Hi friends..I'm new on this site..may i knw where i cna get study material for cat prep. here..??
Okay. Am a newbie on this awesome site!I've got a doubt in the following example, which is seemingly easy but I haven't yet understood the concept properly!
If A=1^1*2^2*3^3*.........*100^100,then how many zeroes will be there at the end of A?
Thanks in advance puys!! :cheerio:
i guess it should be 1050. whats the answer dude???
hey puys this thread is really going blank..post some good questions on number system here....
Okay. Am a newbie on this awesome site!
If A=1^1*2^2*3^3*.........*100^100,then how many zeroes will be there at the end of A?
Thanks in advance puys!! :cheerio:
The total Number of 0's would depend on the number of 5's in the multiplication.
The series has 5^5 * 10^10 * 15^15 .... 100^100
5^5 has five 5's
10^10 has 10 5's and so on..
The rule is for every multiple of 5 except 25, 50, 75, 100
25^25 = 5^50 has 50 5's
50^50 = 2^50 * 5^100 has 100 5's and so on..
Total no of 5's = (summation till 20)*5 + 25 + 50 + 75 + 100 = 1050 + 250 = 1300.
Puys help me out with this problem
10000! = (100!)^k * p
Here p and k are integers,what can be the maximum value of k????
The total Number of 0's would depend on the number of 5's in the multiplication.
The series has 5^5 * 10^10 * 15^15 .... 100^100
5^5 has five 5's
10^10 has 10 5's and so on..
The rule is for every multiple of 5 except 25, 50, 75, 100
25^25 = 5^50 has 50 5's
50^50 = 2^50 * 5^100 has 100 5's and so on..
Total no of 5's = (summation till 20)*5 + 25 + 50 + 75 + 100 = 1050 + 250 = 1300.
thanks dude..i ws on the right line calculating the number of 5's..did a mistake in calculating the number of 5's for 25,50,75,100...
Puys help me out with this problem
10000! = (100!)^k * p
Here p and k are integers,what can be the maximum value of k????
The number of 100! in 10000! is given by the number of times the maximum prime number occurs in the same.
The maximum prime number in 100! is 97.
It occurs 103 times in 10000!.(100 times = 9700 and 3 times after that)
Hence the value of k would be 103.
The number of 100! in 10000! is given by the number of times the maximum prime number occurs in the same.
The maximum prime number in 100! is 97.
It occurs 103 times in 10000!.(100 times = 9700 and 3 times after that)
Hence the value of k would be 103.
Exponent of 97 in 10000! is 104, as 97 has two 97's
But if we check for the exponent of 2, then it comes out to be 9995 in 10000! and 97 in 100!
= 103
So, answer will be 103
What is the largest positive integer 'n' for which there is a unique integer 'k' such that 8/15 a) 49
b) 56
c) 98
d) 112
e) 168
Exponent of 97 in 10000! is 104, as 97 has two 97's
But if we check for the exponent of 2, then it comes out to be 9995 in 10000! and 97 in 100!
= 103
So, answer will be 103
What is the largest positive integer 'n' for which there is a unique integer 'k' such that 8/15 a) 49
b) 56
c) 98
d) 112
e) 168
is it 112 ??
take 15*13 = 195 as common denominator
for 112 make denom as 210,209,208
104
pl. cnfm
is it 112 ??
take 15*13 = 195 as common denominator
for 112 make denom as 210,209,208
104
pl. cnfm
I second it.
When denominator is 208 the fraction is equal to 7/13 and when 210 it is 8/15. From now on for any values of n(113 to 16
there would two possible values of n+kYup, 112 is correct.
8/15 => 13/7 => 6/7
Now, LCM of 7 and 8 is 56
=> 48/56 => 96/112
Here we can see that k can be only 97, now of we increase n then k will have 2 values.
So, 112 is maximum possible value of n for which k will have a unique value
Find the remainder when x^81 + x^49 + x^25 + x^9 + x is divided by x^3 - x
yes 103 is the correct answer.. but i am not clear with the solution as 97 occurs 104 times in 10000! and why we need to find out the exponent of 2 in 10000!... Please explain the answer
@chillfactor: yes 103 is the correct answer.. but i am not clear with the solution as 97 occurs 104 times in 10000! and why we need to find out the exponent of 2 in 10000!... Please explain the answer
Try to realize one thing in 100! we have 25 prime numbers so we need to write it in that form and we need to maximize the power of 100! in 10000!. Now if we consider the last two primes 97 & 89 we can easily understand that if we have 'k' number of 97s we must have 'k' or more number of 89 or any other prime numbers in that matter. So for the prime numbers lower than 97 & more than 50 there would be more than 'k' multiple of that prime number in 10000!.
Now number of 97 in 10000! is 103+1=104
But for prime numbers less than 50 it occurs more than once in 100!. So we need to see whether 10000! has that many prime numbers or not. We know with decrease in the value of prime number the number of multiples (or power) in 100! with increase so we only try to find number of power of 2(the smallest prime) in 100! which is 50+25+12+6+3+1=97.
So we need to find the number of 2^97 (needed for 100!) in 10000!
Now number of 2 in 10000!=5000+2500+1250+625+312+156+78+39+19+9+4+2+1=9995
Now number of 2^97 in 10000!=9995/97=103 (4 remainder )
So there can be at most 103 number of 100! .
Hence k=103
Thanks for giving a perfect solution....N you saved me from mugging up the things
Try to realize one thing in 100! we have 25 prime numbers so we need to write it in that form and we need to maximize the power of 100! in 10000!. Now if we consider the last two primes 97 & 89 we can easily understand that if we have 'k' number of 97s we must have 'k' or more number of 89 or any other prime numbers in that matter. So for the prime numbers lower than 97 & more than 50 there would be more than 'k' multiple of that prime number in 10000!.
Now number of 97 in 10000! is 103+1=104
But for prime numbers less than 50 it occurs more than once in 100!. So we need to see whether 10000! has that many prime numbers or not. We know with decrease in the value of prime number the number of multiples (or power) in 100! with increase so we only try to find number of power of 2(the smallest prime) in 100! which is 50+25+12+6+3+1=97.
So we need to find the number of 2^97 (needed for 100!) in 10000!
Now number of 2 in 10000!=5000+2500+1250+625+312+156+78+39+19+9+4+2+1=9995
Now number of 2^97 in 10000!=9995/97=103 (4 remainder )
So there can be at most 103 number of 100! .
Hence k=103
good one thanks a lot
A four-digit number in duo-decimal system is also a four-digit number in hexadecimal system. How many such numbers exist?
a)18464
b)17769
c)16640
d)16639
P.S. also explain what is a duo-decimal system.
a)18464
b)17769
c)16640
d)16639
P.S. also explain what is a duo-decimal system.
A four-digit number in duo-decimal system is also a four-digit number in hexadecimal system. How many such numbers exist?
a)18464
b)17769
c)16640
d)16639
P.S. also explain what is a duo-decimal system.
Duo-decimal system is Base 12.
Since the number N is a 4-digit number both in base 12 and base 16, we can say that
16^3 - 1
=> N can take (12^4 - 16^3), i.e, 16640 different values
can someone start a thread for approach concept...
i mean how to approach a problem and what should we think when we approach a particular number system problem...because thatts the most important thing i suppose...its an urgent necessity as it will make our roots go strong and we can tackle any problem?
i mean how to approach a problem and what should we think when we approach a particular number system problem...because thatts the most important thing i suppose...its an urgent necessity as it will make our roots go strong and we can tackle any problem?
plzzz help me......my QA sux........hav always been terrified by QA......... number system particularly................plzz tell me which topics to start with in number systems...
im afraid QA will play spoil sport for me in CAT 2011.
can someone start a thread for approach concept...
i mean how to approach a problem and what should we think when we approach a particular number system problem...because thatts the most important thing i suppose...its an urgent necessity as it will make our roots go strong and we can tackle any problem?
evem iam looking forward to this one..........:lookround::lookround:
E(17) = 16
(38^16!)^1777 mod 17=
38^16! mod 17 = 1
remainder = 1
I didnt get wat is E(17) can u explain the method..:oops: