Finding a Six-Digit NumberDate: 05/12/2005 at 20:29:00 From: Caroline Subject: Finding a 6-digit number I'm thinking of a 6-digit number. The sum of the digits is 43. And only two of the following three statements about the number are true: (1) it's a square number, (2) it's a cube number, and (3) the number is under 500,000. Date: 05/13/2005 at 11:34:08 From: Doctor Vogler Subject: Re: Finding a 6-digit number Hi Caroline, Thanks for writing to Dr. Math. The important thing to know about the sum-of-digits function is that for every positive integer n, the sum of digits of n is congruent to n mod 9. What does that mean? See Mod, Modulus, Modular Arithmetic http://mathforum.org/library/drmath/view/62930.html for an introduction to modular arithmetic. Essentially, it means that if you subtract the sum of digits from n, then you will always get something divisible by 9. That is, if you divide n by 9, and you divide the sum of digits by 9, you will get the same remainder. To see why, see the proofs in the links in Divisibility Rules http://mathforum.org/dr.math/faq/faq.divisibility.html for the rule for divisibility by 9. Or search our archives at http://mathforum.org/library/drmath/mathgrepform.html for the phrase sum of digits or the phrase casting out nines or something similar. Now, what's the point of this? Well, since your number has sum of digits 43 (which is 7 mod 9, i.e. the remainder is 7 when you divide by 9), it must also happen that your number is 7 mod 9. Well, this can't happen for a cube. Every cube is either 0, 1, or 8 mod 9. To see this, notice that every integer can be written as either 3k, 3k+1, or 3k+2 for some integer k. (Divide by 3; then k is the quotient and 0, 1, or 2 is the remainder.) Now expand (3k)^3 = 27k^3 (3k+1)^3 = 27k^3 + 27k^2 + 9k + 1 (3k+2)^3 = 27k^3 + 54k^2 + 36k + 8 and all of the terms except the 1 and 8 are divisible by 9. So you can't get 7 mod 9 with a cube. Therefore, you must be looking for a square less than 500,000. Well, there aren't all that many squares less than 500,000. In fact, there are 708 of them (from 0^2 to 707^2). Furthermore, only the numbers that are 4 mod 9 or 5 mod 9 will have squares that are 7 mod 9, which means that you only need to check about 2/9'ths of those numbers. What's more, 43 is a pretty high sum of digits for a six-digit number to have, especially since the first number can't be more than a 4. You see, 499999 only has a sum of 49, so you can only go down by 6. That means that any number with a 0, 1, or 2 in one of the last five places will be too small. And so on. So now look at all of those squares. It turns out that exactly one of them has a sum of digits that is exactly 43. Which one? If you have any questions about this or need more help, please write back and show me what you have been able to do, and I will try to offer further suggestions. - Doctor Vogler, The Math Forum http://mathforum.org/dr.math/ |
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