One More..

A convex polyhedron has for its faces 12 squares, 8 regular hexagons, and 6 regular octagons. At each vertex of the polyhedron one square, one hexagon, and one octagon meet. How many segments joining vertices of the polyhedron lie in the interior of the polyhedron rather than along an edge or a face?

Guys

what is 1 + 3/2 + 6/4 + 10/8 + 15/16 + .........infinite terms..

a_tired_soul Says

is the answer 8

Guys

what is 1 + 3/2 + 6/4 + 10/8 + 15/16 + .........infinite terms..

adding one to both sides of the given equation we get

(a+1)(b+1)=3*5^2*7; (b+1)(c+1)=3*7^2; (c+1)(d+1)=3*5*7

since the last 1st & 3rd eqns have one 7 each & middle eqn has 2 7s there

fore each c+1 & b+1 have one factor of seven in them.so c+1=21 or 7

if c+1=21 =>c=20=4*5 & d=4 b=6 a=74(not possible) as 8! does not have factor of 37.

so c+1=7=>c=6; b=20=4*5;d=14=2*7;a=24=8*3

clearly this satisfies all the condition. hence a-d=24-14=10

DISCLAIMER-accept the above solution at your own peril. i m still a novice.

Solve:

For positive integers a,b,c and d, have a product of 8! and satisfy
ab+a+b=524,
bc+b+c=146, and
cd+c+d=104

What is a-d?

hmmm...
achcha luk at this....

131^4%300=(300-169)^4%300.
expansion parr,
odr than last term, all will be div by 300...
so,
-169 remains..
300-169=131.
????
wats rong here?

131^4 %300

Hmm this might not b acceptable but.lets c:

131^2 = 17161 = 300 * 57 + 61

So 131 ^4 %300 = 61 ^2 %300 = 3721 % 300 = 121 :)

1331^4 % 300 is wat we need.
So 1331 = 300*4 +131
So (300*4+131)^4 % 300 = 131 ^ 4 % 300

Got it chiya?

hey ppl i was juss goin thru the thread......
can anyone plz temme how 1331^4 becumz 131^4?

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.

:)

Using Appo. theorum

AB^2+AC^2=2(x^2+BD^2)

Now BD=CD=5

AB^2+AC^2=2(x^2+25)

Using 1 AB=3a AC=4a

25*a^2=2x^2+50

1st option is not enuf


using 2 option also we have 25a^2=100

a=2 ...

Using 2nd option alone it can be solved x=5

(1) DS ( can be answd using one statement alone? either statement? both the statements together? question cannot be answd ?

what is the length of median to BC in triangle ABC ? (given that BC = 10)
I. AB : AC = 3:4
II. triangle ABC is a right angled @ A. )

catasp2005 Says

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 polygon, is inscribed in the circle. Let L1(n) and L2(n) denote the perimeters of the corresponding polygons of S1(n) and S2(n).
Then {L1(13) + 2p}/L2(17) is

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
(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.

As u guys are REAL FAST...am throwing in one more question...ANSWER BOTH PLz....

A money lender lends out Rs. 9 on the condition that the loan is payable in 10 months by 10 equal installments of Re 1 each. Find the rate % SI.

* And its not 10%..the only hint i got is...Dont consider it at a flat period!!

Soln: 26.67%

See that the no. of years n = 10/12

So, PNR/100 = SI
1=(9x(10/12)xr)/100.

Finding, r => 2/15 i.e, 13.333%????

Bhars.

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

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)

hey u guys...can u tell me which class notes are the best for quant...
i have joined IMS and i am doing the BRMs but lotsa my frnds have joined other classes....so if someone could tell me wot notes are the best then i could do those tooo

hi
i have some doubt regarding your solution
first of all if u r taking no. of terms to be n then in case of log m u r taking n but in the case of log n why r u taking as n-1
and in the last step of ur solution if u r taking n/2 as common then i think the final result would be n/2{logm(n+1)-logn(n-1)}
also - sign would be there in between what do u say

hi
i have some doubt regarding your solution
first of all if u r taking no. of terms to be n then in case of log m u r taking n but in the case of log n why r u taking as n-1
and in the last step of ur solution if u r taking n/2 as common then i think the final result would be n/2{logm(n+1)-logn(n-1)}
also - sign would be there in between what do u say

ABC is a quater circle and a circle is inscribed within it.AC=1cm.find the radius of the smaller circle

p.s. sorry it's zipped

we make use of property log(n/m)=logn-logm


Now above series can be written as
Let number of terms be N

(logm + 2 log m +3log m+.......Nlogm ) -(log n + 2 log n ................(N-1)logn)

logm(1+2+3...........N) + logn(1+2................N-1))


=logm.N.(N+1)/2 + logn.(N-1).N./2

=N/2{ logm.(N+1)/2 + logn.(N-1)/2}

1. What is the sum of `n`terms in the series ?
log m + log( m^2/n )+ log( m^3/n^2 )+ log( m^4/n^3 )+........

mETHOD is ok..ans is sqrt(2) - 1..
but tell me how can yu say that the radius of outer circle is 1 nad not the length f the arc AC.
i mean it says AC = 1.
so AC can be the radius or the quarter ccircle both..
isnt it?

pls calirify..

I think u r rt ...

C attached figure

OD=AD=r (radius of smaller circle)

OA=sqrt(2)*r

OA+OF=R(radius of bigger circle)=1


hence R=sqrt(2)*r+r=>r=1/sqrt(2)*r+1

r=sqrt(2)-1

Is the answer (sqrt 2 - 1)???

I just completed the full circle with similar circles in each quarter.

Please see the attached file. My diagram is not to scale.

Radius of outer circle is 1. I made a square of side 2 r (r is radius of smaller circle). So, the diagonal of the square is sqrt (8r).

Hence sqrt(8r^2) + 2r = 2 (diameter of outer circle).

Therfore r = sqrt(2) - 1

Please let me know if I am wrong somewhere.

santosh_pawar Says

i am getting the answer as 10(sqrt 2 - 1)

Is the answer (sqrt 2 - 1)???

I just completed the full circle with similar circles in each quarter.

ABC is a quater circle and a circle is inscribed within it.AC=1cm.find the radius of the smaller circle

p.s. sorry it's zipped

ABC is a quater circle and a circle is inscribed within it.AC=1cm.find the radius of the smaller circle

p.s. sorry it's zipped

hi is it

n/2{(n+1) log m - (n-1) log n}

1. What is the sum of `n`terms in the series ?
log m + log( m^2/n )+ log( m^3/n^2 )+ log( m^4/n^3 )+........

anupam will return Says

Stupid me.. thanx a ton yoss.. :)