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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 
questions about any of this.

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

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