@techgeek2050 said:The number of divisors of the form 4n + 1 , n>=0 , of the number (10^10)*(11^11)*(13^13) is .... ?approach plz.
is it 1792 ?
@techgeek2050 said:The number of divisors of the form 4n + 1 , n>=0 , of the number (10^10)*(11^11)*(13^13) is .... ?approach plz.
ans 924?
@techgeek2050 said:The number of divisors of the form 4n + 1 , n>=0 , of the number (10^10)*(11^11)*(13^13) is .... ?approach plz.
We just have to consider (5^10)(11^11)(13^13)
5 and 13 are of form (4n + 1) but 11 is of form (4n + 3)
Now, any no of form (4n + 1) can have any exponent of a prime factor of form (4k + 1) and only even exponent of a prime factor of form (4k + 3)
So, no of such factors = 11*6*14 = 924 factors
5 and 13 are of form (4n + 1) but 11 is of form (4n + 3)
Now, any no of form (4n + 1) can have any exponent of a prime factor of form (4k + 1) and only even exponent of a prime factor of form (4k + 3)
So, no of such factors = 11*6*14 = 924 factors
@techgeek2050 said:The number of divisors of the form 4n + 1 , n>=0 , of the number (10^10)*(11^11)*(13^13) is .... ?approach plz.
will be 924
4n+1 can be 5,13,121
from given question we hv,
5^10*121^5*13^13
=> 11*6*14
=>924
@Asfakul said:Find the largest value of y/x if x ˛ + y ˛ - 2x + 4y - 19 = 0.Dont have OA please discuss .
i guess isko aise solve karte hai (x-1)^2*(y+2)^2=24
So now to find the maximum of y/x denomminator must be as low as possible and numerator should be max possible. For x(min) clearly you can see x has to be 2. if you take 1,0,-1 or any other value either the multiplication or the fraction doesnt yield the required result. Now when x=2 the x wala part of the multiplication is 1. so y+2=rt(24)=2rt(6)=>y=2(rt(6)-1)
so maximum of y/x=rt(6)-1. Please some1 check my approach and tell me if this is correct
So now to find the maximum of y/x denomminator must be as low as possible and numerator should be max possible. For x(min) clearly you can see x has to be 2. if you take 1,0,-1 or any other value either the multiplication or the fraction doesnt yield the required result. Now when x=2 the x wala part of the multiplication is 1. so y+2=rt(24)=2rt(6)=>y=2(rt(6)-1)
so maximum of y/x=rt(6)-1. Please some1 check my approach and tell me if this is correct
@chillfactor said:We just have to consider (5^10)(11^11)(13^13)5 and 13 are of form (4n + 1) but 11 is of form (4n + 3)Now, any no of form (4n + 1) can have any exponent of a prime factor of form (4k + 1) and only even exponent of a prime factor of form (4k + 3)So, no of such factors = 11*6*14 = 924 factors
sir yeh thoda detail me explain kar do.. have no idea of the logic used here.
@chillfactor said:We just have to consider (5^10)(11^11)(13^13)5 and 13 are of form (4n + 1) but 11 is of form (4n + 3)Now, any no of form (4n + 1) can have any exponent of a prime factor of form (4k + 1) and only even exponent of a prime factor of form (4k + 3)So, no of such factors = 11*6*14 = 924 factors
sir , i just have a question
if all the co-prime factors (like 5, 11, 13 here) are in some form , say 4k+1, then will all divisors be in 4k+1 form?
for example, if n = (7^2)*(11^2)
7,11 are of 4k+3
but not all factors are of form 4k+3.
@techgeek2050 said:The number of divisors of the form 4n + 1 , n>=0 , of the number (10^10)*(11^11)*(13^13) is .... ?approach plz.
if u want 4n+1 numbers then u dont need 2^k as 2n can never yield a rem of 1 with 4
now left numbers are 5^10*11^11*13^13
rem of 5 by 4 is 1
rem of 13 with 4 is 1
rem of 11 with 4 is -1 so u need all even powers of 11
so 10c1+5c1+13c1+50+65+130+650+1 = 924
@iLoveTorres said:i guess isko aise solve karte hai (x-1)^2*(y+2)^2=24So now to find the maximum of y/x denomminator must be as low as possible and numerator should be max possible. For x(min) clearly you can see x has to be 2. if you take 1,0,-1 or any other value either the multiplication or the fraction doesnt yield the required result. Now when x=2 the x wala part of the multiplication is 1. so y+2=rt(24)=2rt(6)=>y=2(rt(6)-1)so maximum of y/x=rt(6)-1. Please some1 check my approach and tell me if this is correct
bhai multiplication nai hoga equation me.
it's a circle with center (1,-2)
@iLoveTorres said:sir yeh thoda detail me explain kar do.. have no idea of the logic used here.
All the prime factors can be categorized as:-
i) of form 2k
i) of form 2k
ii) of form 4k + 1
iii) of form 4k + 3
Now, (4k + 1)^n will always be of form 4p + 1
But in case of (4k + 3)^n, its not so.
(4k + 3)^(2n) is always of form 4p + 1
and (4k + 3)^(2n + 1) is always of form 4p + 3
and (4k + 3)^(2n + 1) is always of form 4p + 3
Now, in the question asked our number is (10^10)(11^11)(13^13), clearly we don;t need any even prime factor other wise number will be of form 2k
So, we need to consider only 5, 11 and 13. As discussed earlier exponent of 5 and 11 can be anything as they are of form (4k + 1) and for 11 exponent should be even. So, for 11 possible exponents are 0, 2, 4, 6, 8, 10
Thats why 11*6*14 such factors
You can check this post also, may be it will help:-
http://pagalguy.com/forums/quantitative-ability-and-di/cat-2010-concepts-fundas-tips-crack-quants-section-t-55747/p-2191164/r-2344929
The latest registration no issued by registration authority is DL-5S-2234. If all the numbers and alphabets before this have been used up, then find how many vehicles have a registration number starting with DL-5??
192234
192225
172227
NOT
@Shray14 said:The latest registration no issued by registration authority is DL-5S-2234. If all the numbers and alphabets before this have been used up, then find how many vehicles have a registration number starting with DL-5??192234192225172227NOT
is it 182235??
@Shray14 said:The latest registration no issued by registration authority is DL-5S-2234. If all the numbers and alphabets before this have been used up, then find how many vehicles have a registration number starting with DL-5??192234192225172227NOT
B 192225
@meow14 said:In how many ways can 6 identical rings be worn in 4 fingers of one hand if any no. of rings can be worn in one finger??a.6^4 b.4^6 c.4!*6! d.84 e.196
84 . Number of ways of dividing n identical things among r persons is n+r-1Cr-1
@Dexian
@Shray14 said:The latest registration no issued by registration authority is DL-5S-2234. If all the numbers and alphabets before this have been used up, then find how many vehicles have a registration number starting with DL-5??192234192225172227NOT
B 192225
plz explain ur approach??
B 192225
plz explain ur approach??
@simplydevesh said:@Dexian@Shray14 said:The latest registration no issued by registration authority is DL-5S-2234. If all the numbers and alphabets before this have been used up, then find how many vehicles have a registration number starting with DL-5??192234192225172227NOTB 192225plz explain ur approach??
S 19th alphabet
9999*18+2233
9999*18+2233