FIITJEE ALGEBRA- Pre RMO A. AP, GP 1. Consider the sequence 1,,,, 5, 5, 5, 5, 5, 7, 7, 7, 7, 7, 7, 7,... and evaluate its 016 th term.. 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?. Five distinct -digit numbers are in a geometric progression. Find the middle term.. A leaf is torn from a paperback novel. The sum of the numbers on the remaining pages is 15000. What are the page numbers on the torn leaf? 5. In a book with page numbers 1 to 0 some pages are torn off. The sum of the remaining pages is 99. How many pages are torn? 6. Let a 1,a,...,a n be an arithmetic progression of positive real numbers with common difference d. Let (i) n ai1 x (ii) n ai y (iii) an an 1 z i1 i1 Express d in terms of x, y, z, n. 7. Find the value of a a a 6... a98 if a 1,a,a,... is an arithmetic progression with common difference 1, and a1 a a... a98 17. 8. What is the smallest positive integer that can be expressed as the sum of nine consecutive integers, the sum of ten consecutive integers, and the sum of eleven consecutive integers? 9. It is given that log6 a log6 b logc c 6, where a, b, and c are positive integers that from an increasing geometric sequence and b a is the square of an integer. Find a + b + c.. Two geometric sequences a 1,a,a,... and b 1,b,b,... have the same common ratio with a1 7,b1 99, and a15 b11. Find a 9 11. The expressions A 1 5 6... 7 8 9 and B 1 5... 6 7 8 9 are obtained by writing multiplication and addition operators in an alternating pattern between successive integers. Find the positive difference between integers A and B. 1. The terms of an arithmetic sequence add to 715. The first term of the sequence in increased by 1, the second term increased by, the third term is increased by 5, and in general, the k th term is increased by the k th odd positive integer. The terms of the new sequence add to 86. Find the sum of the first, last and middle terms of the original sequence. 1. For -1 < r < 1, let s(r) denote the sum of the geometric series 1 1r 1r 1r... Let a between -1 and 1 satisfy s(a)s( a) 016. Find s(a) s( a). 1. The sum of the first 011 terms of a geometric series is 00. The sum of the first 0 terms of the same series is 80. Find the sum of the first 60 terms of the series. 15. In an increasing sequence of four positive integers, the first three terms form an arithmetic progression, the last three form a geometric progression, and the first and fourth terms differ by 0. Find the sum of the four terms. B. Exponents & Basic Number Theory 1. Rama was asked by her teacher to subtract from a certain number and then divide the result by 9. Instead she subtracted 9 and then divided the result by. She got as the answer. What would have been her answer if she had solved the problem correctly? FIITJEE Ltd., Plot No. 7, Sector 1B, Opposite Bal Bhawan International School, Dwarka, New Delhi 1 075, Ph. : 011-80596/6/65 website : www.fiitjee.com
Pre_RMO_. For how many pairs of positive integers (x, y) is x + y = 0? 1 x. Suppose that x x x1 x1 5,5 6,6 7,..., 16 17, 17 18. What is the product x1x...x 1?. If a b c, b c d,c d a and abcd 0, then what is the value of a b c d? b c d a 5. What is the smallest positive integer k such that and n, with n > 1? n k( 5 ) a for some positive integers a 6. If real numbers a, b, c, d, e satisfiesa 1 b c d e 5 a b c d e, 7. If What is the value of (x ) x, what is the value of a b c d e? x x 8 x x? 8. If x y 985 and x y 7, what is the value of xy? 9. A man walks a certain distance and rides back in hours. How many hours would it take him to walk both ways? hours, he could ride both ways in. A pen costs Rs. 11 and a note book costs Rs. 1. Find the number of ways in which a person can spend exactly Rs. 00 to buy pens and notebooks. 11. Find the number of positive integers that divide 999 but not 998. 1. If p, q, r are positive integers such that 1. Show that 008 y x 008! 1 has no positive integral solutions. 1. Find all non-negative integral solutions of p q r 6, then find the value of p q r. p q r, where p q. n n n n 15. Prove that for any natural number n, the expression. A 90 80 6 61 is divisible by 1897. 16. Let p(n) (n 1)(n )(n 5)(n 7)(n 9). What is the largest integer that is a divisor of p(n) for all positive even integers n? 17. Let S(M) denote the sum of the digits of a positive integer M written in base. Let N be the smallest positive integer such that S(N) = 01. What is the value of S(5N 01)? 18. Let Akbar and Birbal together have n marbles, where n > 0.Akbar says to Birbal, If I give you some marbles then you will have twice as many marbles as I will have Birbal says to Akbar, If I give you some marbles then you will have thrice as many marbles as I will have. What is the minimum possible value of x for which the above statements are true? 19. What is the maximum possible value of k for which 01 can be written as a sum of k consecutive positive integers? 0. A natural number k is such that k 01 (k 1). What is the largest prime factor of k? 1. The first term of a sequence is 01. Each succeeding term is the sum of the cubes of the digits of the previous term. What is the 01 th term of the sequence? 1 FIITJEE Ltd., Plot No. 7, Sector 1B, Opposite Bal Bhawan International School, Dwarka, New Delhi 1 075, Ph. : 011-80596/6/65 website : www.fiitjee.com
Pre_RMO_. For natural numbers x and y, let (x,y) denotes the greatest common divisor of x and y. How many pairs of natural numbers x and y with x y satisfy the equation xy x y (x,y)?. For how many natural number between 1 and 01 (both inclusive) is 8n 9999 n an integer?. The digits of a positive integer n are four consecutive integers in decreasing order when read from left to right. What is the sum of the possible remainders when n is divided by 7? 5. Let n be the largest integer that is the product of exactly distinct prime numbers x, y and x + y, where x and y are digits. What is the sum of the digits of n? 6. Positive integers a and b are such that a b a b. b a What is the value of a b? 7. How many two-digit positive integers N have the property that the sum of N and the number obtained by reversing the order of the digit of N is a perfect square? 8. Let S(n) and P(n) denote the sum of all digits of n and the product of all digits of n (when written in decimal form), respectively. Find the sum of all two-digit natural numbers n such that n S(n) p(n). 9. Find the sum of the digits in decimal form of the number (999...9). (there are 1 nines) 0. The five-digit number a9b1 is a perfect square. Find the value of a b b1 a1 1. A natural number a has four digits and a ends with the same four digits as that of a. Find the value of (080 - a).. N is a 50-digit number (in the decimal scale). All digits except the 6 th digit (from the left) are 1. If N is divisible by 1, find the 6 th digit.. A four-digit number has the following properties: (A) it is a perfect square, (B) its first two digits are equal to each other, (C) its last two digits are equal to each other. Find all such four-digit numbers.. Find all 6-digit natural numbers a1a aaa5a 6 using the digits 1,,,, 5, 6, once each such that the number a1a...a k is divisible by k, for 1 k 6. 5. Prove that the product of the first 00 positive even integers differs from the product of the first 00 positive odd integers by a multiple of 001. 6. Prove that for any natural number n <, n(-n) is not divisible by. 7. Find the least natural number whose last digit is 7 such that it becomes 5 times larger when this last digit is carried to the beginning of the number. 8. Find the number of positive integers n for which (A) n 1991 and (B) 6 is a factor of 9. If p and (n n ) 'p ' are prime numbers, then find the number of distinct prime factors of 0. Let p be the greatest prime factor of 9991. Find the sum of the digits of p. C. Basics of Functions 'p '. FIITJEE Ltd., Plot No. 7, Sector 1B, Opposite Bal Bhawan International School, Dwarka, New Delhi 1 075, Ph. : 011-80596/6/65 website : www.fiitjee.com
Pre_RMO_ x x 1. How many non-negative integral value of x satisfy the equation? 5 7 (Here [x] denotes the greatest integer less than or equal to x. For example, [.] = and [-.] = - ).. Find the number of positive integers x such that x x 99 1. Let N be the set of natural numbers. Suppose f: N N is a function satisfying the following conditions. (a) f(mn) f(m) f(n) (b) f(m) f(n) if m n; (c) f(). What is the value of 0 f(k)? k 1. Let f(x) be a one-to-one function from the set of natural numbers to itself such that f(mn) f(m) f(n) for all-natural numbers m and n. What is the least possible value of f(999)? x x 5. Let f(x) sin cos f(n x fx for all real x. for all real x. Find the least natural number n such that 6. Suppose x is a positive real number such that {x}, [x] and x are in a geometric progression. Find n the least positive integer n such that x 0. (Here [x] denotes the integer part of x and {x} = x [x]). 7. Find all real numbers a such that a 5 and a(a {a}) is an integer. (Here {a} denotes the fractional part of a. For example {1.5} = 0.5; {-.} = 0.6 8. Determine the number of integral solutions of the equation: x x x, (where [a] denotes the greatest integer a.) 9. Find the number of non-integral solutions of the equation: the greatest integer a. 9 9 x [x] x [x], where [a] denotes. Let x, y, z be three positive real numbers such that x [y] {z} 1. [x] {y} z 1. {x} y [z] 15.1 Where [a] denotes the greatest integer a and {b} denotes the fractional part of b (for example, [5.] = 5, {.} = 0.). Find value of x. 11. Let S 1... 1988. Find S. 1. Find all solutions to the equation x 8[x] 7 0. 1. 9 Find the last two digits of the number 1. 1. Find the minimum natural number n, such that the equation x n 1989 has integer solution x. FIITJEE Ltd., Plot No. 7, Sector 1B, Opposite Bal Bhawan International School, Dwarka, New Delhi 1 075, Ph. : 011-80596/6/65 website : www.fiitjee.com
Pre_RMO_5 19 0 1 91 15. Suppose r is a real number for which r r r... r 0 0 0 0 = 56. Find [0r]. (For real x, [x] is the greatest integer less than or equal to x) FIITJEE Ltd., Plot No. 7, Sector 1B, Opposite Bal Bhawan International School, Dwarka, New Delhi 1 075, Ph. : 011-80596/6/65 website : www.fiitjee.com
Pre_RMO_6 HINTS & SOLUTIONS A. AP, GP 1.Sol. Let the 016 th term be (k + 1). So, 1 5 7... (k 1) 016 1 5 7... (k 1) (k 1) k 016 (k 1) Also, 016 5 So, k and hence the 016 th term. 1 89.Sol. Sum of the numbers n(n 1) (n 1) (n ) n(n 1) 17(n ) (n 1) (n ) (n ) n n n 6 n n 6 n 68 0 n 7n 7 0 n 18n 96 0 (n).(n).(7) 7 96 169 0 (n 7) 7 So, for maximum possible value of n, We have (n 7) 1 n 5 Now, 1... 17 5 17 51 So, the maximum sum of m & k is 51..Sol. The greatest and least -digit numbers are and 99. If ar 99 r 9.9 r 1.77 (approx.) a Now, 16 and 81 and So, the five numbers are, So, the middle term is 6. 9,,,. a,ar,ar,ar,ar are the numbers then.sol. Let the novel has pages from 1 to n. 15000 < 1 + + +... + n and 1 + + +... + (n-) < 15000 (n )(n 1) n(n 1) 15000 Now, 17 17 1878 15000 and 17 17 15051 15000 Also, 17 175 155 15000 So, the sum of the pages on the torn leaf can be 51 or 5, But the number of pages should be even. So, page numbers on the torn leaf are 11 and 11. FIITJEE Ltd., Plot No. 7, Sector 1B, Opposite Bal Bhawan International School, Dwarka, New Delhi 1 075, Ph. : 011-80596/6/65 website : www.fiitjee.com
Pre_RMO_7 0 1 5.Sol. Sum of the pages torn 99 1 We know that 1 15 5 1. So, at most 6 papers (or 1 pages could have been torn). Case-I: 1 paper torn Then (x 1) (x ) 1, which is not possible Case-II: papers are torn: (x 1) (x ) (y 1) (y ) 1 Then (x y) 95 which is not possible Case-III: papers are torn: (x 1) (x ) ( y 1) ( y ) z 1) (z ) 1 (x y z) 1 9 9 x y z which is possible Case-IV: papers torn: (x 1) (x ) (y 1) (y ) (z 1) (z ) (x y z t) 1 1 89, not possible Case-V: 5 papers torn: x 1) (x ) (y 1) (y ) (z 1) ( z ) (t 1) (t ) (a 1) (a ) 1 (x y z t a) 1 15 86 which is not possible Case-VI: 6 paper torn: (x 1) (x ) (y 1) (y ) (z 1) ( z ) (t 1) (t ) (a 1) (a ) (b 1) (b ) 1 (x y z t a b ) 1 18 8 which is not possible. So, paper (i.e., 6 pages) are torn. (t 1) (t ) 1 6.Sol. n y x (a a )(a a ) n i1 d(a i a i1) i1 n d a; i1 n d. a1 an nd(a a ) ndz n n1 y z d nz i i1 i i1 7.Sol. Let a a a 6... a98 t..(1) (a 1 1) (a 1) (a 5 1)... (a 97 1) t a 1 a a 5... a 97 t 9..() (1) + () gives, a1 a a a a 5... a97 a98 t 9 17 t 9 186 t 9 FIITJEE Ltd., Plot No. 7, Sector 1B, Opposite Bal Bhawan International School, Dwarka, New Delhi 1 075, Ph. : 011-80596/6/65 website : www.fiitjee.com
Pre_RMO_8 8.Sol. Let N be the required number. Now, N is the sum of 9 consecutive integers Say a,a,a,a 1,a,a 1,a,a,a. N 9a i.e., N is a multiple of 9. Again, N is the sum of consecutive integers, say b, b + 1, b +,..., b + 9. N 6 5 5(b 9) which is an odd multiple of 5. Again, N is the sum of 11 consecutive integers, Say, a 5,a,a,a,a 1,a,a 1,a,a,a,a 5. N 11a i.e., N is a multiple of 11 Hence, N is a multiple of 9, 11, 5 and also odd N 9 11 5 95 9.Sol. log6 a log6 b log6 c 6 6 abc 6 But ac b ( a,b,c are in G.P.) b 6 6 b 6 6 a 1,,9,16,5 a 5,,7,0,11 b 6 But c a a So, a must be a factor or 6 a = 7 (only possibility) 6 6 c 1 8 7 a b c 7 6 8 111.Sol. Let the common ratio be r. a b Now, 15 11 1 a r b r a r b () 1 1 1 1 r 1 1 b 99 11 a 7 8 11 11 a9 a 1.r a 1(r ) 7 7 6 9 11.Sol. B A 1 6.. 6 8 9 ( 8) ( 6... 8) 8 (1... 19) 19 0 8 760 8 7 1.Sol. Let the sequence have n terms that add to 715. 1 5... (n 1) 86 715 11 n 11 n 11. So, let the first term, middle term and last terms be a, m and l respectively Now, n a l 715 715 a l 65 10 11 FIITJEE Ltd., Plot No. 7, Sector 1B, Opposite Bal Bhawan International School, Dwarka, New Delhi 1 075, Ph. : 011-80596/6/65 website : www.fiitjee.com
Pre_RMO_9 Also, a l m m 65 a l m 195 1.Sol. S(r) 1(1 r r...) 1 1 1 r Now, S(a)S( a) 016 1 1 016 1 a 1 a 11 1 a 016 1 1 1 1 a a...(1) 1 1 1 1 S(a) S( a) 1 a 1 a 1 a 1 a 1 (1 a)(1 a) 1 a 1 6 1 1 1.Sol. Let the terms be a 1,a,a,... Also, let the sum of the first n terms be S n. If the common ratio of the G.P. is r, then, 011 a a.r ; k011 k0 k a a.r k 0 011 Now, S a 00..(1) 011 k k 1 0 011 S S a a 0 011 k k 011 k 01 k 1 011 011 011 011 180 a.r r a k k k 1 k 1 011 011 9 180 r 00 r..() 60 011 Again, S S a a 60 0 k k 0 k 0 k 1 011 011 0 0 S 80 a.r r a 60 k k k1 k 1 011 S 80 (r ) 00 60 9 S60 80 00 80 81 80 16 5 (a d) 15. Sol. Let the four terms be a d,a,a d,(a d). a (a d) Now, (a d) 0 a a ad d a ad 0a d ad 0a 0 FIITJEE Ltd., Plot No. 7, Sector 1B, Opposite Bal Bhawan International School, Dwarka, New Delhi 1 075, Ph. : 011-80596/6/65 website : www.fiitjee.com
Pre_RMO_ d d a 0 d ( d) So, d is a multiple of and d is a multiple of d. Also, d <. So, d,6 or 9. d 7,, or 1 But is not a multiple of 7. So, d = 6 or 9 6 1 Case-I: If d = 6, then a ( 6) Then terms are -,, 9, 7. But all the terms are positive. So, this case is not possible. 81 Case-II: If d = 9, then a 7 1 So, the four terms are 18, 7, 6, 8 = 19. B. Exponents & Basic Number Theory 1. Let the certain number be x. x 9 Now, x 19 9 18 18 Required answer 15 9. y 0 x So, for every positive multiple of less than 0, we have exactly one solution. So, there are solutions.. 18 17 (16 ) 16 x1 x1 x1 (x 1 ) x1 x1x x...x1 x1 7 (x1x x...x1 x 1 ) So, 7 x 1 x x...x 1 x 1.... a b c (c d) c d b c d (d a) d a ( d) d c d a d ( d) d a b c d b c d a d d d d d d d d 1 1 1 1 5. n k(7 6 15) a n k 16 a n k 6 a So, least positive integral value of k is 1, with n = and a = 6. 6. a e,b e,c e,d e 1 Now, e 5 a b c d e a b c d (e ) (e ) (e ) (e 1) FIITJEE Ltd., Plot No. 7, Sector 1B, Opposite Bal Bhawan International School, Dwarka, New Delhi 1 075, Ph. : 011-80596/6/65 website : www.fiitjee.com
Pre_RMO_11 7. If e 8 e d 1,c 0,b 1,a Hence, a b c d e 1 0 1 x x (x ).(x ) x x 16 (x ) 16 a a 16, where a x a a (a ) 16 a a a x x 8 16 x x x x 56 58 8. x y 985...(1) x y 7...() (1) + () gives (1) () gives xy 6 8 8 x x 6 158 79 x = 6 y 9 51 y 1 9 y 8 9. Let w and r be respectively the time taken by the man to walk and ride the certain distance one way. So, r =.5 hours r = 1.5 hours. Also, w + r = hours =.75 hours. w =.75 1.5 =.5 hours Time required to walk both ways = w = 5 hours.. Let the number of pens purchased be x and the number of notebooks purchased be y. So, 11x 1y 00 By hit and trial we find that y = 5 is the least natural number that satisfy the above equation. 00 65 So, y 5, gives x 85 11 Hence, y 5 11k and x 85 1k where k is a non-negative integer. Now, x 0 85 1 k 0 k 85 k 6 1 So, for k = 0, 1,,,, 5, 6 we have solutions i.e., there are 7 ways possible 11. 999 999 999 5 998 998 998 5 If a number divides 999 but not 998, then obviously it should be of the form where 0 a 999, 0 b 999 and a, b are integers. a 999 So, 00 cases for 5 And 00 cases for 999 5 b 999 999 But 5 is included in both cases. So, required number of positive integers 00 00 1 1999. a 999 5 or 999 b 5, 1. Let p q r r p r q r ( 1) 6 7 FIITJEE Ltd., Plot No. 7, Sector 1B, Opposite Bal Bhawan International School, Dwarka, New Delhi 1 075, Ph. : 011-80596/6/65 website : www.fiitjee.com
Pre_RMO_1 pr qr Now 1 is odd. r pr qr and 1 7 p5 q5 r 5 and 7 q5 pq r 5 and ( 1) 9 pq Again 1 is odd q5 pq r 5; and 1 9 r 5; q 8 and p 11,q 8,r 5 p q r 8 p8 1. 008! 008 007 006... 1 So, 008 is a multiple of 1 x 008 1 y 008! Must be a multiple of 1 ( 7) So, x is a multiple of both & 7 Let x = 1 k y 008 008 Then 1 (1k) 008! 1 So, y must be greater than 008 y 008 But 1 (1k) 008! And 008! has much fewer times multiple of and 7. So, y cannot be greater than 008. Which leads to the contradiction that y is greater than 008. So, there is no positive integral solution. 1. Let q p t, where t 0 p p t r p (1 t ) r But if t 0, then we have an odd factor of r which is greater than 1, which is not possible. So, the only possibility is t = 0 i.e., p q r p p p i.e., r p 1 Hence, the solutions are p q k, r k 1, where k is any non-negative integer. 15. 1897 7 71 So, we need to prove that A is divisible by both 7 and 71. n n n n Now, A (90 80 ) (6 61 ) = (Multiple of 90 80) - (Multiple of 6 61) = Multiple of 0 Multiple of 0 = Multiple of 7 Multiple of 7 = Multiple of 7 n n n n Again, A (90 6 ) (80 61 ) = (Multiple of 9 Multiple of 5 = Multiple of 71 Multiple of 71. = Multiple of 71. Hence proved. 16. As n is even, so each of (n + 1), (n + ), (n + 5), (n + 7), (n + 9) are odd. So, their product will contain at least one multiple of and exactly 1 multiple of 5. So, the largest integer that is a divisor of p(n) is 5 = 15. 17. 01 6 9 So, N 6999...999 times N 7 00 0...00 0 1 zeroes FIITJEE Ltd., Plot No. 7, Sector 1B, Opposite Bal Bhawan International School, Dwarka, New Delhi 1 075, Ph. : 011-80596/6/65 website : www.fiitjee.com
Pre_RMO_1 5N 01 5000...000 5 01 zeroes zeroes 50 00...000 008 So, S(5N 01) 5 0 0 8 18 18. Let Akbar and Birbal respectively have k and n-k number of marbles So, (k p) (n k p) (where p is the number of marbles Akbar gives to Birbal) n (k p)...(1) Let Birbal give q marbles to Akbar. So, (n k q) (k q) n (k q)..() From (1) & (), we have n to be a multiple of both and. So, the least possible value of n is 1. 19. Let 1... (n 1) 01 1... n We observe that 6 6 1... 6 016 So, 5... 6 6 01 i.e., maximum value of k is 61. Alternative Method 01 = 671 11 61 61 Also, 1 65 61 611 So, keeping as the middle term and taking terms prior to and after, we have...,1,,,,5,... 0terms 0terms Since 0 1, So this is the maximum possible i.e., the maximum value of k is 61. 0. 0 1600, 1 1681, 176, 189, 196, 5 05 So, 01 5 k =. Hence, the largest prime factor of k is 11. 1. Let P n denote the n th term of the sequence. P 01 1 P 0 1 8 0 1 6 7 P 7 7 70 P 7 0 7 0 70 So, we observe that P P P... 70 01 th term in the sequence is 70.. Obviously (x, y) x y xy y y y x But if x y, then (x, y) = 1 So, x = is not possible Case-I : If x = 1, then, y = y + 1 + 1 FIITJEE Ltd., Plot No. 7, Sector 1B, Opposite Bal Bhawan International School, Dwarka, New Delhi 1 075, Ph. : 011-80596/6/65 website : www.fiitjee.com
Pre_RMO_1 Which is not possible. Case-II : If x = then y = + y + (x, y) y = + (, y) So, y = and y = and the only possibilities Hence the ordered pairs can be (, ) and (, ) i.e., possible pairs.. 8n 9999 n p(say) 8n p 9999 np p n 9999 p 8 But n 01 p 9999 p 8 p p 8 9 7p 16 p 1 or p 1 gives n = 1111, which is permissible. p = gives n = 1999.8, which is not permissible. So, there is only one natural number n that satisfies the given conditions.. Let the digits from right to left be (a + ), (a + ), (a + 1) and a. So, a 9 a 6 i.e., a = 0, 1,,,, 5, 6 Now, the given number N = (a + ) 00 + (a + ) 0 + (a + 1) + a (a 1)(00 0 ) 00 0 a 0 7(a 1) 7 56 8 a So, possible remainders are 8 + 0, 8 + 1, 8 +,..., 8 + 6 So required sum = 8 + 9 + 0 + 1+ + + = 17 5. Since x & y are single digit prime numbers. So, they can be,, 5 or 7. But x + y is also prime. So, y cannot be or 5 y = or 7 Case-I : If y =, then x can be or 7 But the maximum product will occur when x = 7. In this case n 7 7 15 Case-II : If y = 7, then x can be So, in this case, n = 7 7 < 7 7. So, the maximum value of n is 15. So, sum of the digits of n = 1 + 5 + + = 1 6. The only possibility for which a b becomes a positive integer is a = b b a a b 11 a b 1 a b 1 1 7. Let N ab a b And n ba b a N n 11(a b) But N + n is a perfect square. So, a + b = multiple of 11 FIITJEE Ltd., Plot No. 7, Sector 1B, Opposite Bal Bhawan International School, Dwarka, New Delhi 1 075, Ph. : 011-80596/6/65 website : www.fiitjee.com
Pre_RMO_15 But 0 < a + b 9 + 9 = 18 So, a + b = 11 is the only possibility. So, a can take any value from to 9 i.e., 8 possibilities. The 8 possible ordered pairs are (, 9), (, 8), (, 7), (5, 6), (6, 5), (7, ), (8, ), (9, ). 8. Let n ab a b n S(n) P(n) gives, a b a b ab 9a ab b 9,a any non-zero digit. Required sum (1 5 6 7 8 9) 9 9 9 81 50 81 51 9. (999...9) 1 times nine 1 1 1 1 ( 1) ( ) ( ) 1 6 1 1 ( 1) ( 1) 1 999...9 999...9 6 times nine 1 times nine 999...9 999...9 7 000...0 6 times nine 11 times nine 1 times zero 999...9 7 000...0 999...9 11 times nine 11 times nine 1 times zero So, required sum of the digits 9 11 7 0 11 9 1 = 16 0. The perfect square ends in 1, So its square root will end in 9 or 1. Again, 10 19600 and 180 00 So, if k a9b1, then 10 k 180 So, we have possible values for k As 11, 19, 151, 159, 161, 169, 171 and 179 11 (10 1) 19600 80 1 19881, not possible 19 (150 1) 500 00 1 01, not possible 151 (150 1) 500 00 1 801, not possible 159 (160 1) 5600 0 1 581, not possible 161 (160 1) 5600 0 1 591, a possibility 169 (170 1) 8900 0 1 8561, not possible 171 (170 1) 8900 0 1 91, not possible So, the only possibility is a9b1 = 591 i.e., a 5,b a b 5 5 16 1 b 1 a 1 1 1. Last four digits of a is a a a Multiple of 000 a(a 1) Multiple of 5 and So, two cases arise :- Case-I : If a is a multiple of 5, then (a - 1) should be a multiple of. So, the last two digits of a can be 5. Then, let a 0k 5, where k is a two-digit number. (0k 5) (0k 5) multiple of 000 000k 5000k 65 0k 5 = Multiple of 000 FIITJEE Ltd., Plot No. 7, Sector 1B, Opposite Bal Bhawan International School, Dwarka, New Delhi 1 075, Ph. : 011-80596/6/65 website : www.fiitjee.com
Pre_RMO_16 900k + 600 = Multiple of 000 9k + 6 = Multiple of 0 (50 1)k + 6 = Multiple of 0 Which is possible only if k = 6, 6, 06, etc. But k is a -digit number. So, this case is not possible Case-II : If a 1 is a multiple of 5 and a is a multiple of. So, the last two digits of a can be 76. Let a = 0p + 76, where p is a -digit number. (0p 76) (0p 76) Multiple of 000 000p 1500p 5776 0p 76 = Multiple of 000 150p + 5700 = multiple of 000 151 p + 57 = Multiple of 0 0p + 50p + p + 57 = Multiple of 0 So, p is odd So, the last two digits of 50 p + p + 57 = 00 i.e., last two digits of p 50 9 Hence, a 976 And so, 080 a = 70. Let x be the 6 th digit. N = 111...1 x 111...1 5 times 1 times 1 But we know that 01 is a multiple of 1. So, 111111 is also a multiple of 1. So, 111...1 is also a multiple of 1. 6k times one Now, N 111...11x 111...1 So, N 1x 000...0 times one times one times zero is a multiple of 1. So, 1x is a multiple of 1. Hence, x = is the only possibility i.e., the 6 th digit of N is.. Let N aabb be the required perfect square. Now, N =10a + 11 b = 11(0a + b) p = 11 (0a + b), where N = p So, 0a + b = multiple of 11 99a + (a + b) = multiple of 11 a + b = multiple of 11 But 0 < a + b 9 + 9 = 18 So, a + b = 11 is the only possibility. Again, p 11(0a b) = 11(99a + (a + b) = 11(99a + 11) = 11 (9a + 1) So, 9a + 1 = k for some natural number k (k - 1) (k + 1) = 9a a = 7 is the only possibility. b = 11 a = Hence, the four-digit number is 77 = 88.. Obviously a 5 = 5. Also, a, a and a 6 are multiples of, and 6 respectively. So, the even digits occupy the even places. And hence the odd digits the odd places. So, 1 and occupy the 1 st and rd place respectively or vice-versa. Case-I : a1a aaa5a6 1a a5a6 FIITJEE Ltd., Plot No. 7, Sector 1B, Opposite Bal Bhawan International School, Dwarka, New Delhi 1 075, Ph. : 011-80596/6/65 website : www.fiitjee.com
Pre_RMO_17 Now, in this case 1a is a multiple of and a, or 6. So, a. Now, we that a should be divisible by and a = or 6. So, a 6. Hence, in this case the only possibility is 165. a a a a a a a 1a 5a Case-II: 1 5 6 6 Now, a1 is divisible by and a =, or 6 a = Again, 1a is divisible by and a or 6 a 6 and hence a6 So, in this case, the only possibility is 165 Hence, the required numbers are 165 and 165. 5. Let N = A B, where A 6 8... 1996 1998 000 and B 1 5 7... 1995 1997 1999 Now, 001 667 9 B 1 5...... 9... 1999 is a multiple of 9 001. Again A 6... 6... 58... 000 is a multiple of 6 6 58 = multiple of 9 = multiple of 9 = multiple of 001 Since, A and B are both multiples of 001, So, N = A B is also a multiple of 001. 6. If HCF (n, ) = 1, then HCF ( n, ) = 1 i.e., in this case n(-n) and have no common factor and hence n( n) cannot be divisibly by. If HCF (n, ) = a, then a is a factor of ( 5 7 11) and let b a Then n( n) a(ab a) a (b 1) is a multiple of a but not b So, n( n) cannot be a multiple of b (which is a factor of ) Hence n( n) is not divisible by if n <. 7. Let N be the k digit number required Also, let N = a1aa...a k 1 ak But ak 7(given) Also, ak 1 5 (since placing 7 on the left most make it a multiple of 5) N a,a,a,...a 57 1 k New number formed is 7a 1,a,...a k 5 N 7 a a a...a 50 Now, 1 k N 7 a1a a...a k5 N 7 k 7 7a 1 a a...a k 5 N 7 k 7 5N 9 N 7 k1 7N 1 k 7 FIITJEE Ltd., Plot No. 7, Sector 1B, Opposite Bal Bhawan International School, Dwarka, New Delhi 1 075, Ph. : 011-80596/6/65 website : www.fiitjee.com
Pre_RMO_18 k1 1 should be a multiple of 7. So, least value of k + 1 is 6 6 1 999999 Least value of N 1857 7 7 8. 9. n n (n 1)(n ) = product of two consecutive integers = always a multiple of. If n k 1 type then n (k 1) is a multiple of. If n k, type then n + 1= (k + 1) is a multiple of. But for n k type neither n + 1(= k + 1) nor n + (= k + ) are multiple of. So, we just need to find the number of natural numbers n such that n 1991 which are not multiples of. So, required number of numbers 1991 66 18 p is prime. So, p cannot be even i.e. p. Again, if p k 1 or k, then p 9k 6k 1 or 9k 1k p (k k 1) or (k k ) Which cannot be prime. p =. p 7 0, which has distinct prime factors,, 5. 0. 9991 000 9 0 97 Both 97 and are primes. So, p = Sum of the digits of p is. C. Basics of Functions x x 1.Sol. Let k 5 7 (k 0 as x is non-negative) x x So, k k 1 and k k 1 5 7 5k x 5k 5 and 7k x 7k 7 7k x 5k 5 k 5 i.e., k 0, 1 or. If k 0, then x 0,1,,, If k 1, then x 7,8,9 If k, then x 1 i.e., there are 9 non-negative integral values of x that satisfy x x.sol. Let k 99 1 (k 0 as x is positive) x x So, k k 1 and k k 1 99 1 i.e., 99k x 99k 99 and 1k x 1k 1 i.e., 1k x 99k 99 i.e., k 99 i.e., k 0,1,,,...,9 For k = 0, x can take 1,,,..., 98 x x 5 7 FIITJEE Ltd., Plot No. 7, Sector 1B, Opposite Bal Bhawan International School, Dwarka, New Delhi 1 075, Ph. : 011-80596/6/65 website : www.fiitjee.com
Pre_RMO_19 For k = 1, x can take 1,,..., 197 For k =, x can take 0, 0,..., 96 For k =, x can take 0, 0,..., 95 and so on. So, the total number of solutions 98 (97 95 9... 1) 1 (99 97 95... 1) 1 500 99.Sol. f(m 1) f(m).f(1) f(1) 1 f( ) f() f() f() f() ( f() f() f()) Also, f(6) f() 6 f(5) 5 f(8) f() 8 f(7) 7 f(1) f(7) 1 f(8) 8; f(9) 9,f(, f(11) 11, f(1) 1; f(1) 1 f(0) f() 0 f(1) 1; f(15) 15,f(16) 16;f(17) 17,f(18) 18;f(19) 19 So, 0 1 f(k) k 0 0 k 1 k 1.Sol. f(999) f(9 111) f(9) f(111) f( ) f( 7) f() f() f() f(7 f() f(7) So, the minimum value of f(999) 1 x 5.Sol. Period of sin and x cos are and respectively. 5 LCM of and 5 LCM of and 5 LCM of and HCF of and 5 6 So, the least natural number n is 6. 6.Sol. {x} x [x] Obviously, x = 0 if [x] = 0. But x > 0 (given) So, [x] 1 Now, {x} ([x] + {x}) = [x] {x} = [x] ([x] {x}) [x] [x] 1 (1 {x}) So, if [x], then {x} (1 0) Which is not possible as So, [x] = 1 {x} 1 FIITJEE Ltd., Plot No. 7, Sector 1B, Opposite Bal Bhawan International School, Dwarka, New Delhi 1 075, Ph. : 011-80596/6/65 website : www.fiitjee.com
Pre_RMO_0 {x}x 1 1 {x}(1 {x}) 1 {x} {x} 1 0 1 1 1 5 {x} 1 But {x} 0 So, 1 5 {x} 1 5 x [x] {x} 0 6 5 5 x 0 9 5 6 5 7 5 x 0 8 9 5 5 7 1 5 x 0 7 1 5 1 5 x 9 7 5 68 5 8 17 5 0 7 1 5 5 11 68 5 5 x 1 5 99.51 0 11 So, x 0 So, x 11 7.Sol. a 5 [a] a {a} Now, a(a {a}) an integer {a} {a} {a} an integer 16 {a} {a} an integer 18 ({a} {a} 1) an integer {a} 1 But 1 1 {a} 1 (1 {a}) an integer <(1+{a}) < 8 (1 {a}),,5,6,7 5 6 7 1 {a},,,, 5 6 7 {a},,,, p i.e., a, where p,,5,6,7 8. Sol. Let x 6k r, where k is any integer and r = 0, 1,,,, 5 Case-I: x 6k 0, then FIITJEE Ltd., Plot No. 7, Sector 1B, Opposite Bal Bhawan International School, Dwarka, New Delhi 1 075, Ph. : 011-80596/6/65 website : www.fiitjee.com
Pre_RMO_1 k k 6k k 0 i.e., x = 0 Case-II: x 6k 1, then k k 6k 1 k 1,i.e.,x 7 Case-III: x 6k, then (k 1) (k 1) 6k k 0 i.e., x = Case-IV: x 6k, then (k 1) (k ) 6k k = 0 i.e., x = Case-V: x 6k, then (k + ) + (k + ) = 6k + k = 0 i.e., x = Case-VI: x 6k 5, then (k ) (k ) 6k 5 k = 0 i.e., x = 5 So, there are 6 integral solutions. 9. Sol. 1 1 x [x] 9 [x] x x [x] x [x] 9 x[x] x[x] 9 ( x [x] given) Now, [x] x[x] x [x] 9 x (if x 0) [x] 9 or - (if x < 0) 9 9 x or 9 9 But x 9 So, [x] x[x] 9 x = -9.. Sol. x [y] {z} 1....(1) [x] {y} z 1....() {x} y [z] 15.1...() (1) + () + () gives (x y z).6 i.e., x y z 1....() () (1) gives y [y] z {z} 8.1 {y} [z] 8.1 [z] 8,{y} 0.1 () () gives, {x} [y] 7 {x} 0,[y] 7 () () gives [x] {z} 6. [x] 6.,{z} 0. x [x] {x} 6 y [y] {y} 7.1 z [z] {z} 8. 11.Sol. Suppose, [ p] k Then k p (k 1) FIITJEE Ltd., Plot No. 7, Sector 1B, Opposite Bal Bhawan International School, Dwarka, New Delhi 1 075, Ph. : 011-80596/6/65 website : www.fiitjee.com
Pre_RMO_ So, the number of times [ p] k is k 1. So, the value of 1,,,... will occur, 5, 7,... times. But 1988 But 196. So, will occur 1988 196 + 1 = 5 times So, required sum (1 5 7... 87) 5 i(i 1) 5 i1 i i 5 i1 i1 87 5 6 5868 96 5816 1.Sol. x 8[x] 7 0 x 8x 7 (x 1)(x 7) (x 1)(x 7) 0 1 x 7 1 [x] 7 So, x 7 8[x] x 8[x] 7 Considering the cases [x] 1,,,,5,6,7 we have the following table [x] x 8[x] 7 x [x] 1 1 1 1 9 17 17 5 5 5 5 5 6 1 1 6 7 9 7 7 Since, the value of [x] is not the same in some cases. So, they are discarded. So, the possible values of x are 1,, 1,7. 1.Sol. 9 9 7 7 k(say) 1 1 1 7 As, 0 1 1 9 7 So, k 1 1 1 1 ( ) 1 6 1 1 6 1 8 So, the last two digits of k is 08. 1. Sol. Obviously x cannot be negative or 0. x 1 n Also, 1989 n Considering the cases n,5,6,7,... FIITJEE Ltd., Plot No. 7, Sector 1B, Opposite Bal Bhawan International School, Dwarka, New Delhi 1 075, Ph. : 011-80596/6/65 website : www.fiitjee.com
Case-I: If n =, we have, 5.07... 1989 Now, 000 5 and 6 So, n Case-II: If n = 5, we have, 5 50.765... 1989 5 Now, 000 50 and 5 51 So, n 5 Case-III: If n = 6, we have, 6 50.765... 1989 1666.67 1960.78 6 Now, 199.0... 50 and 6 1988.07... 50 So, n 6 Case-IV: If n = 7, we have 7 507.65... 1989 7 7 Now, 1989.58 507 & 1989.65 and x 506 506 So, x 507 and x 506 satisfies the condition with the minimum value of n as 7. 19 15. Sol. Let r k 0 91 19 7 Obviously, r r k 1 0 0 0 There are total 7 terms and 56 7 7 5 So, k 7. Also, there are 5 terms equal to k + 1 (= 8). And 8 terms equal to k (=7) 19 7 19 8 So, r 7 0 0 and r 0 0 = 8 56 57 i.e., 7 r 8 and 8 r 9 0 0 7 7 r and r 0 0 0 r lies between 7 and 7 [0r] 7 Pre_RMO_ FIITJEE Ltd., Plot No. 7, Sector 1B, Opposite Bal Bhawan International School, Dwarka, New Delhi 1 075, Ph. : 011-80596/6/65 website : www.fiitjee.com