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 Subject: mathematics Category: Science > Math Asked by: steeprock-ga List Price: \$30.00 Posted: 11 Feb 2003 16:13 PST Expires: 13 Mar 2003 16:13 PST Question ID: 160223
 ```1. Find the smallest positive integer m such that m is not a square and yet in the decimal expansion of the square root of m, the decimal point is followed by at least four consecutive zeros. 2. Find the number of nonegative integers n such that 2003+n is a multiple of n+1. 3. Let f be a function which assigns to each positive integer n a positive integer f(n). We suppose that f(ab) = f(a)f(b)-f(a)-f(b)+2 for all positive integers a, b and that f(c!)=c!+1 for all c (> or = to) 10^10. Show that f(n)=n+1 for all n.```
 ```steeprock, I answered the first two questions using the Perl programming languages to do brute force calculations of the answers, and included a mathematical proof for the third. Perl is a great tool for quick simulations; if you would like to learn more about it, please check out the official Perl site at http://www.perl.org 1. Using the following Perl program: #!/usr/bin/perl \$counter = 1; \$square_root = sqrt \$counter; \$square_root_int = int \$square_root; while ((\$square_root == \$square_root_int) || (\$square_root - \$square_root_int < 0.0001)) { \$counter++; \$square_root = sqrt \$counter; \$square_root_int = int(\$square_root); } print"Number: \$counter\nSqrt : \$square_root\n"; #end of code I was able to detemine that the answer is 100000001. (The square root of 100000001 is 10000.00005) 2. Using the following Perl program: #!/usr/bin/perl \$counter = 0; for (\$counter=0;\$counter<=2003;\$counter++) { if (int((2003+\$counter)/(\$counter+1)) == (2003+\$counter)/(\$counter+1)) { print "\$counter (" . (2003+\$counter) . " / " . (\$counter+1) . " = " . (2003+\$counter)/(\$counter+1) . ")\n"; } } #end of code The answers for nonnegative integers n where 2003+n is a multiple of n+1 were determined as follows: 0 (2003 / 1 = 2003) 1 (2004 / 2 = 1002) 6 (2009 / 7 = 287) 10 (2013 / 11 = 183) 12 (2015 / 13 = 155) 13 (2016 / 14 = 144) 21 (2024 / 22 = 92) 25 (2028 / 26 = 78) 76 (2079 / 77 = 27) 90 (2093 / 91 = 23) 142 (2145 / 143 = 15) 153 (2156 / 154 = 14) 181 (2184 / 182 = 12) 285 (2288 / 286 = 8) 1000 (3003 / 1001 = 3) 2001 (4004 / 2002 = 2) 3. First, for the case where n=c!, we see that f(c!) = c!+1 becomes f(n)=n+1, so that part is trivial. What we need to do is prove f(n)=n+1 where n<>c!, that is, where n cannot be represented by a factoral. To do this, first we copy the other assumption we were given for this problem: f(ab) = f(a)f(b) - f(a) - f(b) + 2 Since this is given to be the case for all positive integers a and b, let us choose a and b to be numbers that can be represented as factorals of other numbers. So we will assume that a=x! and b=y!. We can now substitute these values into our equation as follows: f(ab) = f(x!)f(y!) - f(x!) - f(y!) + 2 Using f(c!)=c!+1, we can thus expand like this: f(ab) = (x! + 1)(y! + 1) - (x! + 1) - (y! + 1) + 2 Calculate (x! + 1)(y! + 1) to get: f(ab) = x!y! + y! + x! + 1 - (x! + 1) - (y! + 1) + 2 Eliminating the remaining parentheses by multiplying out by -1 gives us: f(ab) = x!y! + y! + x! + 1 - x! - 1 - y! - 1 + 2 Collecting terms on the right side of the equation, x! and y! cancel out and the numbers reduce to one, giving us: f(ab) = x!y! + 1 Now we can replace x! with a and y! with b to give: f(ab) = ab + 1 Because multiplying an integer by 1 returns the original integer, any integer can be represented as the product of two other integers. So n can be represented as the product of two integers a and b, and replaced in the equation as such: f(n) = n + 1. Therefore, f(n) = n + 1 for all positive integers n.``` Clarification of Answer by hailstorm-ga on 11 Feb 2003 18:56 PST ```Sorry, part 1 is incorrect! Please give me a couple of minutes to post the correct answer.``` Clarification of Answer by hailstorm-ga on 11 Feb 2003 19:10 PST ```steeprock, My apologies for the original error. Though the Perl code as listed above was correct, my own program had a typo which resulted in the incorrect answer. The correct answer for part 1 is 25000001, the square root of which is 5000.0000999999990000000199999995...``` Clarification of Answer by hailstorm-ga on 12 Feb 2003 13:52 PST ```steeprock, I would like to revisit the third part of this problem, after consulting a bit with fellow GA Researcher mathtalk. Consider the first assumption we are given in the third part of the question, about a function f(n) that satisfies: f(ab) = f(a)f(b) - f(a) - f(b) + 2 for all positive integers a,b. If one considers h(n) = f(n) - 1, the statement becomes simply: h(ab) = h(a) h(b) Moreover the second assumption we are given: f(c!) = c! + 1 for all c > 10^10 can be rewritten: h(c!) = c! We are asked to prove f(n) = n + 1 for all (positive integers) n, or what is equivalent, that h(n) = n. It is not enough to have on the first assumption, because f(n) = 2 identically would satisfy that condition. We need to work the second assumption into the proof somehow. First we establish h(1) = 1, like this: For sufficiently large c, h(c!) = h(1 * c!) = h(1) h(c!). Since h(c!) = c! is nonzero, it follows that h(1) = 1. Now we go after h(n) for n > 1. In this case we can always find a sufficiently large power n^k > 10^10 + 1, so that: h((n^k)!) = h(n^k) * h((n^k - 1)!) (n^k)! = h(n^k) * (n^k - 1)! n^k = h(n^k) = h(n)^k Now take the k'th root of both sides (which is a well-defined operation is we restrict ourselves to the positive numbers) and we have the desired result: n = h(n) for any n > 1. This also emphasizes the importance of thinking long and hard over your proofs before submitting them as an answer.```
 ```hailstorm, i'm afraid your answer to part 3 is also incorrect. you only proved that f(n)=n+1 for all n that can be represented as a a product of two factorials n=x!y! if a and b are not factorials, then you may not rely on f(a)=a+1 and f(b)=b+1 being true. also, you are only given f(c!)=c+1 for c at least 10^10. as for your answer to part 2, it may be correct (i didn't check) but it is slightly incomplete: for a full proof of correctness of your result, you need to mention that if n is greater than 2003 then n+2003 cannot be a multiple of n+1 (since n+1 is greater than (n+2003)/2, the largest possible divisor of n+2003), which is why you can limit the variable \$counter in your perl program to go only over the values 0 through 2003. also, the final answer to steeprock's question (assuming your results are correct) is 16. regards, dannidin```