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I hope you know that 5^2n divided by 13 gives remainder 12.
here, 23^15 = 5^30 = 5^2x15 = 5^2n.
Thus the remainder in this case will be 12.
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See... It depends on what you want to do. If you have more interest in ece, which i think is a more popular branch, then you should go for ece, and amritha is a pretty good institute. However, if you like stuff about buildings and construction, then you should opt for nit. It depends on your liking.
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46 ppl qualified from chennai. SOmeone got AIR 26 in chennai.
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Lol.... Either this magazine is real stupid, or this is a typo
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I did the calculations in my mind.. so dont mind my mistakes.
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I will always love my parents. They come before anything else for me.!
Nice article dude
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For all such sums, the answer is 2^n. As the expert said above, you have to put a =b=1.
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we are given the two roots. Since they are the roots of the equation, each of them must satisfy the equation.
Therefore, putting 2 in place of x, we get :
8 + 4q + 22 - p = 0 => 4q-p = -30
Also,
27 + 9q + 33 - p = 0 => 9q-p = -60
solving the two equations in p and q, we get :
q = -6, p = 10.
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How much percent did u get in your boards?
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Just out of curiosity, If we dont even register for this off campus crap, will we be considered for the main counselling?
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This years results in IIT, AIEEE, and bitsat were not very much upto the mark for me. I expected ranks of under 3k. They were below expectations in all the three. I have decided to do iit once more. Can you suggest some books, which explain the basic concepts clearly, and also give training in problems?
For all the three, ie, PCM. I am basically writing the three to improve my ranks.
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I had a rank of 1k in sastra tanjavur. I didnt apply for kumbakonam. I didnt go for the tanjavur counselling. Waiting for others atm.
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Same here mate. Looks like the traffic is too much
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