Question

A number is equal to the cube of the sum of its digits .Can you guess it. How ???​ It is urgent. ​

Answer 1

Answer:

An Armstrong number of three digits is an integer such that the sum of the cubes of its digits is equal to the number itself.

There are just four numbers, after unity, which are the sums of the cubes of their digits:

For example, 371 is an Armstrong number since 3³ + 7³ + 1³= 371.

The other are:

153, 370, 407

Hope it helps

;)

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Answer 2

Answer:

Hey mate, here is your answer.

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Hey mate, I give the answer that I know,

So please don't consider it Spam.

Hope it help.......

If it is helpful to you, please like it and mark it Brainliest, it's my Humble Request to you.

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Step-by-step explanation:

An Armstrong number of three digits is an integer such that the sum of the cubes of its digits is equal to the number itself.

There are just four numbers, after unity, which are the sums of the cubes of their digits:

For example, 371 is an Armstrong number since 3³ + 7³ + 1³= 371.

The other are:

153, 370, 407

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I have a number, I don't know how large or small, but if I cube the digits of the number and sum them, the sum is equal to the number itself. In other words, $$\sum_{k=1}^n{a_k^3}=\sum_{k=1}^{n}{a_k 10^{k-1}}$$ I know a couple of answers just by guess and check, but I'd rather have a concrete way. I know by some simple analysis that the number can't be a two digit number since $$a^3+b^3=10a+b \Rightarrow a^3-10a=b-b^3 \Rightarrow a(a^2-10)=b(1-b^2)$$ So $b^2$ can either be 0 or 1, but that produces either a $0$ or $1$ for a and that doesn't work (the trivial solutions are of course $0^3=0$ and $1^3=1)$. So the number must be at least three digits. Also, since $$9^3+9^3+9^3+9^3+9^3=5(729)=3,645$$ The number can at most be 4 digits. So from here, $$a^3+b^3+c^3=100a+10b+c$$or$$a^3+b^3+c^3+d^3=1,000a+100b+10c+d$$ By guess and check I got 153 as one answer $(1^3+5^3+3^3=1+125+27=153)$. But I know there are 11 more solutions and I'm not guessing them all. Outside writing a program to extract them, how can I solve the two equations above to get he remaining solutions (not sure if the question I'm solving includes the trivial solutions or not,so there might only be 9 more)?