[great_timer]

Originally Posted by niyant14

Hi
can sumbdy plz. solve tis ques 4 me:

Q)f(x) is a 6 degree term with coefficient of x^6 equal to 1.
f(1)=7.
f(2)=10
f(3)=13
f(4)=16
f(5)=19
f(6)=???

Tis came in this year's cat.

Is the answer 22?

[great_timer]

An easy one to start with

What is the remainder when 2^60/5?

[shashankmishra]

Originally Posted by ChAK_de_fatte

What is the smallest number which when divided by 10 leaves a remainder 9, when divided by 9 leaves a remainder 8, when divided by 8 leaves a remainder 7 and so on till when divided by 2 leaves a remainder 1?

In each case Remainder is -1, So

Smallest such Number

LCM(10,9,8,7 .........2) - 1 = 2520 - 1 = 2519

[shashankmishra]

Originally Posted by shanks4mba

Let me post a question
Find the ratio of the 17th terms of 2 APs if the ratio of the sum upto n terms of the 2 APs is (3n+2):(4n-13)

options (1) 4:7 , (2) 7:6 , (3) 101:119, (4) 26:55, (5) 50:53

Happy solving

Alternate Approach:

Since Sum to n terms = n/2[2a + (n-1)d] = (n^2)d/2 + n(a - d/2)

Ratio = (3n + 2)/(4n - 13) = (3n^2 + 2n)/(4n^2 - 13n)

So d1/2 = 3 , (a1 - d1/2) = 2 => a1 = 5 , d1 = 6

d2/2 = 4 , (a2 - d2/2) = -13 => a2 = -9 , d2 = 8

Ratio of 17th Term = 101 : 119

[shashankmishra]

Originally Posted by ChAK_de_fatte

First Koshchin..

The following expression is valid for a particular value of x. Find the value of 5x?


will post OA+Solution tomorrow morning

A Wild Shot



a^{ [ log(b6)x]^(log(b'a')4) = log(b4)256 or

a^k = 4 = 2^2 , So a = 2 , k = 2

(log(b6)x ) ^ (log(b2)4 ) = 2

[log(b6)x ] ^ 2 = 2

x = 6^(root2) => 5x = 5*[6^(root2)]

And if the [log(base a)4] is power of x then ans becomes = 5x = 30

Waiting for OA (Since I know this is wrong )

[k_himanshu]

Originally Posted by shashankmishra

A Wild Shot

a^{ [ log(b6)x]^(log(b'a')4) = log(b4)256 or

a^k = 4 = 2^2 , So a = 2 , k = 2

(log(b6)x ) ^ (log(b2)4 ) = 2

[log(b6)x ] ^ 2 = 2

x = 6^(root2) => 5x = 5*[6^(root2)]

And if the [log(base a)4] is power of x then ans becomes = 5x = 30

Waiting for OA (Since I know this is wrong )

my approach goes here: a^[log(b6) x] ^log (base a) 4 = log (b4) 256
or, a^[ log (b a) x / log (b a) 6 ] ^ log(b a) 4 = 4
or, x ^ [1/ log (b a) 6 ]^ log(b a) 4 = 4
or, x ^[log(b 6) a] ^ log(b a) 4 = 4
or, [log(b 6) a] ^ log(b a) 4 = log (b x) 4
or, log(b a) 4 = log ( b log (b 6) a ) log (bx) 4

wiz.... 4= log (b x) 4
or, x^4 = 4
or, x= 4^1/4

so 5x= 5* 4^1/4

wats the final answer ..even i am not sure about it !! hehhe

[harshkumar1]

Originally Posted by shanks4mba

Let me post a question
Find the ratio of the 17th terms of 2 APs if the ratio of the sum upto n terms of the 2 APs is (3n+2):(4n-13)

options (1) 4:7 , (2) 7:6 , (3) 101:119, (4) 26:55, (5) 50:53

Happy solving

Sum of 17 terms - Sum of 16 terms = 17th Term

Lets assume that Sum of AP1 = (3n+2)k
Sum of 17 terms of AP1 = 53k
Sum of 16 terms of AP1 = 50k
17th Term of AP1 = 3k

Similary 17th Term of AP2 = 8k

Hence the ratio should be 3:8

But none of the options match
Can someone of you point at the mistake i am doing.

[k_himanshu]

Originally Posted by great_timer

an easy one to start with

what is the remainder when 2^60/5?

=2^60 %5
=2^(6*10) %5
=(64)^10 %5
=(65-1)^10 %5
=(-1)^10 %5

=1

[great_timer]

Originally Posted by k_himanshu

=2^60 %5
=2^(6*10) %5
=(64)^10 %5
=(65-1)^10 %5
=(-1)^10 %5

=1

Correct

Shorter approach

Using Euler theorem

Euler number of 5=4

2^4k mod 5=1...using Euler theorem

[great_timer]

An important problem wrt Numbers

A number n has 48 factors. Find the maximum number of ways n can be expressed as a product of two co-primes.

a) 12 b) 16 c) 20 d) 24 e) None of these