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An A.P P consist of n terms .from the progressions three different progression P1 P2, P3are created such that P1 is obtained by 1st,4th ,7th ....terms of P ,P2 has the 2nd.5th ,8th ...terms of P and P3 has 3rd,6th ,9th ...terms of P.It is found that of P1, P2 and P3 two progressions have the property that their average is itself a term of the original progression P.what are the possible values of n?(with solution please)
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?
1st term = a
2nd term = a+ d
3rd term = a + 2d
4th term = a + 3d
a(a+3d) / (a+d)(a+2d) = 2/3
Also, 4a + 6d = 20 => 2a + 3d = 10
Solve two equations to determine the unknowns.
abhilas SaysAn A.P P consist of n terms .from the progressions three different progression P1 P2, P3are created such that P1 is obtained by 1st,4th ,7th ....terms of P ,P2 has the 2nd.5th ,8th ...terms of P and P3 has 3rd,6th ,9th ...terms of P.It is found that of P1, P2 and P3 two progressions have the property that their average is itself a term of the original progression P.what are the possible values of n?(with solution please)
n = 4k+1; k=1,2...
for starters let us suppose that n=3k.
P = a1, a2, a3,............., a(n)
p1 = a1, a4, a7,.............., a(n-2)
p2 = a2, a5, a8,..............., a(n-1)
p3 = a3, a6, a9,................, a(n)
average of p1 =
average of p2 =
average of p3 =
let a1=a; a2=a+d; a3=a+2d, and so on.
on substituting we observe that n=4k+1
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
f(x)= ax+bx+1
Now since f(5)=f(k)=0,
Therefore, k and 5 are distinct roots of f(x) [ A quadratic equation has 2 roots).
Hence, Dividing f(x) by 'a' we get
x+(b/a)x+(1/a)
Now we know that in any quadratic equation of the form g(x)=x+(b/a)x+(c/a), the product of the roots = c/a
Applying this in f(x) we get,
k*5=1/a
Now a=> 1/a=> 5kHence k
Can someone answer this :-
What will be the remainder when 51^1375 is divided by 7?
Can we bring this form to (mx-1)^n/a in any case?
If yes, please explain by that form.
Can someone answer this :-
What will be the remainder when 51^1375 is divided by 7?
Can we bring this form to (mx-1)^n/a in any case?
If yes, please explain by that form.
Is tha ans 2?
Use Euler's Theorom
Is tha ans 2?
Use Euler's Theorom
I'm sorry i just quoted random figures. You know better.
Moreover, i'm not aware of euler's theorem. What is it btw?
Can someone answer this :-
What will be the remainder when 51^1375 is divided by 7?
Can we bring this form to (mx-1)^n/a in any case?
If yes, please explain by that form.
let me tell u an approach for these kind of questions...............
Whenever we get this kind of question i.e with a prime no. as a denomitor lets say like this x^a/p where p is the prime no. Then x^p-1/p will always give a remainder 1
now lets apply this approach to your question..............
51^1375/7 now using the above approach we can write this as...... (51^6)^229*51/7
again we can write this as............ (1)^229*51/7 which finally gives......
51/7 i.e remainder=2
i hope this helps..........
P.S-this approach will work only with questions with prime numbers as denominators.
Can someone answer this :-
What will be the remainder when 51^1375 is divided by 7?
Can we bring this form to (mx-1)^n/a in any case?
If yes, please explain by that form.
51^1375/7
Now,
(49+2)^1375/7
We know that 7^2=49. Thereby we can eliminate 49 as it is a factor of 7.
So now our problem becomes
2^1375/7
=>*2/7
Now,
the above statement may be re-written as
(7+1)^458*2/7
Hence, the remainder becomes 2.
51^1375/7
Now,
(49+2)^1375/7
We know that 7^2=49. Thereby we can eliminate 49 as it is a factor of 7.
So now our problem becomes
2^1375/7
=>*2/7
Now,
the above statement may be re-written as
(7+1)^458*2/7
Hence, the remainder becomes 2.
nice one but we can use the approach for the prime no.s the as the denominator which i gave earlier........it is much shorter because it holds for all prime no.s as denominator.....why make things unecessarily long
aditya5921 Saysnice one but we can use the approach for the prime no.s the as the denominator which i gave earlier........it is much shorter because it holds for all prime no.s as denominator.....why make things unecessarily long
Mine was just an alternative one.. Gave it because the guy said that he didn't know about euler's method, but nothing beats euler theorem when it comes to solving such problems;)
Rahul Mehra SaysMine was just an alternative one.. Gave it because the guy said that he didn't know about euler's method, but nothing beats euler theorem when it comes to solving such problems;)
yeah we have got a lot of things to thank about to Mr. Euler........i just felt that mine was a bit shorter .........but ur method is like an antibiotic.........works for all....,...
P.S-computer sc. would have been impossible without Mr. Euler
N is a set of all the natural nos less than 500 which can be written as sum of two or more consecutive natural nos .Find d max no of elements possible in N?
a) 250
b) 492
c) 493
d) none of these
a) 250
b) 492
c) 493
d) none of these
Can someone answer this :-
What will be the remainder when 51^1375 is divided by 7?
Can we bring this form to (mx-1)^n/a in any case?
If yes, please explain by that form.
since 51 and 7 are coprime we can use euler theorem and the euler totient of 7 is 6 and therefore 51^6 /7 the rem is 1
the above question can be written as (51^6)^229*51/7
ie 51/7 the rem is 2..learn euler theorem it will reduce the efforts required to solve such problems...
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
this is the question of CL procmock3
solve it as follows :
substituting x=5 in the given eqn gives us
f(5)=25a+5b+1=0
f(5)=25a+5b=-1
now 5b=-25a-1 but since a0 as our answer
hope this helped
N is a set of all the natural nos less than 500 which can be written as sum of two or more consecutive natural nos .Find d max no of elements possible in N?
a) 250
b) 492
c) 493
d) none of these
My take on this one is (d) None of these.
Only powers of two cannot be expressed:
So, 1, 2, 4, 8, 16, 32, 64, 128, 256 cannot be expressed. Rest can be expressed.
Reasoning is like this:
Say, we add two numbers: 1+2=3
Then 2+3 = 5 .. All odd numbers can be represented.
1+2+3=6
2+3+4=9 etc. All multiples of 3 can be represented.
1+2+3+4+5=15
2+3+4+5+6=20 etc. All multiples of 5 can be represented.
Similarly, all multiples of all odd numbers can be represented.
So, basically all multiples of all prime numbers except two can be represented.
So, all numbers except powers of two can be represented.
since 51 and 7 are coprime we can use euler theorem and the euler totient of 7 is 6 and therefore 51^6 /7 the rem is 1
the above question can be written as (51^6)^229*51/7
ie 51/7 the rem is 2..learn euler theorem it will reduce the efforts required to solve such problems...
51^1375/7
Now,
(49+2)^1375/7
We know that 7^2=49. Thereby we can eliminate 49 as it is a factor of 7.
So now our problem becomes
2^1375/7
=>*2/7
Now,
the above statement may be re-written as
(7+1)^458*2/7
Hence, the remainder becomes 2.
Thank you guys for helping me out with this.
Btw, where can I find Euler's method?
I couldn't find it in Arun sharma. Please post a link if it's available on the internet.
P.S- There were many formulae available under euler's formula when i tried to find out. He certainly has done many things, it seems!
Double post.
Btw, where can I find Euler's method?
u can find some help here I hope...
http://www.pagalguy.com/forum/quantitative-questions-and-answers/55747-cat-2010-concepts-fundas-tips-13.html#post2629058