Finding a Desired Perfect CubeDate: 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/ |
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