PRMO Official Test / Solutions

Transcription

1 Date: 19 Aug 2018 PRMO Official Test / Solutions ANSWERKEY Let be the number of pages in the 1st volume Number of pages in 2nd volume & Number of pages in 3rd volume (given) Largest prime factor is 1

2 2. 8 Closest integer of (mod ) So, possible values are or Number of pairs will be Total possible value are 2

3 4. 12 For base the equation can be written as Also, 3

4 6. 18 Given -----(1) (2) Adding Possible value of & are and Sum of 4

5 7. 14 Let be perpendicular bisector to In using cosine rules Closest integer of is 5

6 8. 80 In In In Multiplying we get 6

7 9. 81 Cases : maximum value of 7

8 Let number of cups with handle & let number of cups without handle and 8

9 (mod 3) (mod 3) (mod 3) Hence for {1,1,1} there are 2 possibility for {1,1,-1} ther are possibility for {1, -1, -1} there are possibility for {-1, -1, -1} there are 2 possibility Hence total possibility for But can be selected in ways possibilities. 9

10 Area of (1) For Angle bisector, we have Length of median of right angle triangle 10

11

12 (1) (2) (3) by equation (1) (2) & (3) are pythagoras triples (mod ) If mod (mod ) (mod ) not possible (mod ) (mod ) (mod ) as is the least pythagoras triples 12

13 ) Cases : 2) Cases : Required value 13

14 Let : In...(1) In...(2) Equation (1) = equation (2) & Alternate Solution : In ( G is altitude) In & ; 14

15 (1) Without loss of generality assume Assume & Hence possibilities are or as all are not equal. These do not satisfy equation (1) Hence or or or or & or or or or & as only solution 15

16 Total number of digit 7 are 33 16

17 Let; be a three digit number Now; product of digits for three digit number But, for three digit number product of digits does not exceed 729. Similarly, we can show that 4 or more digit numbers are also not possible. Now: consider a 2 digit number here product of digits is but 99 does not satisfy given condition. But '20' does not satisfy the condition for 2 digit number Using given condition Two digit number is 17. Which satisfy the condition. Also, no one digit number satisfies the given condition. sum of all possible numbers =

18 is parallel to & of it is similar to units. 18

19 22. 6 Sum of available numbers Now, if we want to make disjoint sets with equal sum, then the sum should be divisor of. Now, Total divisors But divisors less than should be neglected as minimum sum of set will be at least Also, the divisor is not a possibility as we want proper subset. Thus, we are le with and We can verify these values that disjoint set can be made. Eg: for sum subsets: we can make {every other remaining number} Using Cauchy-Schwarz inequality we have: Thus, using in the given inequality is satisfied for to for : we get counter example with hence 19

20 Let :...(1) Let : Using in equation (1) Number of Possibilites Total We check the numbers in the sequence error, find first such solution as for divisibility by 13 and 15 and by trial and satisfying all three conditions. Hence sum of squares of digits is 20

21 Case - 1 : above arrangement shows selections where no two square have common side. There are total 61 units square. ways of selecting squares. Case 2 : Instead we can start with selecting 5 square in first row. This given us 6 rows of 5 selections and 5 rows of 6 selection with totals to 60 unit squares. Total ways 21

22 First row can be filled with in ways ways Consider one of the case as shown above: for second row we have following cases: (i) : For which third and fourth row we have only one possibility. (ii) : For which third and fourth row can be filled in ways. (iii) : For which third and fourth row can be filled in ways Total possibilities Sum of digits of is 24 22

23 In In Also : All are non negative integers & number of terms in should be less than or equal to Also, but Maximum degree of can be three is a cubic polynomial By simple trial and error: & 23