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Finding a Desired Perfect Cube

```Date: 04/26/2007 at 21:38:08
From: Shelly
Subject: Positive perfect cube

What is the smallest positive cube that ends with the digits 2007?

I do not know how to begin to solve the problem.

```

```

Date: 04/27/2007 at 01:23:39
From: Doctor Greenie
Subject: Re: Positive perfect cube

Hi, Shelly --

Very cool problem!!!

I had an idea from the start on how to solve the problem, and it
ended up working out.  But it was a lot of work.  On the other hand,
it was a lot of good mental exercise, and even rather fun......

Also note that there may be much more sophisticated and more efficient
ways, that I don't know about, to answer a problem like this.

The general idea is to determine what the requirements are for the
cube of an integer to end in "7"; then to determine the requirements
for the cube of an integer to end in "07"; then to end in "007"; and
finally to end in "2007".

In solving this problem, we will repeatedly use the binomial
expansion of (a+b)^3:

(a + b)^3 = a^3 + (3)(a^2)b + 3(a)b^2 + b^3

(1) perfect cubes ending in "7"

Looking at the cubes of the integers 1 through 9, we see that in
order for the cube of an integer to end in "7", the integer must end
in "3".  So our answer must be an integer of the form

10x + 3

(2) perfect cubes ending in "07"

We know the integer we are looking for must be of the form 10x+3.
So (using the expansion of (a+b)^3 as shown above)....

(10x + 3)^3 = 1000x^3 + 3(100)(3)x^2 + 3(10)(3^2)x + 3^3
= 1000x^3 + 900x^2 + 270x + 27

In order for this expression to end in "07", we must have

270x + 27

ending in "07".  (The first two terms of the expression contribute
nothing to the tens digit.)  In order for that to happen, the "270x"
must have final digits "80" (since the last term will add 27 to that);
for that to happen, x must have final digit "4".

Now we know the last two digits of our number must be "43"; so the
integer is of the form

100x + 43

(3) perfect cubes ending in "007"

We know the integer we are looking for must be of the form 100x + 43.
So....

(100x + 43)^3 = (10^6)x^3 + (3)(10^4)(43)x^2 +
(3)(100)(43^2)x + 43^3

= (10^6)x^3 + (3)(10^4)(43)x^2 +
554700x + 79507

The first two terms in this expansion do not contribute to the
hundreds place.  So in order to get the desired "0" in the hundreds
place, 7 times x must end in "5" (since there is already a 5 in the
hundreds place in the last term waiting to be added), so "x" must be "5".

So now we know that the last three digits of our number must
be "543"; so the integer is of the form

1000x + 543

(4) perfect cubes ending in "2007"

We know the integer we are looking for must be of the form
1000x + 543.  So....

(1000x + 543)^3 = (10^9)x^3 + (3)(10^6)(543)x^2 +
(3)(10^3)(543^2)x + 543^3

= (10^9)x^3 + (3)(10^6)(543)x^2 +
884547000x + 160103007

Again the first two terms do not contribute to the thousands place.
So in order to get the desired "2" in the thousands place, 7 times x
must end in "9" (since there is already a 3 in the thousands place in
the last term), so "x" must be "7".

Now we should have a number whose cube ends in "2007"; that number is

7543

Checking this with a calculator, we find

7543^3 = 429172932007

And our method assures us that 7543 is the smallest integer whose
cube ends in the digits "2007"; so the smallest perfect cube which
ends in the digits "2007" is

429,172,932,007

I hope this helps.  Please write back if you have any further

- Doctor Greenie, The Math Forum
http://mathforum.org/dr.math/
```
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College Number Theory
High School Number Theory
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