The Official CAT 2007 Quant Thread

[STALWART]

Try this one ...


49 ........... infinity
48^
47^
46^
45^
7^

What wud be the second last digit of the expansion ...... ?

.... Sayonara

[mohit_ranka]

Originally Posted by STALWART

Try this one ...


49 ........... infinity
48^
47^
46^
45^
7^

What wud be the third last digit of the expansion ...... ?

.... Sayonara

dint get the question bro... can you please explani it clearly

[the_phantom]

Originally Posted by STALWART

Try this one ...


49 ........... infinity
48^
47^
46^
45^
7^

What wud be the third last digit of the expansion ...... ?

.... Sayonara

is it 7 bhai..

P.S: hows life and work at eil? long time since i logged onto pg..

[iyervani]

Originally Posted by STALWART

Hi the sum of this series wud be something which will end in digit 1 . Just think about the cyclicity of last digit of multiples of 9 follow .... 9 , 8 , 7 , 6 , ........ , 0 , 9 , 8 ....... again the whole pattern is followed .
Sum of frst 90 terms like this will end up in having last digit " 0 " .
But then 9 are still left which gives 1 as last digit .
==>> 0+1 = 1

So , Answer = 1


...... Sayonara

very late to ask smthing on this ques which has been discussed very very earlier in the thread...and seems quite simple...but genuine doubt....
1 to 1000 -- 9 Nines
1001 to 200 -- 9 Nines
.
.
9000 to 10,000 -- 9 Nines
so 9 Nines in each interval ...and the number of intervals is 10.
So total number of 9's is 90....
Am i missing out smthing?????

[sgx100]

Some questions on graphs :

1. Draw f(x^2) = |x^2-2x|
2. Max (5+x,x^2-4,5-x)

y = f(x) is given then draw the following


3. |y| = f(x)
4. x = f(y)


Cheers
sgx

[STALWART]

@ mohit , phantum and all ......

Hey guys .... sincere apology for that .. actually the last question posted by me was incorrect as the answer was indeterminable . Actually that shud have been this ...



49 ........... infinity
48^
47^
46^
45^
7^

What wud be the second last digit of the expansion ...... ?

here
7 is raised to the power 45 first
then that 45 is raised to the power of 46
then that 46 is raised to the power 47 .... and like this till infinity . Hope it is clear now.

@ Phantum ....Hey life is pretty cool at E.I.L . .. yaar Govt Job ... no deadlines ... no work load .. what else .... but there is no internet too .. thats y I have become irregular over here .

.... Sayonara

[mohit_ranka]

Originally Posted by STALWART

@ mohit , phantum and all ......

Hey guys .... sincere apology for that .. actually the last question posted by me was incorrect as the answer was indeterminable . Actually that shud have been this ...



49 ........... infinity
48^
47^
46^
45^
7^

What wud be the second last digit of the expansion ...... ?

The question is 7 ^ 45 ^ 46 ^ 47 ^ 48 ^ 49 ^ .......

or 7 ^ (45 ^ @#$@%)

now if you notice, any power of 5 results in 25 as last two digits...

so, it becomes 7^25

Now,

7 ^ 1=07
^ 2=49
^ 3=43 (7 ^ 3 = 343)
^ 4=01
^ 5=07

so, 7 has a cyclinicity of 4(= 5-1),for two digits....

so the last two digits of 7^25, will be same as the last two digits of 7^1 ( ^1=25 % 4), that is 07.

Hence the last two digits of the term is 07

and as the question asks for the second last digit, it has to be 0

Hope its correct...

[STALWART]

Originally Posted by mohit_ranka

The question is 7 ^ 45 ^ 46 ^ 47 ^ 48 ^ 49 ^ .......

or 7 ^ (45 ^ @#$@%)

now if you notice, any power of 5 results in 25 as last two digits...

so, it becomes 7^25

Now,

7 ^ 1=07
^ 2=49
^ 3=43 (7 ^ 3 = 343)
^ 4=01
^ 5=07

so, 7 has a cyclinicity of 4(= 5-1),for two digits....

so the last two digits of 7^25, will be same as the last two digits of 7^1 ( ^1=25 % 4), that is 07.

Hence the last two digits of the term is 07

and as the question asks for the second last digit, it has to be 0

Hope its correct...

Ahan .... Rite .. man ... U got the crux of the prob !!


Actuallly .... 45 * 45 ...... any number of times will leave the remainder 1 when divided by 4 .

7^(4k+1)

So now observing the cyclicity in the patterm method of powers of 7 .... we observe that last digit comes out to be 7 while second last comes out to be 0 .

.... Sayonara

[mohit_ranka]

Originally Posted by STALWART

Ahan .... Rite .. man ... U got the crux of the prob !!


Actuallly .... 45 * 45 ...... any number of times will leave the remainder 1 when divided by 4 .

4^(4k+1)

So now observing the cyclicity in the patterm method of powers of 7 .... we observe that last digit comes out to be 7 while second last comes out to be 0 .

.... Sayonara

4 ^ (4k + 1) ?? you sure meant 7 ^ (4k + 1)......


PS - I understand it is a typo (that too if it is), just posted this post, so that others do not get confused...

[jha ji]

Originally Posted by STALWART

@ mohit , phantum and all ......

Hey guys .... sincere apology for that .. actually the last question posted by me was incorrect as the answer was indeterminable . Actually that shud have been this ...



49 ........... infinity
48^
47^
46^
45^
7^

What wud be the second last digit of the expansion ...... ?

here
7 is raised to the power 45 first
then that 45 is raised to the power of 46
then that 46 is raised to the power 47 .... and like this till infinity . Hope it is clear now.

@ Phantum ....Hey life is pretty cool at E.I.L . .. yaar Govt Job ... no deadlines ... no work load .. what else .... but there is no internet too .. thats y I have become irregular over here .

.... Sayonara

i am also getting 7 as the ans