When I post a solution of a weekly challenged problem, I can not see my solution in the proper view. I post the solutions after writing them in 'LATEX'. But they come to me in plane text. How can I clarify my fault? Does anybody else also face such a problem or this problem is happening just with me? How should I write the solutions? Can the other users see my solutions in proper view?............. Please help me.....a link of my solution is....https://brilliant.org/assessment/s/algebra-and-number-theory/2937020/ ......you can check one of my written solution in 'LATEX' through this link.

Calvin: Type it as such \( \frac{a}{b} \).

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How can I write a fraction in latex? Is it \frac {a}{b} ? But, this format does not show the fraction a/b. I got this text from the Formatting guide! What is the right format?......please help me.

Calvin: Type it as such \( \frac{a}{b} \).

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As several others have mentioned, you need to place your Latex code within \ ( \ ) (with the spaces removed. So, instead of \frac{a}{b}, it will appear as \( \frac{ a} {b} \).

I'd add a line to your comment and your post, so that you can view how to write it properly, by choosing the edit option to see what I typed.

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Thank you very much. I used \ ( \ ) (with the spaces removed) like ''\(frac{a}{b}\)'' but not like \(\frac{a}{b}\). I shall try this in the solutions. THANKS.

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oh! Very very thanks. I got it. That helped me so much, Mursalin.

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You're welcome!

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Yeah, I submit solutions through that 'Formatting guide'. That 'Formatting guide' is just a guide to write in 'LATEX'. I always follow that. Can you please go through the above link of my solution, to check if there is any problem of my solution writing? I am not understanding why this thing is happening!

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That link's not going to work for everyone (for instance: me). Try copying your solution and paste it as a comment on this thread.

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Recurrence Relation a

{n} - a{n-1}=n(a{n-1} - a{n-2}), a{n-1} - a{n-2}=(n-1)(a{n-2} - a{n-3}), a{n-2} - a{n-3}=(n-2)(a{n-3} - a{n-4}), . . a{3} - a{2}=3(a{2} - a{1})=3(3-1)=3*2, a{2} - a{1}=3-1=2; So, we can easily find that, a{n} - a{n-1} = n!, a{n-1} - a{n-2}=(n-1)!, a{n-2} - a{n-3}=(n-2)!, . . a{3} - a{2}=3!, a{2} - a{1}=2!; By adding these (n-1) equations, all the a{i} 's will be canceled out, where i is a positive integer and 2 \leq i \leq (n-1). So, we shall get, a{n} - a{1} = \displaystyle \sum{j=2}^n j! \Rightarrow a{n}= \displaystyle \sum{j=1}^n j! [where a{1}=1, given]. Note that, 11 divides \displaystyle \sum{j=1}^10 j! and for all positive integer k \geq 11; 11 also divides k!. So, when n \geq 10, a{n} is divisible 11. So, for 10 \leq n \leq 1000, there are 991 such a{n}'s. Now we just need to find, how many a{n}'s are divisible by 11, for 1 \leq n \leq 9. It is very easy to calculate a{n}, where 1 \leq n \leq 9. Then, we shall find that only a{4} and a{8} are divisible by 11. So, the total number of such a_{n} is (991+2) = 993 [answer]Log in to reply

\(2+3\).

Hope this helps!

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Did you check out the

Formatting guide, which is to the left of thePreviewandPostbuttons?Log in to reply