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chandhuru
 2 years ago
Armstrong number is a number obtained summing the cubes of each number.....
consider the number to be abc
so took the eqn as
100a + 10b + c = a^3 + b^3 + c^3
how to solve this eqn???
chandhuru
 2 years ago
Armstrong number is a number obtained summing the cubes of each number..... consider the number to be abc so took the eqn as 100a + 10b + c = a^3 + b^3 + c^3 how to solve this eqn???

This Question is Closed

Jemurray3
 2 years ago
Best ResponseYou've already chosen the best response.1There are an infinite number of possible solutions to that equation, but I suspect that's not quite what you're looking for...

chandhuru
 2 years ago
Best ResponseYou've already chosen the best response.0I want to find the solution... in what way can i approach it ??

Jemurray3
 2 years ago
Best ResponseYou've already chosen the best response.1Having googled "Armstrong Number", a better way to phrase that question is "An Armstrong number is a 3digit number such that the sum of the cubes of each of its constituent digits is equal to the number itself. How can I find an armstrong number?"

Jemurray3
 2 years ago
Best ResponseYou've already chosen the best response.1Are you trying to find a general expression for all Armstrong numbers, or are you trying to simply find one Armstrong number?

chandhuru
 2 years ago
Best ResponseYou've already chosen the best response.0yaa i know it is a 3 digit number.... i am trying to find the general expression and i want to find the total number of armstrong numbers present.

Jemurray3
 2 years ago
Best ResponseYou've already chosen the best response.1I realize that you know the definition. My point was that I did not, so what I said would have been a better way to ask the question in the first place. Incidentally, you are restricting your attention to 3 digit numbers, then? Because the general concept of an Armstrong number can be applied to a number with any number of digits. If you are restricting yourself to 3 digit numbers, I don't think you can express a general relationship between the digits but it would be straightforward to systematically solve for them. There are only a handful.

chandhuru
 2 years ago
Best ResponseYou've already chosen the best response.0if i can find the solution for this eqn ... then same procedure can be used to find it 4 or 5 or 6 digit number...

chandhuru
 2 years ago
Best ResponseYou've already chosen the best response.0@hartn can u help me ???

Jemurray3
 2 years ago
Best ResponseYou've already chosen the best response.1You're asking for THE solution to an equation with MANY solutions. That is the primary problem here.

chandhuru
 2 years ago
Best ResponseYou've already chosen the best response.0@Jemurray3 cant we find that several numbers???

Jemurray3
 2 years ago
Best ResponseYou've already chosen the best response.1Yes, but I'm saying that you'd probably have to do it algorithmically rather than try to find a general expression for the digits. For instance, rearranging the equation, \[100aa^3 + 10bb^3 + cc^3 = 0\] if a = 1 and b = 1, \[99 + 9 = c^2 c \] which doesn't yield an integer c... but going through this process, you might find that 153 works out just fine.

chandhuru
 2 years ago
Best ResponseYou've already chosen the best response.0yaa i found that one... i solve it like dw:1348380805812:dw i dont no to proceed further.. if i go by ur method it will be trial and error method.. so i am trying for some different approach.

Jemurray3
 2 years ago
Best ResponseYou've already chosen the best response.1If you code a program you could do it in about ten seconds...

Jemurray3
 2 years ago
Best ResponseYou've already chosen the best response.1def Armstrong(): for i in range(1,9): for j in range(0,9): for k in range(0,9): if i*i*i + j*j*j + k*k*k == 100*i + 10*j + k: print(100*i+10*j+k)

Jemurray3
 2 years ago
Best ResponseYou've already chosen the best response.1I'm serious. This is a perfect example of when a computer program would be wonderful. The above code took a little over 10 seconds to write and produced the four 3digit Armstrong numbers.
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