Please discuss the questions
Got the answer of Q3. as 837 Q4. {1,2,4} and Q6. 43
1Don't you think that the paper was too easy. I mean no doubt people might have done even all correct.
The cutoffs gonna be high.
I got
Ans3→657
Ans4→{2,3,6} (1,2,4 is surely wrong 11 +14=54 which can't be k2 for any integer k
Ans6→43
Ans1→ simple use of formula area=12ab sinθ
Ans2→ solving eqn. in a,b,c by cramers rule and bit of determinants at the end
Ans5→quad AEMF actually becomes a rhombus so you automatically prove it
13 is definitely 729 and how can 4 be 1,2,4 1/4+1/2=3/4 is not an integral multiple of 1!
first can be solved by taking ratios of areas of triangles in the hexagon as products of their lengths
even i solved 2 by cramers rule there must be a simpler method
1on dividing into groups of hundreds you'll get 9 each not 11 wich gives 9into9into9=729
lol if ne1 is interested here are those 729 nos
1116, 1124, 1132, 1148, 1156, 1164, 1172, 1188, 1196, 1212, 1228, 1236, 1244, 1252, 1268, 1276, 1284, 1292, 1316, 1324, 1332, 1348, 1356, 1364, 1372, 1388, 1396, 1412, 1428, 1436, 1444, 1452, 1468, 1476, 1484, 1492, 1516, 1524, 1532, 1548, 1556, 1564, 1572, 1588, 1596, 1612, 1628, 1636, 1644, 1652, 1668, 1676, 1684, 1692, 1716, 1724, 1732, 1748, 1756, 1764, 1772, 1788, 1796, 1812, 1828, 1836, 1844, 1852, 1868, 1876, 1884, 1892, 1916, 1924, 1932, 1948, 1956, 1964, 1972, 1988, 1996, 2116, 2124, 2132, 2148, 2156, 2164, 2172, 2188, 2196, 2212, 2228, 2236, 2244, 2252, 2268, 2276, 2284, 2292, 2316, 2324, 2332, 2348, 2356, 2364, 2372, 2388, 2396, 2412, 2428, 2436, 2444, 2452, 2468, 2476, 2484, 2492, 2516, 2524, 2532, 2548, 2556, 2564, 2572, 2588, 2596, 2612, 2628, 2636, 2644, 2652, 2668, 2676, 2684, 2692, 2716, 2724, 2732, 2748, 2756, 2764, 2772, 2788, 2796, 2812, 2828, 2836, 2844, 2852, 2868, 2876, 2884, 2892, 2916, 2924, 2932, 2948, 2956, 2964, 2972, 2988, 2996, 3116, 3124, 3132, 3148, 3156, 3164, 3172, 3188, 3196, 3212, 3228, 3236, 3244, 3252, 3268, 3276, 3284, 3292, 3316, 3324, 3332, 3348, 3356, 3364, 3372, 3388, 3396, 3412, 3428, 3436, 3444, 3452, 3468, 3476, 3484, 3492, 3516, 3524, 3532, 3548, 3556, 3564, 3572, 3588, 3596, 3612, 3628, 3636, 3644, 3652, 3668, 3676, 3684, 3692, 3716, 3724, 3732, 3748, 3756, 3764, 3772, 3788, 3796, 3812, 3828, 3836, 3844, 3852, 3868, 3876, 3884, 3892, 3916, 3924, 3932, 3948, 3956, 3964, 3972, 3988, 3996, 4116, 4124, 4132, 4148, 4156, 4164, 4172, 4188, 4196, 4212, 4228, 4236, 4244, 4252, 4268, 4276, 4284, 4292, 4316, 4324, 4332, 4348, 4356, 4364, 4372, 4388, 4396, 4412, 4428, 4436, 4444, 4452, 4468, 4476, 4484, 4492, 4516, 4524, 4532, 4548, 4556, 4564, 4572, 4588, 4596, 4612, 4628, 4636, 4644, 4652, 4668, 4676, 4684, 4692, 4716, 4724, 4732, 4748, 4756, 4764, 4772, 4788, 4796, 4812, 4828, 4836, 4844, 4852, 4868, 4876, 4884, 4892, 4916, 4924, 4932, 4948, 4956, 4964, 4972, 4988, 4996, 5116, 5124, 5132, 5148, 5156, 5164, 5172, 5188, 5196, 5212, 5228, 5236, 5244, 5252, 5268, 5276, 5284, 5292, 5316, 5324, 5332, 5348, 5356, 5364, 5372, 5388, 5396, 5412, 5428, 5436, 5444, 5452, 5468, 5476, 5484, 5492, 5516, 5524, 5532, 5548, 5556, 5564, 5572, 5588, 5596, 5612, 5628, 5636, 5644, 5652, 5668, 5676, 5684, 5692, 5716, 5724, 5732, 5748, 5756, 5764, 5772, 5788, 5796, 5812, 5828, 5836, 5844, 5852, 5868, 5876, 5884, 5892, 5916, 5924, 5932, 5948, 5956, 5964, 5972, 5988, 5996, 6116, 6124, 6132, 6148, 6156, 6164, 6172, 6188, 6196, 6212, 6228, 6236, 6244, 6252, 6268, 6276, 6284, 6292, 6316, 6324, 6332, 6348, 6356, 6364, 6372, 6388, 6396, 6412, 6428, 6436, 6444, 6452, 6468, 6476, 6484, 6492, 6516, 6524, 6532, 6548, 6556, 6564, 6572, 6588, 6596, 6612, 6628, 6636, 6644, 6652, 6668, 6676, 6684, 6692, 6716, 6724, 6732, 6748, 6756, 6764, 6772, 6788, 6796, 6812, 6828, 6836, 6844, 6852, 6868, 6876, 6884, 6892, 6916, 6924, 6932, 6948, 6956, 6964, 6972, 6988, 6996, 7116, 7124, 7132, 7148, 7156, 7164, 7172, 7188, 7196, 7212, 7228, 7236, 7244, 7252, 7268, 7276, 7284, 7292, 7316, 7324, 7332, 7348, 7356, 7364, 7372, 7388, 7396, 7412, 7428, 7436, 7444, 7452, 7468, 7476, 7484, 7492, 7516, 7524, 7532, 7548, 7556, 7564, 7572, 7588, 7596, 7612, 7628, 7636, 7644, 7652, 7668, 7676, 7684, 7692, 7716, 7724, 7732, 7748, 7756, 7764, 7772, 7788, 7796, 7812, 7828, 7836, 7844, 7852, 7868, 7876, 7884, 7892, 7916, 7924, 7932, 7948, 7956, 7964, 7972, 7988, 7996, 8116, 8124, 8132, 8148, 8156, 8164, 8172, 8188, 8196, 8212, 8228, 8236, 8244, 8252, 8268, 8276, 8284, 8292, 8316, 8324, 8332, 8348, 8356, 8364, 8372, 8388, 8396, 8412, 8428, 8436, 8444, 8452, 8468, 8476, 8484, 8492, 8516, 8524, 8532, 8548, 8556, 8564, 8572, 8588, 8596, 8612, 8628, 8636, 8644, 8652, 8668, 8676, 8684, 8692, 8716, 8724, 8732, 8748, 8756, 8764, 8772, 8788, 8796, 8812, 8828, 8836, 8844, 8852, 8868, 8876, 8884, 8892, 8916, 8924, 8932, 8948, 8956, 8964, 8972, 8988, 8996, 9116, 9124, 9132, 9148, 9156, 9164, 9172, 9188, 9196, 9212, 9228, 9236, 9244, 9252, 9268, 9276, 9284, 9292, 9316, 9324, 9332, 9348, 9356, 9364, 9372, 9388, 9396, 9412, 9428, 9436, 9444, 9452, 9468, 9476, 9484, 9492, 9516, 9524, 9532, 9548, 9556, 9564, 9572, 9588, 9596, 9612, 9628, 9636, 9644, 9652, 9668, 9676, 9684, 9692, 9716, 9724, 9732, 9748, 9756, 9764, 9772, 9788, 9796, 9812, 9828, 9836, 9844, 9852, 9868, 9876, 9884, 9892, 9916, 9924, 9932, 9948, 9956, 9964, 9972, 9988, 9996,
1quad AEMF does not become a rhombus
first i also assumed that by mistake but it is false
AM need not bisect EF
1the more important fact in 5 was to prove that EF bisects AM and is perpendicular to it and it was easy to prove it using properties of triangles
1last sum answer is 43 anyways it has to be a silly mistake so check it sol is dead easy
first take
k^2<n<(K+1)^2
therefore
k<√n<k+1
by the prop of greatest integer func root n has to be k
now take n=k^2+1
root n =k
thereofre a=k+1/k
now for n=k^2+2/k therefore a(n+1) is greater or equal to
we can similarly prove for all numbers except for
n=k^2+2k which is geater than k^2+2k+1
so we have now proved this all numbers except k^2-1 are the naswers
so 1775 is the last answer 44^2-1 and 45^2-1 is greater than 2010 so the answer has to
43 as 0 cant be included this is the most efficient proof
14)
sol simple
c=kab/a+b thereofre just use simple hit and trial to get the answer
1EF has to bisect AM thats the definition of reflection
but AEMF does not need to be a rhombus
what property of triangle have you applied?
1Here is my proof for the 3rd ques
you can make no.s using 9 digits (1-9)
to find nos divisible by 4
there are 18 such possible combinations
like 12, 16, 24 ...
so you have 9*9*18 such nos
and there are 89 pairs of such 3 digit nos
like 112, 128... 992
so there are 9*89 such numbers
required numbers is
9*(9*18 - 89)
=657
well I dont know if the answer is correct but thats how i did it
13 was 729, 4 was 2,3,6, 6 was 43....verify?
much easier than last year this time....silly mistakes in drawing diagrams ruined my proofs
13)
answer too simple
1st we find all numbers between 1000-99999 divisible by 4 and not containg any 0
thats simple the method i did was if we take last digit to be 2
tens digit can be 1 3 5 7 9 for 4 2468
6 13579 for 8 2468 as the last 2 digits has to be divisible by 4 ...... now we find the answer 1458
now the best part
every number n divisible by 4 has to be of form 4(2k) or 4(2k+1) now since half the numbers are of the form 4(2k) and 4(2k+1) (this can be proved easily as there are 2250 numbers divisible by 4 and there are even number of terms with 0 removed so among the removed numbers half will be of the form 8k )
therfore the answer is 729
1i should have gone for the rmo everyhting ould have been relaxed for me and i am easily solving 4 of it
1Answer of Q4 is 1,2,4 only
Have a look
1 + 1/2 = 2+1/2
= 3/2
= 6 * 1/4
That's an integal multiple of the third number.
And varun.tinkle in th third question it isn't given that alpha is root. It is just given that for x = alpha P1 = P2 = P3
1it doesnt matter actually if u just change some concepts and assume it the method will be same
1got 4 correct
2) subtracted the equations in cyclic order to get three equation with alpha as root and applied common root and discriminant......
3)729...... simplest of the paper
4)assumed first multiple to be 1 and second to be 2..... and solved .. got the numbers as 2,3,6......
6)43......
1will they give ne marks if i wrote the first few steps correctly? anyone getting more than 4 correct?
1even I got 4 correct
messed up the 5th and the 3rd one
I hope thats enough for a 12th guy to get through
36I tried only 2 and 5
Since i m in 10th so was not able to face the questions 3, 4 and 6
I m not sure bout question 5 also.....
I did 2 in this way..
P1(x) = ax2 - bx - c;
P2(x) = bx2 - cx - a and,
P3(x) = cx2 - ax - b
Now, for α, P1(α) = P2(α) = P3(α)
Case 1. when P1(α) = P2(α)
then, aα2 - bα - c = bα2 - cα - a
=> α2(a - b) + α (c - b) + (a - c) = 0
=> α2(a - b) + α (c - b) + (a - c) = 0.α2 + 0.α + 0
On comparing the coefficients we get,
a - b = 0 , c - b = 0, a - c = 0
=> a = b , b = c, a = c
=> a = b = c
similarly when P2(α) = P3(α) and P3(α) = P4(α) then,
a = b = c
Hence, a = b = c proved....
5. I proved BC = EF
1guys,
i present here the solution 0f q no 5
please tell if it is right or wrong
we had to prove that m lies on BC
that is easy to prove
if we prove <amb +<amc = 180 we are done
in tri AMB, AMB + BAM + ABM = 180
in tri AMC, AMC + CAM + ACM = 180
adding we get, AMB + AMC + 60 + (180 - 60) = 360
=> AMB + AMC = 180
=> B,C,M ARE COLLINEAR
