PRMO 07-8_0-08-07 (SOLUTIONS) [ ] PRMO 07-8 : QUESTIONS & SOLUTIONS. How many positive integers less than 000 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? Soln. (8) Let n be the positive integer less than 000 and s be the sum of its digits, then 3 n and 7 s. 3 n 3 s s Also n < 000 s 7 s Clearly, n must be a 3 digit number. Let,, 3 be the digits, then + + 3 = () where 9,0,3 9 3 8 3 For = 3, 4,..., 9, the equation () has,, 3,..., 7 solutions total possible solution of equation () 7 8... 7 8. Suppose a, b are positive real numbers such that a a b b 83, a b b a 8. Find 9 (a b). 5 Soln. (73) a a b b 83 () a b b a 8 () From () & (), 3 3 a b 3 a b a b 83 3 8 a b 3 79 a b 9 (3) From (), a b a b 8 8 a b 9 From (3) & (4), (4) 9 9 (a b) a b a b 5 5 9 8 8 5 9 = 73 Mentors Eduserv: Parus Lok Comple, Boring Road Crossing, Patna- Helpline No. : 9569668800 754405993/4/6/7
[ ] (SOLUTIONS) PRMO 07-8_ 0-08-07 3. A contractor has two teams of workers : team A and team B. Team A can complete a job in 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? Soln. (6) Team A s one day work Team B s one day work 36 Work done by team A in 6 days Work done by team B in days 6 36 8 4 Remaining work after team A left 8 9 4 / 9 No. of days required for team B to complete the remaining work 6 / 36 4. Let a, b be integers such that all the roots of the equation Soln. (5) integers. What is the smallest possible value of a + b? ( a 0)( 7 b) 0 are negative Let < 0 be the roots of + a + 0 = 0 and 0 be the roots of + 7 + b = 0, then 0, a b, 7 possible values of (, ) are ( 0, ), ( 0, ), ( 5, 4) and that of (, ) are ( 6, ), ( 5, ),...,( 9, 8). Smallest value of a = ( 5 4) = 9 and that of b = ( 6) ( ) = 6 smallest value of a + b = 9 +6 = 5. 5. Let u, v, w be real numbers in geometric progression such that u > v > w. Suppose u 40 = v n = w 60. Find the value of n. Soln. (48) u, v, w are in G.P. v uw v uw u 40 = v n = w 60 40 60 40/60 /3 u w w u u 40 n u v uw n u u n/ u /3 5n/6 5n 40 n 48 6 Mentors Eduserv: Parus Lok Comple, Boring Road Crossing, Patna- Helpline No. : 9569668800 754405993/4/6/7
PRMO 07-8_0-08-07 (SOLUTIONS) [ 3 ] 6. Let the sum Soln. (83) 9 9 n n 9 n n(n )(n ) written in its lowest terms be p q (n ) n n(n )(n ) n(n )(n ) 9 n n(n ) (n )(n ).....3.3 3.4 9.0 0. 7 p 0 0 q q p 0 7 83 7. Find the number of positive integers n, such that n n. Soln. (9) n n () n n squaring both sides, n n n n 0 60 3600 n n n 9 Mentors Eduserv: Parus Lok Comple, Boring Road Crossing, Patna- Helpline No. : 9569668800 754405993/4/6/7. Find the value of q p. number of positive integers satisfying () is 9. 8. A pen costs and a notebook costs 3. Find the number of ways in which a person can spend eactly 000 to buy pens and notebooks. Soln. (7) Let the person buys pens and y notebooks, then + 3y = 000 () 000 3y (00y) (y ) = (9 y) (y + ) y Let y + = (k ), k y k 6 (97 k) (k ) 98 3k But > 0 98 k k 7 3 for each k {,,...,7}, we get a unique pair (, y) = (98 3k, k 6) satisfying equation () Hence 7 ways are possible.
[ 4 ] (SOLUTIONS) PRMO 07-8_ 0-08-07 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.) Soln. (60) Each possible route gives a permutation of (B, C, D, E) taken two or more at a time and vice-versa. required no. of ways = P + P P 4 4 4 4 3 4 4 4 4 4 60 0. 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 eactly 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? Soln. (48) The rooms can be selected in only two possible ways as follows : G G G G G G or G G In each case guests can be accommodated in 4 ways. required number of ways 4 = 48. 3. Let f() sin cos for all real. Find the least natural number n such that f(n ) f() for all 3 0 real. Soln. (60) 3 f() sin cos 3 0 period of sin 3 6 3 3 0 0 period of cos 0 3 3 0 period of f() = L.C.M. of 6, 3 least value of n satisfying f(n ) f() is 60.. 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 4 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? Soln. (4) Let the number of boys and girls in the class be 4 and 3 respectively, then according to question, 4 8 = (3 4) 4 8 9 84 96 9 88 04 0 (9 34)( 6) 0 34 or 6 9 But N, 6 total no. of students in class = 4 + 3 = 4. Mentors Eduserv: Parus Lok Comple, Boring Road Crossing, Patna- Helpline No. : 9569668800 754405993/4/6/7
PRMO 07-8_0-08-07 (SOLUTIONS) [ 5 ] 3. In a rectangle ABCD, E is the midpoint of AB; F is a point on AC such that BF is perpendicular to AC; and Soln. (4) FE perpendicular to BD. Suppose BC 8 3 Let BAC Since E is mid point of hypotenous AB of right AE = FE = BE EFA FAE. Find AB. AFB, therefore and FEB EAF EFA EBD 90º BEF 90º But FAE CAB DBA 90º 30º D A E F C B 90º in ABC, BC tan AB AB BCcot 8 3 cot 30º = 4. 4. Suppose is a positive real number such that {}, [] and are in a geometric progression. Find the least positive integer n such that n > 00. (Here [] denotes the integer part of and {} = [].) Soln. (0) A/Q, [] = {}, > 0 Let [] = m and {} = f, where m then m f (m f) ( m f) m mf f 0 m 0 f 0 0 m 0 f f 0 m m f 5 m But f 0 m f 5 m f 5 m But 0 f 5 0 m 5 0 m 5 m ( m 0) W and 0 f, f 5 Mentors Eduserv: Parus Lok Comple, Boring Road Crossing, Patna- Helpline No. : 9569668800 754405993/4/6/7
[ 6 ] (SOLUTIONS) PRMO 07-8_ 0-08-07 5 5 m f Now n > 00 n 5 00 5 3 and 3 8 8 n 8 9 0 5 5 Now 00 least value of n is 0. 5. Integers,, 3...n, where n >, are written on a board. Two numbers m, k such that < m < n, < k < n are removed and the average of the remaining numbers is found to be 7. What is the maimum sum of the two removed numbers? Soln. (5) A/Q, ( 3... n) (m k) 7 n n(n ) (m k) 7(n ) n(n ) m k 7(n ) But 5 m k n 3 n 33n 68 5 n 3 n 33n 68 n 33n 58 0 and n 37n 74 0 (n or n 3) and (3 n 34) 3 n 34 n 3 m k 8 n 33 m k 34 and n 34 m k 5 maimum value of m + k is 5. 6. Five distinct -digit numbers are in a geometric progression. Find the middle term. Soln. (36) Let the numbers be a, ar, ar, ar 3, ar 4, where a N, 0 a 99 and r Q, r. 4 4 00 ar 00 r 0 r a p Let r in lowest form q 4 4 then ar q a 4 q or 3 ( q a 00) For q =, a 6k, k Mentors Eduserv: Parus Lok Comple, Boring Road Crossing, Patna- Helpline No. : 9569668800 754405993/4/6/7
PRMO 07-8_0-08-07 (SOLUTIONS) [ 7 ] 4 4 ar k.p 00 p 3 ( r, p q) and 4 k 3 00 k For q = 3, a 8m, m 4 4 ar m p 00 p 3 but p > q = 3 no value of p in possible. 3 Hence only possible solution is a = 6 and r Middle term ar 6 36 4 7. Suppose the altitudes of a triangle are 0, and 5. What is its semi-perimeter? Soln. (Bonus) Here h a = 0, h b =, h c = 5 9 a h b h c h a b c 0a b 5c 60 (say) 30,a 6,b 5 and c 4 a b c 5 s s(s a)(s b)(s c) 5 3 5 7 30 30 4 8 5 7 7 5 8 60 s 7 7 8. If the real numbers, y, z are such that of + y + z? Soln. () + 4y + 6z = 48 () y + 4yz + z = 4 () From () & (), + 4y + 6z (y + 4yz + z) = 0 {( y) (y 4z) (4z ) } 0 y 4z Now from (), 3 48 6 y z 4 6 = 6 + 4 + = 4y 6z 48 and y + 4yz + z = 4 what is the value Mentors Eduserv: Parus Lok Comple, Boring Road Crossing, Patna- Helpline No. : 9569668800 754405993/4/6/7
[ 8 ] (SOLUTIONS) PRMO 07-8_ 0-08-07 9. Suppose,, 3 are the roots of the equation 4 a b c. Find the value of c. Soln. (36) Let 4th root be, then sum of roots = 3 0 6 Product of roots 3 ( 6) c c 36 0. What is the number of triples (a, b, c) of positive integers such that (i) a < b < c < 0 and (ii) a, b, c, 0 form the sides of a quadrilateral? Soln. (73) a b c 0 a,b,c,0 form a quadrilateral iff a + b + c > 0 i.e. a b c Now a b c c c c 3c 3 c 5 5 c 9 For c = 5, a b 6 and a b 4 which has only two solutions (, 4) & (3, 4). For c = 6, a b 5 and a b 5 which has 5 C = 8 solutions. For c = 7, a b 4 and a b 6 which has 6 C = 4 solutions. For c = 8, a b 3 and a b 7 which has 7 C = solutions. For c = 9, a b and a b 8 which 8 C = 8 solutions. Total no. of solutions is 73.. Find the number of ordered triples (a, b, c) of positive integers such that abc = 08. Soln. (60) abc = 08 = 3 3 number of triplets (a, b, c) = number of ways of distributing two s and three 3 s among a, b, c = + C 3+ C = 6 0 = 60.. Suppose in the plane 0 pairwise nonparallel lines intersect one another. What is the maimum possible number of polygons (with finite areas) that can be formed? Soln. (36) Let there be n(n ) lines in the plane. Now introducing a new line will cut each of the eisting n lines and hence will form n new polygons at most. Maimum possible number of polygons formed with the 0 lines 8 9... 8 36 Mentors Eduserv: Parus Lok Comple, Boring Road Crossing, Patna- Helpline No. : 9569668800 754405993/4/6/7
PRMO 07-8_0-08-07 (SOLUTIONS) [ 9 ] 3. Suppose an integer r, a natural number n and a prime number p satisfy the equation Soln. (47) Find the largest value of p. n 7 44 p n (7 )( 6) p Let m = 6, then (7m + 40) m = p n () If m 7m 40, then (7m + 40) m < 0 (8m 40)(6m 40) 0 0 m 5 m 6 3 But m = 6 does not satisfy () m 7m 40 But both must be powers of p or. m 7m 40 m 40 m is power of a prime or m,, 4, 8, 5 Of these only m = or 8 satisfy (). For m =, 47 = p n p 47,n For m = 8, 8 = p n p, n 7 n 7 44 p. Largest value of p = 47. 4. Let P be an interior point of a triangle ABC whose side lengths are 6, 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. Soln. (Bonus) Clearly PKBT, PMCL and PSAN are parallelograms. Let PT = KB =, PM = LC = y, PK = BT = z and KL = MN = ST = PTM ~ ABC y 6 65 6 Again, NKP ~ ABC y 78 65 78 Adding () & (), 6 78 65 6 78 78 3 78 78 6 5(56 4 ) 6 5 56 () () 78 K 78 l N l l y A l l y S 65 65 l z l z L P y y B z T 6 l M l z C 6 Mentors Eduserv: Parus Lok Comple, Boring Road Crossing, Patna- Helpline No. : 9569668800 754405993/4/6/7
[ 0 ] (SOLUTIONS) PRMO 07-8_ 0-08-07 5 56 30 6 But must be less than 6, hence no solution is possible. 5. Let ABCD be a rectangle and let E and F be points on CD and BC respectively such that area (ADE) = 6, area (CEF) = 9 and area (ABF) = 5. What is the area of triangle AEF? Soln. (30) Let AD = and AB = y ar ( ADE) 6 3 DE DE 6 D 3/ 6 E 9 C F 50/y Similarly, 50 BF y A y B 3 CE AB DE y and Now ar ( CEF) 9 50 CF BC BF y 3 50 y 9 y 600 y 3 50 8 y 600 y 00 y A 00A 600 0 (A y) (A 0)(A 80) 0 A 0 or 80 But A = y = ar ( ABCD) 0 A 80 ar ( AEF) 80 (6 5 9) 30 6. 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 Soln. (75) chords AB and CD is (m n)/k, where m, n, k are positive integers with gcd(m, n, k) =. What is the value of m + n + k? Draw OE AB and OF CD. AB CD Clearly EB 3, FD 4 OE 5 3 4 and OF 5 4 3 OEB ~ DFO Let EOB ODF, then BOD AOC 80º ( 90º ) = 90º A C Mentors Eduserv: Parus Lok Comple, Boring Road Crossing, Patna- Helpline No. : 9569668800 754405993/4/6/7 O 4 3 E F 3 4 5 5 B D
PRMO 07-8_0-08-07 (SOLUTIONS) [ ] Now area of portion between the chords = (area of minor sector BOD) ar( AOB) 5 6 4 4 m 5,n 48 and k 5 5 48 4 m n k 75 7. Let be a circle with centre O and let AB be a diameter of. Le P be a point on the segment OB Soln. (0) different from O. Suppose another circle with centre P lies in the interior of. Tangents are drawn from A and B to the circle intersecting again at A and B respectively such that A and B are on the opposite sides of AB. Given that A B = 5, AB = 5 and OP = 0, find the radius of. Let radius of be R and that of be r From figure, ADP ~ AAB DP AP A B AB r R 0 () 5 R Again, BPE ~ BAB A 5 R B O D E A r 0 R 0 r P 5 B D PE AB BP BA r R 0 () 5 R Dividing () by (), R 0 3 R 0 R 0 8. Let p, q be prime numbers such that n 3pq n is a multiple of 3pq for all positive integers n. Find the least possible value of p + q. Soln. (8) Let p q. For p, n 3pq 6q 6q mod3 n n 3 p 3 pq is odd. Let 3 pq = k +, then 3pq k k n n n n n((n ) ) 3pq n(n ) n n and 3 n(n ) 3pq 3 n n n n n ((n ) n ) (n n) Now, 3pq 3q p 3q 3q But using fermat s lettle theorem, Mentors Eduserv: Parus Lok Comple, Boring Road Crossing, Patna- Helpline No. : 9569668800 754405993/4/6/7
[ ] (SOLUTIONS) PRMO 07-8_ 0-08-07 p (n 3q p ) n 3q 3q p n n 3q i.e. p n(n ) n p but p n(n ) n so, p 3q Similarly q 3p. The pair with least possible sum satisfying above conditions is (, 7). least value of p + q = + 7 = 8. 9. For each positive integer n, consider the highest common factor h n of the two numbers n! + and (n + )!. For n < 00, find the largest value of h n. Soln. (97),3,...,n each divides n but does not divide n any common factor of n and n must be a factor of n + also. But any factor of n + other than and itself does not divide n. so either n, n or n + is a prime. But if n + is prime, then by Wilson s theorem, n n n, n n. and so largest value of h n = largest prime less than 00 = 97 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 si products so obtained, two products are 96 and 576, determine the square root of the maimum possible area of the trapezium to the nearest integer. Soln. (3) Let, y, z, w be areas of the four triangles as shown in figure. then ar( ADB) ar( ACB) y w y w Also AE ar( ADE) ar( AEB) EC ar( DEF) ar( BEC) y z w y z,y, are in G.P. y z Let y = zr and = zr, where r. To make area of trapezium ABCD maimum, we take zy = z r = 576 and yw = z r = 96 ( z y ) z r 96 9 z r 576 4 r z 6 area of trapezium ABCD = + y + z + w = zr + zr + z 9 z( r) 6 3 4 Answer is 3. A D y z E w C B Mentors Eduserv: Parus Lok Comple, Boring Road Crossing, Patna- Helpline No. : 9569668800 754405993/4/6/7
PRMO 07-8_0-08-07 (SOLUTIONS) [ 3 ] ANSWER - KEY. (8). (73) 3. (6) 4. (5) 5. (48) 6. (83) 7. (9) 8. (7) 9. (60) 0. (48). (60). (4) 3. (4) 4. (0) 5. (5) 6. (36) 7. (Bonus) 8. () 9. (36) 0. (73). (60). (36) 3. (47) 4. (Bonus) 5. (30) 6. (75) 7. (0) 8. (8) 9. (97) 30. (3) Mentors Eduserv: Parus Lok Comple, Boring Road Crossing, Patna- Helpline No. : 9569668800 754405993/4/6/7