-, Kalu Sarai, Sarvapriya Vihar, New elhi , Ph , 66949, Fax 6194

9 PRMO E F ar EF k ar 1 1 k 1 k ar EF 9 (i) 1k 1 ar E 1 ar E ar (ii) 1 ar F 1 ar F 1 ar 1 k 1 k 1 (iii) 1 k solving (i), (ii) & (iii), we get 80, and k Thus, area ( EF ) = area of rectangle [ar(e) + ar(ef) + ar(f)] = 80 ( ) ar(ef) = 0 sq. units 6. Let and be two parallel chords in a circle with radius such that the centre O lies between these chords. Suppose = 6, = 8. Suppose further that the area of the part of the circle lying between the chords and is (m + n) / k, where m, n, k are positive integers with gcd (m, n, k) = 1. What is the value of m + n + k? Sol. 4 P 6 R 8 Radius = units OT = 7 O = 74 RO = O = 106 Q r area (segment P) = ar O area (segment Q) = ar(sector OQ) ar(o) FIITJEE Ltd., FIITJEE House, 9-, Kalu Sarai, Sarvapriya Vihar, New elhi , Ph , 66949, Fax 6194

10 PRMO r area () = rea of circle ar(segment P) ar(segment Q) = = = 4 = 4 48 mn k m+n+k = = 7 7. Let 1 be a circle with centre O and let be a diameter of 1. Let P be a point on the segment O different from O. Suppose another circle is with centre P lies in the interior of 1. Tangents are drawn from and to the circle intersecting 1 again at 1 and 1 respectively such that 1 and 1 are on the opposite sides of. Given that 1 =, 1 = 1 and OP = 10, find the radius of is 1. Sol. Let the radius of bigger circle = R radius of smaller circle = r 1 PN ~ 1 r R 10 R 1 M r R 10 PM ~ 1 1 R ividing equation (i) by (ii) r / R 10 we get, r / 1 R 10 P N 1 R 10 R 10 R 0 = R+10 R = 0 8. Let p, q be prime numbers such that n pq n is a multiple of pq for all positive integers n. Find the least possible value of p + q. n Sol. Using Euler's Totient function a 1mod n pq pq1 pq n n pq n n 1 it is easy to see that p,q have got to be both odd and neither can be, p, q are pair wise relatively prime (pq) = (p1)(q1) pq 1 pq n n 1 n pq pq 1 if FIITJEE Ltd., FIITJEE House, 9-, Kalu Sarai, Sarvapriya Vihar, New elhi , Ph , 66949, Fax 6194

11 PRMO p 1 q 1 (pq 1) since p.q are odd pq 1 p 1 (pq 1) p 1 q 1 q 1 (pq 1) q 1 p 1 oth should hold simultaneously the least p+q is = 8 (for p = 11, q = 17) 9. 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. Sol. n 1 will be divisible by (n+1) if (n+1) is a prime number (by Wilson Theorem) so, H..F of ( n 1, n 1) = n 1 if (n + 1) is prime H..F of n 1, n 1 1 if (n+1) is not prime so, only possibility for maximum H..F when n = 96 for n = 96 oth 96 1 and 97 will be divisible by onsider the areas of the four triangles obtained by drawing the diagonals and of a trapezium. The product of these areas, taken two at a time, are computed. If among the six products so obtained, two products are196 and 76, determine the square root of the maximum possible area of the trapezium to the nearest integer. 1 Sol. learly 1 1 Now 6 products are 1, 1, 1, 1, 1, 1 reduces to cases only ase-i: When and 1 = 76 gives = 169 ase-ii: When 1 76 and 1 = 196 gives = < 169 ase-iii: When 1 = 196 and 1 = 76 gives Hence, max value of O 1 FIITJEE Ltd., FIITJEE House, 9-, Kalu Sarai, Sarvapriya Vihar, New elhi , Ph , 66949, Fax 6194