Number System

[chetna]

the cat,

here is ur prob

y= x^(-4)+x^2
then wht is the min val of y??

the solution:-
i hope u understood how i got 3*2^(-2/3)
i have differentiated it...n all that stuff what we used to do in 11th n 12th class to find the minima.

In case of considering the integers:-

i simply made the graph...once u make the graph...its very easy to see that u get an integer value only when x=1. or x=-1

all the points near point (1,2) or (-1,2), either the sumation will be less than 2 but not 1...or otherwise more than 2
and the graph can also not have 0 value.

thus u can say that the minimum value is 2.(integer)

[Varun Khullar]

Originally Posted by ernil

err.. what was that again?

1+2+3+4+5+6+7+8+9 =45 .. sum of digits.. .. is divisble by 3 then it is divisible by 3.. repeated dividble by 3 so divisible by 9..
now i get..
37+38+39+40+41+42+43+44 = taking.. only digits its is equal to 1+2+3+4+..
sum of 8 natural numbers.. ..
36 /45... getting 4..
i might be wrong i guess..

[Varun Khullar]

somebody.. solve..

32^32^32^32....till infinity divided by 9. what's the answer..
and .. 31^78/123 find the remainder.. remainder

[rajat_nda]

Originally Posted by Varun Khullar

1+2+3+4+5+6+7+8+9 =45 .. sum of digits.. .. is divisble by 3 then it is divisible by 3.. repeated dividble by 3 so divisible by 9..
now i get..
37+38+39+40+41+42+43+44 = taking.. only digits its is equal to 1+2+3+4+..
sum of 8 natural numbers.. ..
36 /45... getting 4..
i might be wrong i guess..

Please refer to post#995 by @clsuresh

ciao

[dushyant bhatia]

Originally Posted by Varun Khullar

somebody.. solve..

31^78/123 find the remainder.. remainder

123=3*41
31^78 mod 3
= 1^78/3 which leaves the remainder as 1

31^78 mod 41 (eulers no - 40)
= -10^38mod 41
= 100^19 mod 41
=18^19 (100-84)
=18*18^18mod 41
=18*324^9mod 41
=18*(-4)^9mod 41
=18(-4)*16^4mod 41
=-72*256*256 mod 41
=-31*10*10 mod 41
=10*10*10 mod 41
=1000 mod 41
=16

the remianders shall be one of the fol:
16,57,98.

answer is 16 since when divided by 3 gives remainder 1
i hope Im rite..
ne short cuts??
cheerz

[chetna]

31^78/123 find the remainder

yes my answer is also 16 and same method....i m also waiting for any nice shortcut for this

[rajat_nda]

Puy's succour me out...
i did---
31^80=1 mod 123
so, 31^2 x 31^78 = 1 mod 123
by distributive property,
31^2 x 31^78 = (100) x (x) mod 123 = 1 mod 123
or, 100 * (x) = 1 mod 123

after this i've tried my luck..nothin...even knowin d answer 2 b 16,
i can't get d method for gettin 16....

Plz think...coz this seems 2 b the shortest method...if only someone
came up with d next step.....
cmmon ppl.....

[Varun Khullar]

Originally Posted by chetna

31^78/123 find the remainder

yes my answer is also 16 and same method....i m also waiting for any nice shortcut for this

@chetna and the other guys..

think in terms of inverse..
i agree to remainder theorem etc..

we understand.. 31*4 =124 =123+1..

so .. we got.. 31*4 =1mod123
now.. we got 31^-2 which is .. same as 4^2 =16..

[chetna]

Originally Posted by Varun Khullar

we understand.. 31*4 =124 =123+1..

so .. we got.. 31*4 =1mod123
now.. we got 31^-2 which is .. same as 4^2 =16..

wow...great answer varun....good one

[Varun Khullar]

can somebody give me the answer to

32^32^... till infinity.. divided by. 9.. plz..