P is the product of first 30 multiples of 30. N is the total number of factors of P. In how many ways N can be written as the product of two natural numbers such that the HCF of these two natural numbers is 19?
(a)3
(b)4
(c)5
(d)6
i think b)4...
P is the product of first 30 multiples of 30. N is the total number of factors of P. In how many ways N can be written as the product of two natural numbers such that the HCF of these two natural numbers is 19?
(a)3
(b)4
(c)5
(d)6
i think b)4...
ya ans is 4 ......but will u plz explain it....
128^1000=16384^500\153=13^500\153 if you go further you will get a loop
103^125\153=103*52^62\153=again you get 2704\153 reminder will be 103 odd time loop will end on 52 so the ans.will be 52.
note:-use reminder theorem a^r\p=b^r\p where b=reminder of a\p
Originally Posted by prathi s View Post
find the remainder when (38^16!)^1777 is divided by 17
a. 1
b. 16
c. 8
d. 13
I think d ans is 1....
^1777/17=4^16!^1777/17=(17-1)^16!^1777=(-1)^16!^1777=1.....how...???
let's see...if we take ,(-1)^16!^1777=x =>(-1)^16!=x^1/1777 =>1=x^1/1777 =>x=1...also...
wen we r supposed 2 find unit's digit of suppose..33^77^88^99..we use reverse direction to find unit's digit..as...we find unit's digit of 88^99..den dat we put in 77^()...
i hope m correct.... :)
bhavna202 Saysya ans is 4 ......but will u plz explain it....
first 30 multi.of 30 are 30*1,30*2....30*30 so p=30^30*30!=2^56*3^44*5^37*7^4*11^2*13^2*17^1...*29^1 so N will be 57*45*38*5*144
or 2^4*3^5*5^2*19^2 so total 4 pairs (19,19*rest),(19*3^5,19*rest),(19*2^4,19*rest),(19*5^2,19*rest).
bhavna202 Saysya ans is 4 ......but will u plz explain it....
p=30^30*30!=2^56*3^44*5^37*7^4....29^1
n=57*45*38*5*3^3*2^4=2^4*3^5*5^2*19^2
total pairs (19,19*rest),(19*2^4,19*rest),(19*3*5,19*rest),(19*5^2,19*rest).
...........................error............................
Need your help puys!
I am looking for the post where a detailed tutorial on Number System fundas were posted.
Plz help me to find it 
Need your help puys!
I am looking for the post where a detailed tutorial on Number System fundas were posted.
Plz help me to find it
hi,
I would recommend totalgadha notes. ping me your mail id if you do not have it . i will send you ... its good..
we have to find HCF of the nos
lets try for
(a^2-1) & (a^3-1)
we get HCF as (a^1-1)
also for 2 & 3 we have HCF 1
try for
(a^2-1) & (a^4-1)
we get HCF as (a^2-1)
also for 2 & 4 we have HCF 2
for
(a^8-1)= (a^4-1)*(a^4+1)
= (a^2-1)*(a^2+1)*(a^4+1)
& (a^6-1)==(a^2-1)(a^4+a^2+1)....apply (a^3-b^3) formula
we get HCF as (a^2-1)
also for 8 & 6 we have HCF 2
so we conclude this method of taking HCF of powers is applicable for (a^n-1) as 1 can take any power and can be used for expansion using (a^2-b^2) or (a^3-b^3) or (any other) formulae to find HCF of the terms
ohh...!! got it..!! Thanx for that good explanation...!!! its kind of mathematical induction we gotta work with... in exam, we can take small numbers like (2^6-1) and (2^4-1) and observe the behaviour of HCF in order to work it out quickly... i guess my shortcut out of ur xplaination goes well...??
thanx again..!!
😁
Need your help puys!
I am looking for the post where a detailed tutorial on Number System fundas were posted.
Plz help me to find it
here:
http://www.pagalguy.com/discussions/conceptstotal-fundas-25023536
Please tell the approach for these kind of problems
Remainder for (2^164)/164 is ??
I have one :
Find the 7th digit from the right for 25! ?
I have one :
Find the 7th digit from the right for 25! ?
i think its a tedious one... coz u hav 2 find num of 2's, 3's, 4's, and other prime factors of 25(upto 23) and find the unit's digit using cyclicity... this is becoz 25! will contain 6 0's...and yeah dun frget 2 exclude 6 5's and 6 2's frm their individual powers...
phew..!!!
if any 1 got better method..plz post...
add all possible pairs
that results in
4*(a+b+c+d+e) = 408 => a+b+c+d+e=102
always lowest sum given is sum of two lightest item and highest sum is sum of two heaviest item
now a+b=35 and d+e=47, this implies c= 20
a+c= 36 => a= 16
hope thats clear
what will be heaviest weight in this case ? 27??
I have one :
Find the 7th digit from the right for 25! ?
Is the ans 2???
Please tell the approach for these kind of problems
Remainder for (2^164)/164 is ??
Use Eulers remainder theorem...
Euler of 164 is 80, so we can write 2^164 as (2^(2*80))*(2^4) which gives 1*16= 16 as remainder
Originally Posted by deepmp View Post
I have one :
Find the 7th digit from the right for 25! ?
its 5......
how????......
no. of zeros=6......(25!/5+25!5^2)
therefore,it will have 5 as its 7th last digit,,since 25 is a multiple of 5.....
@nitzr7-dude can u plz state d euler's theorem elaborately,,i min wid examples n all..i searched in wikipedia,,but cudnt glean nethng from wat was written dre....thanking u in advance 😃
Originally Posted by deepmp View Post
I have one :
Find the 7th digit from the right for 25! ?
its 5......
how????......
no. of zeros=6......(25!/5+25!5^2)
therefore,it will have 5 as its 7th last digit,,since 25 is a multiple of 5.....
yar...bt when u hav taken 6 zeroes, u hav taken into consideration all 5's in 25!, so hw cn u say that the last digit is 5...??i cn giv u an example...the last non zero digit in 15! is 8...
hw will u ans that...??
plz explain...i can only find one lenghthy method fr the prob...