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11 Jul 2008 13:39:03 IST

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8th-10th grade-Day 1; 10th-12th grade - Day 2; >12th - Day 3

None

1. Find the value of $\log \left ( \sqrt {3 - \sqrt 5} + \sqrt {3+\sqrt 5} \right )$

2. Show that in the decimal expansion of $\left( \sqrt {65} - 8 \right )^{100}$ , no non-zero digit appears until after 121 zeros, including the zero in the ones place.

Now I hope you guys wont wait for an expert to finish the job for you!!

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Srikanth

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11 Jul 2008 14:15:31 IST

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Rahul Duggal

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11 Jul 2008 14:52:26 IST

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Re:8th-10th grade-Day 1; 10th-12th grade - Day 2; >12th - Day 3

Srikanth

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11 Jul 2008 14:59:14 IST

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we can write the given equation as log(a+b)

= log{a(b/a + 1)}

= log a + log (b/a + 1)

putting values of a and b, and solving, we get

0.719/2 + 0.1405

= 0.5

(Sorry, but i am not able to write the equation...something wrong with my browser, i think)

SUNDEEP ALLAMRAJU

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11 Jul 2008 16:01:22 IST

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1)Now that the answer is out,Here's one more approach.

We have,6-2rt5=(rt5-1)² and 6+2rt5=(rt5+1)².

So,

Thus,the answer is (1/2)log10=1/2[If it is to base 10].

Hari Shankar

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11 Jul 2008 17:42:59 IST

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way to go rahul!

man, ur image took hajaar time to load. Why dont you try latexing?

Srikanth

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11 Jul 2008 17:51:35 IST

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sir, is my method correct...because i tried to find an alternate method...though i ended up getting the longest one!

Hari Shankar

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11 Jul 2008 18:29:05 IST

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norton, in such problems they expect that you will not use the log tables. The use of conjugate surds should motivate you to find a way to multiply them or add them and Rahul killed two birds with one stone by squaring.

®µD®A

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11 Jul 2008 19:15:53 IST

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Sorry i couldn't post the answer because I was in school and was followed by tuiton.

$Let (\sqrt{3-\sqrt{5}}+\sqrt{3+\sqrt{5}})=x\\\\\Rightarrow x=10^\frac{1}{2}\\\\\Rightarrow \log x=\log 10^\frac{1}{2}\\\\\Rightarrow \log \left ( \sqrt {3 - \sqrt 5} + \sqrt {3+\sqrt 5} \right )=\frac{1}{2}\ or\ 0.5$

Hari Shankar

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11 Jul 2008 19:24:07 IST

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nice solution rudra. That was the appetizer

Now on to the second one, the main course

abhishek sinha

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11 Jul 2008 20:54:04 IST

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may I use the value of log 2 ( to base 10 ) ?

Hari Shankar

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12 Jul 2008 09:25:19 IST

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This still being Day 2, the 8th-12th grade students are allowed to use $\log_{10} 2 = 0.3010$

abhishek sinha

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12 Jul 2008 13:13:55 IST

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Then I am first proving that the given qty is less than 10^(-120 ) ( i.e. have more than or equal to 121 zeros)

Proof :

Let P= $\sqrt(65)-8$

Then the given qty is P^100

Now we see that P^100 = $(\frac{1}{\sqrt(65) +8})^100$ <(1/16)^100 .................(1)( as sqrt(65)>8)

Now given log 2 = 0.3010

so log 16 = 4*0.3010 >1.20

So log (16^100)>120

i.e. 16^100>10^120

or (1/16)^100<10^(-120) from (1) it follows that the given qty is < 10^(-120)

i.e. have more than or equal to 121 zeros .

Now we have to prove that it has less than or equal to 121 zeros .

Then the problem is solved .

Hari Shankar

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12 Jul 2008 13:52:22 IST

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I thank everybody for their active and enthusiastic participation.

Anand Hegde

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12 Jul 2008 18:17:50 IST

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Quote:

This still being Day 2, the 8th-12th grade students are allowed to use

Hari Shankar

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12 Jul 2008 19:42:31 IST

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thanx anand, but in the end it did not serve the purpose

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