Question

Find the smallest number k such that k(3^3 + 4^3 +5^3 )where a and b r are natural no

Answer 1

Answer:

As the smallest + ve integer is 1, so we try to get such value for ‘a’ & ‘n' such that k possibly becomes one.

K (3^3+4^3+5^3) = a^n ( given)

So, K(27 + 64 + 125) = a^n

=> K x 216 = a^n

=> K = a^n /216

=> K = a^n / 16²

So it's possible to get the value of K as 1, which is the smallest + ve integer , if a^n = 16²

ie, K = 16² / 16² = 1

So, here a^n = 16²

ie, a=16 & n=2……….....(1)

OR K = (4²)² /(4²)²

=> K= 4^4 / 4^4 = 1

So, a= 4 , n= 4…………….(2)

OR K = (2²)^4 / (2²)^4 =1

=> K = 2^8 / 2^8

So, a= 2 , n= 8……………..(3)

Also k= 216/216 = 6^3/6^3

So a= 6, n= 3…………..(4)

Further we get a=24, n= 9……(.5)

a= 72, n= 3 ………..(6)

a= 27, n= 8…………..(7)

a=12, n=18……….(8)

a= 36, n= 6………….(9)

a= 108, n= 2………(10)

This way, here we notice that a, & n are all the factors of 216 for the least natural value of ‘k' ie for k=1

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Answer 2

For the smallest value of k a^n should be smallest.

For the smallest value of k a^n should be smallest.Given a , n and k are positive integers.

For the smallest value of k a^n should be smallest.Given a , n and k are positive integers.Smallest positive integer is 1 so k cannot be less than 1.

For the smallest value of k a^n should be smallest.Given a , n and k are positive integers.Smallest positive integer is 1 so k cannot be less than 1.Putting k=1 we get

For the smallest value of k a^n should be smallest.Given a , n and k are positive integers.Smallest positive integer is 1 so k cannot be less than 1.Putting k=1 we get1×(27+64+125)=a^n

For the smallest value of k a^n should be smallest.Given a , n and k are positive integers.Smallest positive integer is 1 so k cannot be less than 1.Putting k=1 we get1×(27+64+125)=a^n216=a^n

For the smallest value of k a^n should be smallest.Given a , n and k are positive integers.Smallest positive integer is 1 so k cannot be less than 1.Putting k=1 we get1×(27+64+125)=a^n216=a^nOr a^n=6^3

For the smallest value of k a^n should be smallest.Given a , n and k are positive integers.Smallest positive integer is 1 so k cannot be less than 1.Putting k=1 we get1×(27+64+125)=a^n216=a^nOr a^n=6^3Comparing we get a=6 , n=3

For the smallest value of k a^n should be smallest.Given a , n and k are positive integers.Smallest positive integer is 1 so k cannot be less than 1.Putting k=1 we get1×(27+64+125)=a^n216=a^nOr a^n=6^3Comparing we get a=6 , n=3Both a and n are positive integers so assuming k=1 is correct and hence 1 is the smallest value of k.

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