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If 2^29=n (mod 9),then 9-n is the answer.
Noticing that 2^3=-1(mod 9),2^29=(2^3)^9x2^2=(-1)^9x2^2=-4=5 (mod 9).
Therefore the answer is 9-5=4. This can be easily derived using mod. Hope you got it :)

You're welcome :D....but you're 3 yrs elder, so you must refrain from calling me sir, how could you? And this solution is also inspired by an experiment I did with the cyclicity of digital roots of powers of 2. You could check out my easy-peasy problem on that though :)

@Krishna Ar
–
U deserved "sir" :D.....and I solved the problem u talking about....but not with the (mod 9) method
But with cyclic sum of digits of powers of 2

@Poonayu Sharma
–
@Poonayu Sharma - Woah! I'm mind -blown!!!! How did you level up in Algebra and Number Theory so soon? What resources did you use to gain here?

@Krishna Ar
–
I saved the challenging questions for later. ..then I read all the posts or notes related to them..For some.I used Google to search theories related to sum..and then tried to solve them....though I couldn't solve all(surprisingly I was able to solve 3 lvl 5 question) ;P
..I guess you can also do it ...faster than me :D

@Poonayu Sharma
–
What did you use to learn all those algebra question- inequalities , Symmetric bounding,,all that. And sorry, I really dont think I can do it faster :(.

@Krishna Ar
–
Frankly speaking. ...The symmetric bounding question was a hit and trial by me..(I think one should not hesitate to use hit and trial if there are no other means at times.).I got it in the second try..

.whenever u see a problem. ..try to find a note about it .(That's the only thing I used to crack sums..also me and my 2 friends did few sums together)..I remember daniel liu once gave a link to a note which could be used to solve a problem...It was useful then ..If u don't find any such notes...Google it...believe me problems do become easier then..
:D
Hoping ur level soars high :P

@Poonayu Sharma
–
Nope. I do math only as a hobby. I'd wan't to take up medicine as my mother wants me to do it. U wanna go to IIT or somewhere else Mr. Future Cosmologist?

@Krishna Ar
–
I am bad at bio and chemistry right from birth :P..I will do cosmology (wanna go to NASA) whether I get iit or not..currently I'm preparing for iit ...let's see if luck shines upon me and if i get a good rank then ill do iit ...well if not then I'm thinking for PhD in mathematics and then cosmology ( possibly astronaut too) :P

But the way medical ppl will have a tough competition because of u :P

@Poonayu Sharma
–
Nice :). I also loved the way you ended it with a joke :). You want to become an astronaut or cosmologist? I must say there is a great difference between the two! But, whatever it be I'm sure you'd do well in it :P. You are preparing for IIT on your own? I feel ICSE curriculum is much better than CBSE. What do you feel?

@Krishna Ar
–
I want to become both actually :P....I have joined classes here called pace for preparation. .I myself have studied from icse and I feel it's standard is better than other boardr...But while you prepare for iit
..It doesn't matter which board ur from ...All become more or less same

Thanks for ur compliment though ...

And good luck. .. (remember to enjoy life as much as u can now...After 10 ...its hell :( :P)

@Poonayu Sharma
–
YUP. Why do you say life after 10th is hell? In fact, I would consider that heaven. I would have to study only my favorite two subjects. English is passable too. In fact, life now is hell for me!.

@Krishna Ar
–
I used to think the same way bro....You will realise it after u leave ur school and join college or classes ...everything suddenly gets screwed up :P
but if u manage ur time and other things (which I failed to)...u wont have troubles ..
Hope you enjoy it even after school :D

Except there you are given that it's a 9 digit number with distinct digits. The first step in this problem is to show that it is a 9 digit number (using log) and then obviously by the way the problem is stated implies that the digits are distinct

That rises an interesting question. Which powers (or 2) are missing exactly 1 digit (but could have repeats of others)? Unfortunately, I don't think there is an easy way to answer this question.

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This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.

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## Comments

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TopNewestIf 2^29=n (mod 9),then 9-n is the answer. Noticing that 2^3=-1(mod 9),2^29=(2^3)^9x2^2=(-1)^9x2^2=-4=5 (mod 9). Therefore the answer is 9-5=4. This can be easily derived using mod. Hope you got it :)

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Ohh...use of mod didn't strike me :P

I was trying to solve it using logs etc. . :P

Btw...thanks you sir

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You're welcome :D....but you're 3 yrs elder, so you must refrain from calling me sir, how could you? And this solution is also inspired by an experiment I did with the cyclicity of digital roots of powers of 2. You could check out my easy-peasy problem on that though :)

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@Poonayu Sharma - Woah! I'm mind -blown!!!! How did you level up in Algebra and Number Theory so soon? What resources did you use to gain here?

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Next target ....JEE ,CALCULUS AND MECHANICS :P

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.whenever u see a problem. ..try to find a note about it .(That's the only thing I used to crack sums..also me and my 2 friends did few sums together)..I remember daniel liu once gave a link to a note which could be used to solve a problem...It was useful then ..If u don't find any such notes...Google it...believe me problems do become easier then.. :D Hoping ur level soars high :P

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But the way medical ppl will have a tough competition because of u :P

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Thanks for ur compliment though ...

And good luck. .. (remember to enjoy life as much as u can now...After 10 ...its hell :( :P)

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but if u manage ur time and other things (which I failed to)...u wont have troubles .. Hope you enjoy it even after school :D

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Are there other ways too?

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Sorry, I can't fathom any. :(... I'm really bad at intuition :-/

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You can view the question here with a detailed solution discussion.

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Thank you sir :D

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Except there you are given that it's a 9 digit number with distinct digits. The first step in this problem is to show that it is a 9 digit number (using log) and then obviously by the way the problem is stated implies that the digits are distinct

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Yeah I forgot to mention it because I thought that its understood :P

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That rises an interesting question. Which powers (or 2) are missing exactly 1 digit (but could have repeats of others)? Unfortunately, I don't think there is an easy way to answer this question.

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