106Q2. http://targetiit.com/iit-jee-forum/posts/try-this-one-12515.html
Q3. discussed a lot of times (eureka had posted this one i think)
Well ,let's make mincemeats of these --- :)
1 > Do try to find the remainder when 21990 is divided by 1990 .
I have found a way without using binomial but , still it would good to see the other ways too ---- maybe an easier way is there , and I am not sure of my way :)
2 > For each natural number K , let Sk denote the the circle with radius K centimeters and center at the origin O . On the circle Sk , a particle moves K centimeters in the counter - clockwise direction . After completing its motion on Sk , it moves to Sk+1 in the radial direction . The motion of the particle starts at ( K , 0 ) .
The motion of the particle continues in this manner . If the particle , for the first time , crosses the starting position K times on the circle Sn , then locate that circle .
3 > In a certain test , there are n questions . In this test 2n-i students give wrong answers to at least i question ( 1 ≤ i < n ) .
If total number of wrong answers given is 2047 , then find number of questions .
4 > Show that any positive integral power ( greater than 1 ) of a positive integer mr , is the sum of m consecutive odd positive integers . Find the first such integer in terms of m .
5 > Find the number of words that can be made from the letters of the word “ TRIANGLE “ such that the realative order of vowels do not change ?
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3 Answers
106
1@Ashish -- thanks a lot for the links -- by the way you got q.5 wrong -- ans -- 6720
What about this word -- ITRNGLAE -- BY YOUR METHOD ITS OMITTED :o
1seems no one is interested -- well,here are the solutions --- first one first --
We can say that , 2φ(p) = 1 ( mod 199 )
Again , φ( p ) = 198 ( 199 being a prime number )
So 2198 = 1 ( mod 199 )
Hence 2199 = 2 ( mod 199 )
Therefore , 21990 = 210 ( mod 199 )
Now , by cyclicity of the powers of 2 , we can say that 21990 ends with 4 , hence 21990 - 1024 ( i.e. 210 ) has 0 in its unit’s digit .
So we can also say that , 21990 = 1024 ( mod 10 )
Now , g.c.d. ( 199 , 10 ) = 1
So 21990 - 1024 is also divisible by 1990 .
Hence the remainder is 1024 .