[shashankmishra]

Originally Posted by bhupi3149

If the numbers 2^218 and 3^109 are both evaluated, converted into base 12 and written one beside another, how many digits are there, on the whole,in the result?
a)109
b)110
c)218
d)219
e)111

No. of digits in 2^218 in base 12 given by........

x = log(2^218 ) (base 12) + 1

For 3^109

y = log(3^109) (base 12) + 1

Total Digits = log(12^109) (base 12) + 2

So Ans 111

Edited :

Mee too now confused........
As first they are converted and then written then it must be (109 + 2)..........
May be I am wrong.........

Got it Ans must be 110 because of the use of floor function..........

[bhupi3149]

Originally Posted by jaidev_iyer

log(12) 12^109 +1 = 110

You are rite man .... as always.


Can you post the approach ... i mean for such kind of questions ...

Thanks in advance

[shashankmishra]

Originally Posted by Vineetverma

TRY this;

There are 4 identical dice which are similar in all respects. All 4 identical dice are rolled together.In how many different ways are outcomes possible?

1) 126 2) 6^4 3) 4^6 4) 1296 5) None of these


, ,

i have used these smilies to tell u my answere is it correct... :P


EDIT :

2)N is a natural number such that 21^21^21^21 (21 times). Find the remainder when N is divided by 25.
1) 1 2) 17 3) 19 4) 16 5) 21

(2) Given 21^21^21...(21times.........) mod 25

E(25) = 20

So we check
21^21^21...(20 times) mod 20 = 1

So 21^21^21........(21times) = 21^(20k + 1)

So

21^(20k + 1) mod 25 = 21

Whats the OA..........??

[jaidev_iyer]

Originally Posted by shashankmishra

No. of digits in 2^218 in base 12 given by........

x = log(2^218 ) (base 12) + 1

For 3^109

y = log(3^109) (base 12) + 1

Total Digits = log(12^109) (base 12) + 2

So Ans 111

Originally Posted by bhupi3149

You are rite man .... as always.


Can you post the approach ... i mean for such kind of questions ...

Thanks in advance

But, am lil confused after seeing Shashank's post.

[Vineetverma]

Originally Posted by shashankmishra

(2) Given 21^21^21...(21times.........) mod 25

E(25) = 20

So we check
21^21^21...(20 times) mod 20 = 1

So 21^21^21........(21times) = 21^(20k + 1)

So

21^(20k + 1) mod 25 = 21

Whats the OA..........??

srry bro no answers for the first one!!!

YOUR second ans is correct

I am getting 126 for the first one (as indicated by the smilies )

i used 6c4 + 6c1 x 5c1 .......
NOT sure though

edit : coz i felt like editing

[archprateek]

I guess ur first 1 is correct .. 2nd 1 shuld be option 5) 21

[jaidev_iyer]

Originally Posted by bhupi3149

You are rite man .... as always.


Can you post the approach ... i mean for such kind of questions ...

Thanks in advance

No rocket science here buddy,
To find the number of digits in x^y in base "m".
Just find out log(m)(x^y) and add one to it.

P.S.You wudnt even need to practice it.
<You might want to find out why this is happening>

[shyam.nair]

Originally Posted by Vineetverma

TRY this;

There are 4 identical dice which are similar in all respects. All 4 identical dice are rolled together.In how many different ways are outcomes possible?

1) 126 2) 6^4 3) 4^6 4) 1296 5) None of these


, ,

i have used these smilies to tell u my answere is it correct... :P


EDIT :

2)N is a natural number such that 21^21^21^21 (21 times). Find the remainder when N is divided by 25.
1) 1 2) 17 3) 19 4) 16 5) 21

attempted the second one
21^21^21^21 (21 times)
=21^21(euler of 25 is 20..so remainder is one in every case)
=21mod25
=21..

[uhhh...]

Originally Posted by bhupi3149

If the numbers 2^218 and 3^109 are both evaluated, converted into base 12 and written one beside another, how many digits are there, on the whole,in the result?
a)109
b)110
c)218
d)219
e)111

[log(base12)(4^109)] + [log(base12)(4^109)] + 2

where [.] greatest integer function

[log(base12)(4^109)] + [log(base12)(3^109)] <= log(base12)(12^109) = 109

log(base12)(4^109) = i1 + f1 (i1 = integer, 0<f1<1)
log(base12)(3^109) = i2 + f2 (i2 = integer, 0<f2<1).. we know both are not integers

so i1 + i2 + f1 + f2 = 109 an integer

also 0 < f1 + f2 < 2.. so f1+ f2 =1 is the only possible value

so [log(base12)(4^109)] + [log(base12)(3^109)] = i1+i2 = 108

so number of digits = 108+2 = 110

[uhhh...]

Originally Posted by Vineetverma

TRY this;

There are 4 identical dice which are similar in all respects. All 4 identical dice are rolled together.In how many different ways are outcomes possible?

1) 126 2) 6^4 3) 4^6 4) 1296 5) None of these


, ,

i have used these smilies to tell u my answere is it correct... :P

same as number of whole no solutions to

x1 + x2 + .. +x6 = 4 ... 9C5