IIT JEE , AIEEE Forum - Physics , Mechanics

Deedy_Dash

I've started this topic in the algebra page already and want more answers so am spreading it fast
1. A convex polygon has 80 sides. Inside the polygon 50 points are taken, in such a way that no 3 points are collinear. I tis cut into triangles, such that the vertices of the triangle are just the 80 vertices of the polygon and the 50 points within it.

2.In a class, there were 10 students. Each student is asked to write the sum of the ages of the nine other students, giving 9 unique sums:
82,83,84,86,87,88,90,91 and 92
The 10th sum is equal to one of the other sums. Find the ages of the students.

3. Find the remainder of 2^1990 divided by 1990.

4.Find the remainder when 3^333 is divided by 14.

5. Prove a^n + b^n =c^n is not possible for n>2.

Note: I know the answer to 4 of these problems and not the other, but please show me the procedure to do these sums. ( I know that no one can do all of them)

rini

here is the sol of ques 2...

let the sum of all the ages be S.

and individual ages be x1,x2,x3............x10

now,

S -- x1 = 82

S -- x2 = 83

''

S -- x9 = 92

--------------------

adding,

9S -- ( S -- x10) =783

=> 8S + x10 = 783 ------------(1)

now we know S -- x10 = p(where p is one of the given sums)----------(2)

adding (1) n(2)

9S = 783 + p

=> S = 87 + p/9

hense p must be multiple of 9 and hence 90..

there fore S=97

hense find x1,x2... by using privious equations.

plz plz rate me if u got the solutoin and if u liked the method...

nick

4

See 3^333 can be written as
(3^3)^111
or 27^111
this can be written as

(28-1)^111
now as 28 is divisible by 14 the last term in this expansion that is 1^111 will not be divisible by 14,

hence the remainder will be 1.

kghedriu

answer 4..........

well...i think nick is wrong here......

3^333= 27^111

= (28-1)^111

= (27)*(28^111+28^110+...28^2+28+1)

= 28k + 27

= 28k+14+13

= 14p+13..... (p belongs to N)

hence remainder is 13..

Deedy_Dash

Yes, Mr.kghedriu is right. I rated him.
Sorry, nick.
And to all the others, there are still many left

nitin62225

ans2= 15,14,13,11,10,9,7,6,5,7

soln:

let the anknown sum of 9 ppl =x

add up all the given sum---- it comes out to be 783+x

now the added up quantity is 9(sum of all the ages) i.e 9*integer

so( 783+x)/9 should be an integer and also x belongs to one of other sums so x comes out to be 90

and sum of ages =97

so each individual age can be calculated by subtracting the sum of other nine from the total.

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