[krsh.vik]

Originally Posted by gpta_varun

Some important questions

1) There are 8436 steel balls , each of which have radius 1 , stacked in a pile such that 1 ball is at top , then 3 balls in second layer , then 6 in third , then 10 and so on. Number of layers ??

1
3 = 1+2
6 = 1+2+3 ... and so on ..
Hope this will help ..

[catalyst407]

Originally Posted by sastry751

use binomial expansion u are done but bettr u give the options mate here it is i solved it
write 1001 as 1000+1 and 999 as 1000-1 same with the denominator after expanding we are left wiht only even terms in our num and denom 1000^134+1000^132+ ...+1/100^134+100^132+....+1

now it is gm use a.r^n-1/r-1 tell me wht were the options is this is wht u meant or somthng else

i am dead weak and dead slow in quant

i m stuck with with r1=1000^2 r2=100^2 and a=1 and n=68 i tried to solve it further.but no ans
can u explain it?

[catalyst407]

Originally Posted by lcg
hi suresh how to go for rhis question.........

Find the last three digits of 25^63*63^25?

1)625 2)375 3)675 4)325

Plz give an approach which can be used for all such question.

Thanx in advance.

Originally Posted by krgaur.m

The right answer would be 1) 625, because if u take any power of 25 .. 25^n where n not= 0 and 1 the last three digit will be 625 only. Try multipling on calc to verify

but what about 63^25 how can u assume it as 1.could elaborate on that?
rest, 25^63 will always give 625 is right.

[catalyst407]

Originally Posted by maverick37

we can select 3 points from the heptagon in 7c3 way ..
so the ans will be 35

Had the question been unique triangles of heptagon,what would have been the answered?

just thought aloud.is it 28 ?

[vineet.nitd]

Originally Posted by catalyst407

Originally Posted by lcg
hi suresh how to go for rhis question.........

Find the last three digits of 25^63*63^25?

1)625 2)375 3)675 4)325

Plz give an approach which can be used for all such question.

Thanx in advance.




but what about 63^25 how can u assume it as 1.could elaborate on that?
rest, 25^63 will always give 625 is right.

The correct answer should be last 3 digits as 375 ...

63 = -1(mod 8 ) => 63^25 = -1 (mod 8 )= > 63^25 =8x -1
and 25^63 = 1000y + 625
hence , 63^25 * 25^63 = (8x-1)(1000y + 625) = -625 mod1000 =375mod(1000)

[catalyst407]

Originally Posted by vineet.nitd

The correct answer should be last 3 digits as 375 ...

63 = -1(mod 8 ) => 63^25 = -1 (mod 8 )= > 63^25 =8x -1
and 25^63 = 1000y + 625
hence , 63^25 * 25^63 = (8x-1)(1000y + 625) = -625 mod1000 =375mod(1000)

Nice and logical approach.Specially,i liked the way you came to conclusion of 63^25=8x-1
thanku

[krsh.vik]

Originally Posted by vineet.nitd

The correct answer should be last 3 digits as 375 ...

63 = -1(mod 8 ) => 63^25 = -1 (mod 8 )

Is this because
63 = 8x - 1
63^25 = (8x-1) ^ 25
Hence the step :satisfie:

[clsuresh]

Folks, i am bit busy in developing a website for CAT aspirants where they can take online tests for CAT Quant...........and this is the reason why i stopped posting here....I think from today you can expect an active participation from me......

Folks, first i would like to answer this question

Find the last three digits of 25^63*63^25?

1)625 2)375 3)675 4)325

U can think of a simple logic here
Since he is asking about the last three digits and we know that the divisibility rule of 8 depends on the last three digits use that logic here

I mean the remainder obtained when a number is divided by 8 is same as the remainder obtained when the last three digits of that number is divided by 8.

Using this logic here.....

The remainder obtained when 25^63 * 63^25 is divided by 8 is
1^63 8* (-1)^25 = -1 = 7

so the given number leaves a remainder 7 when divided by 8
So now go to ur options and look at an option that gives remainder 7 when divided by 8

Only 375 is the option that leaves remainder 8 when divided by 8
So the answer must be 375.

In case if there is more than one option satisfying this condition what to do????

i have discusses a lot regarding the calculation of last digit, last two digits, last three digits and so on in the early pages of this thread. Please go through them and if u have any doubts u can contact me


Here comes our very good question.........

1.Find the total number of ways in which a student can answer atleast 3
questions out of 200 questions in an exam if each question consists of 4
multiple options.

Folks, the fundamental logic to answer this question is already discussed somewhere in the first 3 pages of this thread...so go through it and try to answer this.......

Regards

[marijuana_user]

Here comes our very good question.........

1.Find the total number of ways in which a student can answer atleast 3
questions out of 200 questions in an exam if each question consists of 4
multiple options.


TOTAL ques -200

No of possible ways of answering-5 (leave d ques or mark A or mark B or mark C or mark D)

total possible ways- 5^200

for at least 3...
total -( none+ways of answeing exactly 1 ques + ways of answering exactly 2 )

none is being answered - only 1 way..i.e leave all d ques..
Exactly 1.. 4* 200 = 800
Exactly 2 - 4 * 200 + 4*199 (doubtful)

hence final answer 5^200 - 1597...

[vineet.nitd]

Originally Posted by clsuresh

In case if there is more than one option satisfying this condition what to do????

Kindly have look few posts before . i have proved why will it only be 375




regards ,
Vineet