Thread: Need to find four-figure number.

Hello

I need to find a four-figure number which can be divided by 7 and which can be written as a sum of square and cube of any natural number..
My English is not perfect, but I hope you will understand what I mean.

I would be very glad if someone could help me.

H. Please write me personal in skype: agripasiga i'wiill try to help solve you problem

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10^3 + 1^2 = 1001 which is divisable by 7 143 times.

By the way: the square and the cube has to be of the same natural number. For example 10^2 + 10^3

It looks to me like this is just a matter of trying! We are looking a number, x, such that has four digits and is divisible by 7. Since 7 is prime, that means that x must be divisible by 7. so for to be four digits itself the other factor in must be between 3 and 30. The only cubes in that interval are , , and so the only possible values for x are 7(1)= 7, 7(2)= 14, and 7(3)= 21.

If x= 7, then which does not have four digits. If x= 14, then . That has four digits and so is a perfectly good answer. If x= 21, then . Both 2940 and 9702 fit the requirements.

Ok then, 14^3 + 14 ^ 2 which divided by 7 is 420.. May I have a thanks?

HallsofIvy Interesting explanation. My answer was 13, because 13^2+13^3 is 2366 and 2366 : 7 = 338, but I had no idea how to explain that. I think your solution is enough good. I don't know how to find the answer in other ways.