RMO 2009

yes govind is correct....... harsh, your method is wrong....

i can say that because i qualified rmo 2009......

any one tell me wha is the cut off in bihar region

why the answer of last question can"t b 13..........

it can be 13 if the book is printed only one side..............................think over it

I solved the question using method of contradiction.

Suppose a² - 3a - 19 is divisible by 289.
Then a² - 3a - 19=289k for some a, k ε Z
then a² - 3a - 19 - 289k=0 is a quadratic in a where a ε Z.

for a to be an integer, discriminant must be a perfect square.

i.e. 1156k + 85 must be a perfect square.
i.e. 17(68k + 5) must be a perfect square.
i.e. 68k + 5 must be divisible by 17.

but that is not possible.

Thus our assumption that a² - 3a - 19 is divisible by 289 is false.

an easy method fr ques 4......

S(odd nos.) = 111 + 113 + 115 + 117 + 119 +131 + ....... + 999
S(odd nos.) = 999 + 997 + 995 + 993 + 991 + 979 + ....... + 111
Adding,
2S(odd nos.) = 1110 + 1110 + 1110 + ........ (125 times)
Hence, S(odd nos.) = (1110 x 125) / 2 = 69375
S(even nos.) = (1088 x 100) / 2 = 54400
S(3-digit nos.) = 900 / 2 (100 + 999) = 494550
Hence, Req. Value = 494550 - (54400 + 69375) = 370775

according to me the answer is 2
proof:-

sum of pages torn is =101

suppose first page folded is n & total number of pages folded is t

so
n + (n+1) + (n+2) + .......... +(n+t-1) =101

nt + ( 1 + 2 + 3 +......(t-1) ) = 101

nt + t(t-1)2 = 101
nt = 101 - t(t-1)2
n = 101t - (t-1)2
now n must be integer & 101 is prime number

so t only can be 1 , 2 , 101 & 202 only (if we put any other positive value of t then n comes in fraction)

when we put t = 1 then n = 101
when we put t = 2 then n = 50
when we put t = 101 then n = - 49
similarly t ≠202

so t only can be 2 because n can be 1 , 2 , 3 .....100
so pages torned are 50,51 only

tell me that my answer is wrong or correct (jindalharshkumar@gmail.com)

i read in class X. in which class/year should i attempt to sit in rmo?
can anybody suggest?

isit possible that the book has page printed only one side
in this case i found the possible answers may reach 13 pages : 1,2,3,4,5,6,7,8,9,10,11,12 & 23
i have written about both cases -when book has pageprinted only one side
and when the book has page printed both side
anone plz tell iam right or wrong

paper was very easy

see people sum of total pages is n(n+1)/2 and sum of remaining pages is 15000 . therefore if rth page is torn then n(n+1)/2 - r(r+1)=15000 solving the above diophantine equation yields no integral solution .. how is this possible....?????

i will also give rmo this year.....can u also plz mail me de book....i hav posted my email address on your mailbox...

5)The fifth one was unsolved here...I was trying this and so posting what i got.. its simple..

1)When we measure the distance of a line from a point the minimum distance is the perpendicular distance and maximum occurs at either of the end points of a line. So distance between two lines is maximum when the two point we choose is either of the four possible pair of points, one of which is end point of one line and the rest is end point of the other line..But none of those distances are greater than 1, according to the condition of the problem. So distance between any two points on the boundary must be smaller than 1.

2)Join X,Y and extend on either sides. Let it intersect the polygon at Z₁ and Z₂.The points Z₁, Z₂ are in opposite directions of both X and Y.. Since Z₁,Z₂ are in the boundary of the polygon 2|Z₁Z₂|≤ 2, Which means (|XZ₁| + |YZ₁|) + (|XZ₂| + |YZ₂|)≤2. So at least one of the summands in the bracket must be ≤ 1. Choose Z = Z_i, if (|XZ_i| + |YZ_i|)≤ 1.

i m an 8th grader and wen i saw these problems i had no idea abt them can anyone help me qualify for inmo next year

6 was probably the easiest ques of the paper.....
it was an simple application of ap.......
u got sum of torn page nos. = 101 after that taking least possible values, n/2(n+1)<101
which gave n=13
it meant at maximum, 6 pages cud be torn....
even no. of pages wer not possible.....
as odd x even = even
for 1,3,5
for 1 page, values wer 50,51...... (even,odd) pair..... which is not possible.... as page nos. of books are (odd,even) pairs......
then same u needed 2 prove fr 5 pages.......
hence only 3 pages satisfy every value........
hence, 3 pages wer torn........

my solution to 4

we are required to find sum of all 3 digit nos. - (sum of all numbers with only even digits + sum of numbers with odd digits)
let us calculate the sum of 3 digit numbers with all even digits.

sum of even digit numbers between 200-300
200+220+240+260+280=1200\\ 202+222+242+262+282=1210\\ 204+224+244+264+284=1220\\ 206+226+246+266+286=1230\\ 208+228+248+268+288=1240
so sum = 6100

sum of even digit numbers between 400-500 can be obtained by adding 200 to each of the above 25 numbers
so sum = 6100+25 \times200=11100
sum of even digit numbers between 600-700 =11100 +25\times200=16100
sum of even digit numbers between 800-900=16100 +25\times200=21100
so sum of even digit numbers = 54400

let us calculate the sum of 3 digit numbers with all odd digits.

sum of odd digit numbers between 100-200
111+131+151+171+191=755\\ 113+133+153+173+193=765\\ 115+135+155+175+195=775\\ 117+137+157+177+197=785\\ 119+139+159+179+199=795
so sum = 3875

sum of odd digit numbers between 300-400=
3875+25 \times 200=8875
sum of odd digit numbers between 500-600 =8875+25 \times 200=13875
sum of odd digit numbers between 700-800 =13875+25 \times 200=18875
sum of odd digit numbers between 900-1000 =18875+25 \times 200=23875
so sum of odd digit numbers = 65500

sum of all numbers with only even digits + sum of numbers with odd digits= 119900
total sum of 3 digit numbers= 449550
so sum of required numbers= 449550-119900=329650

i think there is a calculation mistake but the method is the same