@TootaHuaDil said:12341234..........upto 400 digits.......find the remaindr when it is divided by 101
101 prime ==> any digit repeated (p-1) multiple times ==> rem =1 ..so 1??
@TootaHuaDil said:12341234..........upto 400 digits.......find the remaindr when it is divided by 101
79?
@fireatwill said:if 1st january 2000 is a monday,then how many times between 2001 and 2020 will be the first day of a year fall on monday ?
3 times ?
@TootaHuaDil said:Easy one. Time pass.10000! = (100!)^k , where K is integer. What can be the maximum value of K?
Highest power of 97 in 10000! i.e. 103 ?
@catahead said:30% of the men are more than 25 years old and 80% of the men are less than or equal to 50 years old. 20% of all men play football. If 20% of the men above the age of 50 play football, what percentage of the football players are less than or equal to 50 years?(a) 15% (b) 20% (c) 80% (d) 70%
16/20 => 80% ?
Divisibility Rules. Should be useful for CAT 13 aspirants :)
This is one of the important concept of number system .Questions on remainders and factor are quite frequent in CAT these days . Lets see the divisibility rule of few numbers.
We all perhaps know what is the divisibility rule of 2,4,6,8,3,9 .But do everyone know the logic behind it ?
First case :
Let me start from divisibility of 2
If last digit is divisible by 2 , then it is divisible by 2 ...
why ?
Last digit is nothing but divide by 10 ...10 is perfectly divisible by 2..
Now we can extend this for 4 ..
lets take last digit --> divide by 10 --> 10 /4 --> Not perfectly divisible
Now move to next digit--> divide by 100
100 is perfectly divisible by 4 .. So divisibility rule of 4 is last two digits to be divisible by 4
Now this goes on ... jus try for various numbers like 8 ,16,32
the above logic works for factor of 10^n like powers of 2 and 5 . Other numbers can't be treated this way
Second case
Lets take an example of next number .. take 3
10/3 --> not divisible
100/3 --> not divisible
1000/3 --> not divisible and it goes on ..
But in all three cases , the remainder is 1 .
If remainder is 1 , add the digits
Divisibility rule of 3 will add the digits and then check if its divisible by 3 . This is applicable for numbers which leaves remainder 1 when 10^n is divided by that number
Quick divisibility rule :
Take 33 :
100/33 leaves remainder 1 . So one can frame divisibility rule of 33 as sum of digits taken two at a time from the right hand side of number .
Eg . Is 2112 divisible by 33 ?
12 + 21 = 33 . 33/33 perfectly divisible . Yes .. isn €™t that easy ?
Third case :
Lets take an example of next number .. take 7
10/7 --> not divisible but remainder is not 0 or 1 or -1
100/7 --> not divisible but remainder is not 0 or 1 or -1
1000/7 --> not divisible but remainder is -1
so every three digit we will have -1 remainder ..
if 10^3 has -1 remainder , 10^6 has 1 as remainder...
so its alternative in nature for every three digits . Divisibility rule of 7 will difference of sum of the digits taken at three at a time alternatively from RHS of the number and then check if this resultant divisible by 7
say number is 123130 to check for divisibility by 7 .,..
so according to this concept we have , 130 - 123= 7 which is divisible by 7
Or let €™s say 7123130 -- > (130 + 7) €“ 123 = 14 which is divisible by 7
The beauty of this method is it works well for big number divisors (for instance- 143) .
235378/143 .. is it divisible ?
10^3 / 143 = remainder is -1
By this method ,we can say 378 - 235 = 143.. 143 / 143 is perfectly divisible .That's all you r done
Applying this formula , we can test divisibility of 2,3,4,5,7,8,9,11,27,13,33,37,99,77,143,259(37*7) etc ....
Just work on these lines .. You can easily find the divisibility of all above numbers
Ps : Feel free to revert me back in comments if you have any doubts .
@TootaHuaDil is it 0?
rem[1234/101]=22 for one set of 1234
rem[12341234/101]=44 for 2 sets of 1234
rem[123412341234/101]=66 and so on...
so the prb is: find rem [22(100)/101]=-22=79
@adityaknsit said:Please explain the approach
Read the post just above. You should be able to solve on your own :)
@techgeek2050 said:104 ?my bad... it's 97
Easy one. Time pass.10000! = (100!)^k , where K is integer. What can be the maximum value of k?
Is this Ques right..I mean there are no prime numbers such as 101, 103..(Prime>100) on RHS to neutralize them on LHS?? I think K will hold no value at all..or else the ques is Incomplete..or i have delve 2 mch into it..
@pyashraj said:@TootaHuaDilEasy one. Time pass.10000! = (100!)^k , where K is integer. What can be the maximum value of k?Is this Ques right..I mean there are no prime numbers such as 101, 103..(Prime>100) on RHS to neutralize them on LHS?? I think K will hold no value at all..or else the ques is Incomplete..
it has to be lik 10000! (100!)^k * n ( i guess ) where n is not a prime !!
@gs4890 bro ,if we check 1 january 2000 was a monday
1 jan 2001 will be wednesday
1 jan 2005 will be monday
1 jan 2008 will be thursday
1 jan 2009 will be saturday
1 jan 2011 will be monday
1 jan 2015 will be saturday
1 jan 2016 will be sunday
1 jan 2017 will be tuesday
so the answer can be two and if it is three how is that so?
1 jan 2001 will be wednesday
1 jan 2005 will be monday
1 jan 2008 will be thursday
1 jan 2009 will be saturday
1 jan 2011 will be monday
1 jan 2015 will be saturday
1 jan 2016 will be sunday
1 jan 2017 will be tuesday
so the answer can be two and if it is three how is that so?
@pyashraj said:@TootaHuaDilEasy one. Time pass.10000! = (100!)^k , where K is integer. What can be the maximum value of k?Is this Ques right..I mean there are no prime numbers such as 101, 103..(Prime>100) on RHS to neutralize them on LHS?? I think K will hold no value at all..or else the ques is Incomplete..or i have delve 2 mch into it..
Touche! 
i agree with @ravi.theja
i agree with @ravi.theja @TootaHuaDil said:12341234..........upto 400 digits.......find the remaindr when it is divided by 101
79..............
@pyashraj said:@TootaHuaDilEasy one. Time pass.10000! = (100!)^k , where K is integer. What can be the maximum value of k?Is this Ques right..I mean there are no prime numbers such as 101, 103..(Prime>100) on RHS to neutralize them on LHS?? I think K will hold no value at all..or else the ques is Incomplete..or i have delve 2 mch into it..
The actual question is
10000!=C*(100!)^k
where C is a constant. Find the maximum value of k?
10000!=C*(100!)^k
where C is a constant. Find the maximum value of k?
@bs0409 said:The actual question is10000!=C*(100!)^kwhere C is a constant. Find the maximum value of k?
104??
@anitito123 said:If r is the remainder when each of 1059,1417,2312 is divided by 'k' ,where k is an integer greater than 1,then the value of 'r+k' is :1.164 2.179 3.343 4. 431Please tell approach.
343 ...
@bs0409 said:The actual question is10000!=C*(100!)^kwhere C is a constant. Find the maximum value of k?
Its 103.....had to check the 2s too...
