PRMO Solution 0.08.07. 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?. Suppose a, b are positive real numbers such that a a b b 83. a b b a 8. Find a b 5 3. A contractor has two teams of workers: team A and team B. Team A can complete 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? x ax 0 x 7x b 0 are negative 4. Let a, b be integers such that all the roots of the equation integers. What is the smallest possible value of a + b? 40 n 60 5. Let u, v, w be real numbers in geometric progression such that u v w. Suppose u v w. Find the value of n? 6. Let the sum n n n n written in its lowest term be p q. Find the value of q p. 7. Find the number of positive integers n. Such that n n. 8. A pen costs Rs and a notebook costs Rs 3. Find the number of ways in which a person can spend exactly Rs 000 to buy pens and notebooks.. There are five cities A, B, C, D and E on a certain island. Each city is connected to every other city by road. In how many ways can a person starting form 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) 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 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? x 3x f x sin cos f n x f x for. Let for all real x. Find the least natural number n such that 3 0 all real x.. 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? 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 FE perpendicular to BD. Suppose BC 8 3. Find AB. 4. Suppose x is a positive real number such that x, xandx are in a geometric progression. Find the n least positive integer n such that x 00. (Here [x] denotes the integer part of x and {x} x [x].) 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 maximum sum of the two removed numbers? 6. Five distinct -digit numbers are in a geometric progression. Find the middle term. 7. Suppose the altitudes of a triangle are 0, and 5. What is its semi-perimeter?
8. If the real numbers x, y, z are such that value of x y z? x 4y 6z 48 and xy 4yz zx 4. What is the. Suppose,, 3 are the roots of the equation x 4 ax bx c. Find the value of c. 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 side of a quadrilateral?. Find the number of ordered triples (a, b, c) of positive integers such that abc 08.. Suppose in the plane 0 pair-wise nonparallel lines intersect one another. What is the maximum possible number of polygons (with finite areas) that can be formed? n 3. Suppose an integer x, a natural number n and a prime number p satisfy the equation 7x 44x p. Find the largest value of p. 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. 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) and area (ABF) 5. What is the area of triangle AEF? 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 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? 7. Let be a circle with center O and let AB be a diameter of. Let P be a point on the segment OB different from O. Suppose another circle with centre P lies in the interior of. Tangents are drawn from A and B to the 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. 8. Let p,q be the 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.. 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 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 6 and 576, determine the square root of the maximum possible area of the trapezium to the nearest integer.
Solutions. Considering both the conditions, the sum of digits must be divisible by. The below table consists of Sequences of digits whose sum is. Sequence Number of Permutations (3,, ) (4, 8, ) 6 (5, 7, ) 6 (6, 6, ) (5, 8, 8) (5,, 8) 3 (6, 7, 8) 3 (6,, 7) 3 (4,,7) 8. a a b b 83 a b b a 8 a u ; b v Solving 3 3 u v 83 u v v u 8 3 3 u v 83... i 3u v 3v u 3 8 3u v 3v u 546... ii 3 3 u v 3u v 3v u 7 u v 3 7 u v 3 3 u v R u v uv (u + v) 8 uv 8 8 u u u u 8 8u u 8
u 8u 8 0 6 6 Solve to get a, b a b 365 73 5 5 3. Team A complete the job in days day work of Team A Team B complete the job in 36 days day work of Team B 36 (A+ B) team day work 3 36 36 A s 4 days work 4 3 Work left 3 3 Team (A+B) days work Work left 6 4 3 Team B do work in day 36 Team B do 4 36 4 work in 6 days 4. (x a x 0)(x 7x b) 0 All roots are ve smallest value (a+b)? (x+ ) (x+ )(x+ ) (x+ )0 Here + a here + 7 0 b + a Vice versa 0 as & 0 4 5 min 4 5 Vice Versa b as & 6 6min 5 30>6 3 4 >6 4 3 >6 5 >6 6 >6
7 0 >6 8 >6 For min (a+b) we need min (a) & min (b) min (a+b) +6 5 5. u u v ur w ur u 0 u n r n u 60 r 0 r 0 u -0 6. r u 6 u 40 u n. 5n 6 u n u 6 40 5n 6 n 40.6 5 48 n(n )(n ) n n(n )(n ) n + + (n ) (n ) n n n n n 4 6 8 0 4 6 8 (n ) 3 4 5 6 7 8 0 (n ) 6 8 0 4 6 8 0 n(n )(n ) 6 5 8 0 6 0 4 0 5 7 0 0-7 83 7. n n n I
both sides are +ve squaring both sides n + n + + n(n ) < n+ n(n ) < 0 n n(n ) < 60 n(n ) is always > n but < n+ in decimal representation n+ n+ fraction < 60 n + fraction < 60 n can befromto as 58 fraction 5 60 Values of X 8. P + 3N 000 000 3N P Maximum possible value for N can be 76. Possible values for N for the above equation is in a pattern N 5, 6, 7, 38, 4, 60, 7. When we go through city we need to follow The Same path back No of ways 0 when through cities we have 4p ways i.e when through 3 cities 4p 3 4 when when through 4 cities 4p 4 4 Total we of ways 0++4+4 60
0. Case i Case ii No of ways 4! No of ways 4! Total x4! 48ways. f(x) sin x 3 + cos 3x 0 Period of sin x 3 6 Period of cos 3x 0 0 3 Period of sin x 3 3x + cos 0 LCM of 6, 0 3 LCMof 6 &0 HCFof &3 60 60 f ( 60 + x) f(x) n 60. Number of boys 4x Number of girls 3x According to question 4x 8 3x 4 4x 8 x + 6 84x 0 x 6 84x 4x 8 0 x 88x 04
88 7744 7344 x 8 88 400 x 8 88 0 x 8 3. 08 68 x 6 or (neglect) 0 8 x6 boys 4x 4 girls 3x 8 total 4 in rt d AFB AF y coa ; BF y sin ; In rt d BGE BG + GE BE (ysin ) + (y sin ) (sin ) y Sin +4sin sin y 0 Cos 4sin sin [ sin a sin a cos a ] 6 sin 4 (sin ) 4 sin 300 4 tan BC 8 3 4 3 4 x 3 y AB y y 3 y
AB Y 4 cm 4. given { x } [ x ] x are in GP xr + least +ve integer S.T x n > 00 x { x } r [ x ] { x } r r r+ r 5 r 5 (but it can t be (-ve) as x is + ve always [ x ] can t be (-ve) ).6 As 0 < { x } < if { x } 0 [ x ] 0 0 < [ x ] < r then x 0 0 < [ x ] < [ x ] 5 { x } r 5 x r r 5 r x 0 as 0 n > 00 for n 5 n log 5 n > n > 00 > log.6 log.6 < log.6 log.6 < 00 0.0 log.6 < 0 nearest + ve integer > is n 0 log.6
n(n ) n(n ) (n ) 3 5. 7 n n n n 4n n n 6 7 (n ) (n ) n 3n (n 3)(n ) 7 (n ) (n ) n n 3 7 n < 35 and n > 3 n 3, 33, 34 case-, n 3 n(n ) p 7 (n ) P 8 n(n ) 7 (n ) p Case-, n 33 P 34 Case-3, n 34 p 5 Maximum sum 5 6. Let the terms be a, ar, ar, ar 3, ar 4 Each of the terms is a -digit no. Possible values for r are 3 or 3 Case a 6, r 3 a 6 ar ar 36 3 6 4 Case a 8, r 3 a 8 ar 4 8 36 Case 3 a 8, r 3 Values are not digit no. similarly
Case 4 a 6 r 3 Discard 7. 0 a b 5 c 0a b 5c L.C.M of 0,,5 60 Dividing by 60 a b c 6 5 4 Let a b c k 6 5 4 a 6k, b 5k, c 4k s(s a)(s b)(s c) 5k 5k 5k 5k 6k 5k 4k 5k 3k 5k 7k 5 7k 4 5k 7 6k0 4 k 8 7 48 a, b 7 40 7 3 c 7
0 a b c 7 a b c 60 7 8. x + 4y + 6z 48 By priniciple of home geneity let all 3 terms be equal x 4y 6z 6 x 4 y Z xy + 4yz + zx 4 Put x 4 y z here xy 4yz zx 8 It also follows principle of homogeneity x +y +z 4 + +. x 4 + 0.x 3 + ax + bx c + + 3 + 8-6..3. -36 -c 36 0 0 0. a, b, c, Where a+b+c>0 a < b< c < 0 We start with assuming a, b, then using a +b +c > 0 i.e c 8, c >7 no of cases i.e (,, 8) or (,, ) Now keep a, increase b by b become 3 C > 6 No. of cases 3 i.e (, 3, 7) ; (, 3, 8); (, 3, ) when we continue in the same fashion we get the following we get the following table
a b c No of cases 8, 3 7,8, 3 4 6,7,8, 4 5 6,7,8, 4 6 7,8, 3 7 8, 8 3 6,7,8, 4 4 5,6,7,8, 5 5 6,7,8, 4 6 7,8, 3 7 8, 8 3 4 5,6,7,8, 5 3 5 6,7,8, 4 3 6 7,8, 3 3 7 8, 3 8 4 5 6,7,8, 4 4 6 7,8, 3 4 7 8, 4 8 5 6 7,8, 3 5 7 8, 5 8 6 7 8, 6 8 7 8. 08 a b c 08 3 3 If N x a y b Total solution a+ C b+ C total solution 4 C 5 C 6x0 60
. No of nm- overlapping polygons n 3n (0) 30 00 30 36 3. (7x-) (x-6)p n 7x- p and x -6 p 40 p - 7 p If N,, p is a divisor of 40 p or 5 If p, 40-7. 3.5-7. 3 and 40+56 Z hence not possible If p 5 then 40 5-7.5 3.5 5 7.5 and 5 40 +35 z hence not possible So 0 p 47 p 47 and 4. k a + l a + ma a Common length a ( k +m ) b ( k + l ) c ( l + m ) a( -l ) b (-m) c ( l +m ) 65 ( l) 78 (l m ) 6 ( l + m) 5 ( l) 6( l-m ) (l+m) 5-5l l+m 5-5l 6-6m 7l + m 5 6m-5l - 6-6m l +m l+8m 6 (i) 6l 4 7 l m 5 7l 8 3
5. 5 common length 6 30 3 ab 3 cd 8 ef 50 F b+c a e+d Possible values a, 3, 4, 8 b c,,, 3, 6, 8 d e, 50,, 5, 5, 0 f only a 8 and f 0 are satisfying the condition area of rectangle 8 0 80 ar (AFE) 80-50 30 units 6. here clearly AOB 74 0 COD 06 0 AOC + BOD 80 0 Area b chord AB & CD w Ar ( AOB) + Ar ( COD) + Ar (AOCA)+Ar(BODB) 5.4.6.3.8 4 + 5
48 5 mn k m48 n 5 k m+k+n75 7. Here AA B 0 0 AB B 0 0 In AB B in AA B AB B ~ PXB AA B~ AYP AB' AB A 'B AB PX PB YP AP 5 R 5 R r R 0 r R 0 5(R 0) 5(R 0) r R R 3(R-0) R+0 R 40 R 0 unit 8. 8. n! + is not divisible by,,n (n+)! is divisible by,,,n So HCF n+ Also (n+)! Is not divisible by n +, n+3. Let n! and 00! HCF 00 is not possible as 00 divides! Not possible with composites consider prime i.e. n 7 Now 6! + and 7! Are both divisible by 7 So HCF 7
30. Case a k 576 - a k 3 6-3 a k 6 a k 576 k 4 3 k ( neglect ve) a 576 3 a x a 384 Case ak ak 6 576 K 4 A 6 Case3 a k 3 6 a k 576 k 6 576 k 4 a 3 3 area a (k+) area 6 3 3 area.5 3 area 3 3 4