# MATH...MATH...Ath...TH..TH...Quantitative Report

Hey Junta!!!
Good Morning!!! Hope u are all doing GREAT...am starting this section as a common thread for my queries in MATH...all others are invited to solve..to help around..to post thier own questions....
Hey Mods..if u think..this mail shld belong somewhere instead of a new topic u are ...
Even I don't know what it means... this question is from WebSprint from CL...
The only further information on this is the following four choices. Let me also know if you make meaning of this:
greater than p/4 and less than 1 greater than 1 and less than 2 ...
Whats 'p' for????? M i missinng sumthing or is it not mentioned???
There is a circle of radius 1 cm. Each member of a sequence' of regular polygons S1(n), n = 4, 5, 6,..., where n is the number of sides of the polygon, is circumscribing the circle; and each member of the sequence of regular polygons S2(n), n = 4, 5, 6,... where n is the number of sides of the po...
hey thanks a lot both of u
Another one...got it from the previous post.
(17^24)%300
= (300-11)^12 % 300
= 11^12 % 300
= 1331^4 % 300
= 131^4 % 300
= 121.
:)
could you explain this plzzz. i cudn't understand...
A little doodling on paper gives the following:
17^24 = 289 ^ 12 = (300-11) ^ 12. This when divided by 300 will give no remainder for the terms having 300. For the term with only powers of 11, we have 11 ^12. Now, 11^4 = 14641 = 14640 +1.
So, we have (14640+1) ^3. The terms with 3 and 2 pow...
Ans - 121
(17^24)%300
= (113^ 8 ) %300
= (169^4)%300
= (61^2)%300
= 121
hi! can anyone give me some method to solve this sum plzzz
find the remainder when (17)^24 is divided by 300
1) 121 2) 168 3)241 4) none of these
(32^32^32)%9
= (5^32^32)%9
= (32^32)%6
= (2^32)%6
= 2*(2^31)%3
= 2*2
= 4.
Thanks buddy
32^32^32/9
(33 - 1)^32^32/9
THE SERIES IF EXPANDED BINOMIALLY WILL GIE THE LAST 2 TERMS AS
-33 + 1
aLL THE TERMS BEFORE THESE 2 TERMS ARE HIGHER POWERS OF 33 AND HENCE DIVISIBLE BY 9.
SO THE REM = (-32)%9 = - 5 OR 9 - 5 = 4.
Hi friends I am new to PG community. Since morning I am trying to figure out this prob.
What will be the remainder (32^32^32)/9
Ans: will tell u later.
can u please explain ow the answer is 26.67% and not 13.33%
yes the answer shd. be 13.33%
no that is the last step
what u have dne is reverted it back to last but one step
i think the last step would be like this
{logm^(n+1)-log^(n-1)}^n/2
u had missed out the power sign what do u say
Hi ...my frnd joined TIME lucknow and i borrowed his notes ..they were really gud ..i can't say the same about other coaching institutes .............a
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Hi ........in case of logn the series is starting from 0 and not from 1 ..so nth term will be n-1.................yup i made a mistake that 2 shud not b there ..thx for pointing out my mistake ..i will rectify it ................
hope it helps
bye tc
using property log a + log b = log(ab)
log m + log(m^2/n) + .....
= log (m^1 / n^0 * m^2 / n^1 * m^3 / n^2 * .................* m^n / m^(n-1))
= log ( m^(n(n+1)/2) / n^(n(n-1)/2) )
= log m^(n(n+1)/2) - log n^(n(n-1)/2)
= n/2 ((n+1) log m - (n-1) log n)