Plz solve this:
a number 20 is divided into 4 parts that are in a.p auch that the product of the first and fourth is to the product of second and third is 2:3.find the largest number,
how to get answer in smallest possible way?
(a-3d),(a-d),(a+d),(a+3d)
(a^2-9d^2)*3=(a^2-d^2)*2
3a^2-2a^2=27d^2-2d^2
a^2=(5d)^2
Add all the four terms 4a=20a=5
So a=5,d=1
So 2,4,6,8 are the numbers
Answer=8
dont u think....it will be much simpler and faster if u directly solve it thru assumption method
Hi,
Kindly explain the below using euler's number.
1.Remainder when 30^72^87 divided by 11,
plz explain the logic to solve this kind of Q
2. Remainder when 32^32^32 is divided by 7.
Also,pls explain the euler's number and related formulae.
Hi,
Kindly explain the below using euler's number.
1.Remainder when 30^72^87 divided by 11,
plz explain the logic to solve this kind of Q
2. Remainder when 32^32^32 is divided by 7.
Also,pls explain the euler's number and related formulae.
Euler's number for any number N = (a^p)(b^q)(c^r)... , where a, b, c, ... are prime numbers is given by:-
(N) = N(1 - 1/p)(1 - 1/q)(1 - 1/r).....
Euler's number is also important in another way, as it denotes the number of numbers that are co-prime to N and less than N.
We can also notice that if N is a prime number, then
(N) = N(1 - 1/N) = (N - 1)
Now, as per Euler's theorem
{(a^(n)) - 1} is divisible by n, where (n) is Euler's number of n and n is coprime to a.
Since, (n) = (n - 1) for the case when n is prime, we can say that
If {(a^(n - 1)) - 1} is divisible by n, if n is a prime number and n is coprime to a. This is Fermat's theorem, which is just a sub-part of Euler's theorem.
Remainder when 30^(72^87) divided by 11
Euler's number of 11 is 10
=> 30^10k will leave a remainder of 1 when divided by 11.
So, basically if we can somehow find that what will be the remainder when 72^87 is divided by 10, then we can reduce the problem to a smaller number. Since unit digit of 72^87 is 8, remainder is 8
72^87 = 10k + r
=> 30^(72^87) = 30^(10k + 8 ) = 30^10k * 30^8
Now 30^10k leaves a remainder 1, so we just have to find the remainder by 30^8.
30^8 = (33 - 3)^8
So, we just have to look for remainder by 3^8, which comes out to be 5
In the similar manner, you can try other one.
Euler's number for any number N = (a^p)(b^q)(c^r)... , where a, b, c, ... are prime numbers is given by:-
(N) = N(1 - 1/p)(1 - 1/q)(1 - 1/r).....
Euler's number is also important in another way, as it denotes the number of numbers that are co-prime to N and less than N.
We can also notice that if N is a prime number, then
(N) = N(1 - 1/N) = (N - 1)
Now, as per Euler's theorem
{(a^(n)) - 1} is divisible by n, where (n) is Euler's number of n and n is coprime to a.
Since, (n) = (n - 1) for the case when n is prime, we can say that
If {(a^(n - 1)) - 1} is divisible by n, if n is a prime number and n is coprime to a. This is Fermat's theorem, which is just a sub-part of Euler's theorem.
Remainder when 30^(72^87) divided by 11
Euler's number of 11 is 10
=> 30^10k will leave a remainder of 1 when divided by 11.
So, basically if we can somehow find that what will be the remainder when 72^87 is divided by 10, then we can reduce the problem to a smaller number. Since unit digit of 72^87 is 8, remainder is 8
72^87 = 10k + r
=> 30^(72^87) = 30^(10k + 8 ) = 30^10k * 30^8
Now 30^10k leaves a remainder 1, so we just have to find the remainder by 30^8.
30^8 = (33 - 3)^8
So, we just have to look for remainder by 3^8, which comes out to be 5
In the similar manner, you can try other one.
Thanks dude for the complete concept.
It's really helpful.
Reminder for the second one is 4.Correct it if i am wrong.
If p is a prime number and w, x, y, z are four natural numbers whose sum is less than p, then
(w + x + y + z)p (wp + xp + yp + zp) is always divisible by Add to my
ap 1 bp2 cp dp + 1
(w + x + y + z)p (wp + xp + yp + zp) is always divisible by Add to my
ap 1 bp2 cp dp + 1
If p is a prime number and w, x, y, z are four natural numbers whose sum is less than p, then
(w + x + y + z)p (wp + xp + yp + zp) is always divisible by Add to my
ap 1 bp2 cp dp + 1
Are you sure you have typed the question properly???
is always divisible by
options are
a. p-1
b.p^2
c.p
d.p+1
rest question is same it is just looking clumsy
If p is a prime number and w, x, y, z are four natural numbers whose sum is less than p, then
(w + x + y + z)p - (wp + xp + yp + zp) is always divisible by Add to my
ap - 1 bp2 cp dp + 1
(w + x + y + z)p - (wp + xp + yp + zp)
= (w + x + y + z)p - (w + x + y + z)p
=0
It is divisible by any number
If p is a prime number and w, x, y, z are four natural numbers whose sum is less than p, then
(w + x + y + z)^p (w^p + x^p + y^p + z^p) is always divisible by
options are
a. p-1
b.p^2
c.p
d.p+1
sorry guyz this time pakka no mistakes now plz solve this:w00t:
(w + x + y + z)^p (w^p + x^p + y^p + z^p) is always divisible by
options are
a. p-1
b.p^2
c.p
d.p+1
sorry guyz this time pakka no mistakes now plz solve this:w00t:
If p is a prime number and w, x, y, z are four natural numbers whose sum is less than p, then
(w + x + y + z)^p (w^p + x^p + y^p + z^p) is always divisible by
options are
a. p-1
b.p^2
c.p
d.p+1
sorry guyz this time pakka no mistakes now plz solve this:w00t:
(c) p
according to fermat theorom if p is prime no then (a^p - a) is divisible by p
so in this case it can be written as
( (w+x+y+z)^p-(w+x+y+z)-(w^p-w +x^p-x+y^p-y+z^p-z))/p
thus its divisible by p
What is the OA?
(c) p
according to fermat theorom if p is prime no then (a^p - a) is divisible by p
so in this case it can be written as
( (w+x+y+z)^p-(w+x+y+z)-(w^p-w +x^p-x+y^p-y+z^p-z))/p
thus its divisible by p
plz explain the red part ye kaise kiya
If p is a prime number and w, x, y, z are four natural numbers whose sum is less than p, then
(w + x + y + z)^p (w^p + x^p + y^p + z^p) is always divisible by
options are
a. p-1
b.p^2
c.p
d.p+1
sorry guyz this time pakka no mistakes now plz solve this:w00t:
Its p.
Since p is prime and w, x, y, z all are less than p, we can say that they all are coprime to p
=> w^(p - 1) = 1 (mod p)
x^(p - 1) = 1 (mod p)
y^(p - 1) = 1 (mod p)
z^(p - 1) = 1 (mod p)
& (w + x + y + z)^(p - 1) = 1 (mod p)
=> w^p = w (mod p)
x^p = x (mod p)
y^p = y (mod p)
z^p = z (mod p)
& (w + x + y + z)^p = (w + x + y + z) (mod p) ..............(1)
=> (w^p + x^p + y^p + z^p) = (w + x + y + z) (mod p) ......(2)
From (1) and (2)
(w + x + y + z)^p (w^p + x^p + y^p + z^p) = 0(mod p)
(c) p
according to fermat theorom if p is prime no then (a^p - a) is divisible by p
so in this case it can be written as
( (w+x+y+z)^p-(w+x+y+z)-(w^p-w +x^p-x+y^p-y+z^p-z))/p
thus its divisible by p
plz explain the red part ye kaise kiya
Kucch nehi (w+x+y+z) add kiya aur subtract kiya :w00t:
How can we be certain that it won't be divisible by p^2. I did not get that point.
It might be divisible by p^2, but question say that its always divisible by, that's why it will be p
A function f(x) is defined for all real values of x as f(x) = ax2 + bx + 1. It is also known that f(5) = f(k) = 0, where k is not equal to 5. If a 1.b2.K3.B>0
4.K>0
4.K>0
Find all the values of n which satisfy (n!/4!)^2+7!.5!/4.4!=240(n!/4!).
Please explain the complete approach
1 and 8 are the first two natural no for which 1+2+....+n is a perfect square.Find the 4th such number.
Expecting complete approach.
1 and 8 are the first two natural no for which 1+2+....+n is a perfect square.Find the 4th such number.
Expecting complete approach.
We knew sum of n numbers --n(n+1)/2
By condn
n(n+1)/2=perfect square
n(n+1)=2perfect square..
==>Now
now n and n+1 are co prime in L.H.S of eqn..
and in R.H.S we have 2*perfect square..
If we take it as p^2..then we will get R.H.S=2*p*p which is not possible..
since both are coprime in L.H.S..
So to satisy we must have R.H.S as 2*(perfect sq1)*(perfect sq2)..
means we shud have 2 perfect square..
also in L.H.S..one is odd and other is even..vice versa..also we can say perfect sq2 must be odd..since 2*psq1 is even...
So let analyze..
8*9=2*4*9
by hit n trail..
next one we got..
49*50=2*25*49
and 4th one
288*289=2*144*289..
So 4th number is 288
Find all the values of n which satisfy (n!/4!)^2+7!.5!/4.4!=240(n!/4!).
Please explain the complete approach
Let n! be x
Then simplifying we get..
x^2/4!+7!5!/4=2.5!x...here i have written 240=2*5! to make job easy
=>x^2+7!5!*6=2*4!*5!x
x^2-2*4!*5!x+7!6!=0
x^2-(7!+6!)+7!6!=0
that implies answer is n=7 or6