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11 August 2012

Some tips to Memorize Fast and Remember More

Saturday, August 11, 2012 How to, Learning No comments

You will be able to remember names, facts, dates, history, story lines, outlines, speeches, and more.

You memorize fast and remember more just follow some tips below.

Connect it - make connections with what you already know to memorize fast.
See it - visualize the concept to memorize fast and remember more. Chant it - (Re) Produce the material out loud to memorize faster.
Write it - reproduce the material in writing to remember
Mock it - use catchy puns and phrases to memorize faster.

IMO problems from 1959 to 2011

Sunday, July 31, 2011 Learning, Mathematical Olympiad, Mathematics 1 comment

IMO problems from 1959 to 2011.

English language

Day 1 problem

Saturday, July 30, 2011 Learning, Mathematics, Maths Daily No comments

How many zero are the end of $P(10000,13)$

Solution

The formula of permutation is $P(n, k)=\frac{n!}{(n-k)!}$ . Then $P(10000, 13)=\frac{10000!}{(9987)!}$
so we'll find zero the end of 10000! and 9987!.

there are $n_{1}=\left [\frac{10000}{5} \right ]=2000$ integers less than $10000$ which are multiples of $5$ ,
and $n_{2}=\left [\frac{10000}{5^2} \right ]=400$ which are multiples of $5^2$ ,
and $n_{3}=\left [\frac{10000}{5^3} \right ]=80$ which are multiples of $5^3$ ,
and $n_{4}=\left [\frac{10000}{5^4} \right ]=16$ which are multiples of $5^4$ ,
and $n_{5}=\left [\frac{10000}{5^5} \right ]=3$ which are multiples of $5^5$ ,

so we get $2000+400+80+16+3=2499$ zero are the end of $10000!$

By the same way we get $2493$ zero are the end of $9987!$

So there are $2499-2493=6$ are the end of $P(10000,13)$ .

Please correct me! by leave your comment . Thanks