F 10101: Fill & Download for Free

Oh, that’s a fun one.I believe this problem can be solved quite easily given enough time (not all IMO problems can.) It’s much harder to do under time pressure, which is why it’s a P3; you’d need amazing insight or a lot of luck to figure it out quickly. But figuring it out slowly is actually not that difficult.We have a function [math]f[/math][math][/math], defined for every positive integer [math]n[/math] like this:[math]\displaystyle f(1)=1, f(3)=3[/math][math]\displaystyle f(2n)=f(n)[/math][math]\displaystyle f(4n+1)=2f(2n+1)-f(n)[/math][math]\displaystyle f(4n+3)=3f(2n+1)-2f(n)[/math]The first thing to observe is that this really does define a function for all positive integers [math]n[/math], with no ambiguity or unknown values.We are asked to find how many fixed points [math]f[/math][math][/math] has through [math]1988[/math], meaning how many numbers [math]n[/math] between [math]1[/math] and [math]1988[/math] satisfy [math]f(n)=n[/math]. As always, this [math]1988[/math] is clearly just a red herring. We need to generally understand the fixed points, which likely means we need to understand what this function does in general.Staring at the recursive definition doesn’t necessarily make anything stand out, so we simply do the most natural thing: calculate some values of [math]f[/math][math][/math]. I always prefer to write [math]a\to b[/math] instead of [math]f(a)=b[/math]: it’s clearer, and the name [math]f[/math][math][/math] is useless when there’s only one function around.So, we quickly compute:[math]1 \to 1[/math][math]2 \to 1[/math][math]3 \to 3[/math][math]4 \to 1[/math][math]5 \to 5[/math][math]6 \to 3[/math][math]7 \to 7[/math][math]8 \to 1[/math][math]9 \to 9[/math][math]10 \to 5[/math]Danger lurks at this point: having made 10 experiments you may think you’ve sampled enough, and you may conclude that [math]f(n)=n[/math] holds for all odd [math]n[/math]. If you stop now and try to prove this, though, you will quickly run into trouble. Let’s keep going.[math]11 \to 13[/math] (oh!)[math]12 \to 3[/math][math]13 \to 11[/math][math]14 \to 7[/math][math]15 \to 15[/math][math]16 \to 1[/math][math]17 \to 17[/math][math]18 \to 9[/math][math]19 \to 25[/math][math]20 \to 5[/math]At this second milestone of 20 numbers computed, another danger lurks: the most reasonable theory at this point is that [math]f(n)=n[/math] occurs exactly around powers of [math]2[/math], that is when [math]n=2^k \pm 1[/math]. Right? This perfectly explains why [math]11[/math], [math]13[/math] and [math]19[/math] are out while all other small oddies are in.Do not succumb to temptation. Persevere. Do one more. Just one more.[math]21 \to 21[/math]Aha! There goes that theory out the window.So, what’s going on here?The fixed points we have so far are [math]1,3,5,7,9,15,17[/math] and [math]21[/math]. It’s not too clear what these numbers are. The even numbers clearly can’t be fixed points, but the way they map is relevant to how the odd numbers behave. What’s the order in this chaos?The inspired observation you need is to stop looking at these numbers in base 10. Looking at the rules defining [math]f[/math][math][/math], and how calculating it involves doubling and quadrupling, it may seem natural to look at all of these values in binary.[math]1\to 1[/math][math]10\to 1[/math][math]11\to 11[/math][math]100 \to 1[/math]Actually, let’s just look at the more interesting ones.[math]1011 \to 1101[/math] (that’s [math]11\to 13[/math])[math]1101 \to 1011[/math] (that’s [math]13\to 11[/math])[math]10011 \to 11001[/math] (that’s [math]19 \to 25[/math])Do you see it now?Yes! The function simply reverses the binary digits of its input. That’s why [math]19[/math] goes to [math]25[/math], and we can easily confirm that [math]25[/math] indeed goes back to [math]19[/math]. That’s also why [math]21[/math], which is [math]10101[/math][math][/math] in binary, is a fixed point.Noticing this simple description of the function [math]f[/math][math][/math] is the only really hard part of this problem. Gathering the data for this observation is purely mechanical, and proving it – which we shall now proceed to do – is pretty straightforward, too. Realizing it takes experience and observation, but we get many clues.To prove that [math]f[/math][math][/math] really reverses in binary, we can just do this: declare [math]g[/math] to be the function that actually reverses the binary digits, and show that [math]g[/math] satisfies the same conditions that [math]f[/math][math][/math] does. Since these conditions determine [math]f[/math][math][/math] uniquely, it must be that [math]f=g[/math].So if [math]g[/math] is, by definition, the bit-reversal function, then obviously [math]g(1)=1[/math] and [math]g(3)=3[/math]. It’s also clear that [math]g(2n)=g(n)[/math]: the binary representation of [math]2n[/math] is the same as that of [math]n[/math], with a single [math]0[/math] attached to the end. Reversing [math]2n[/math] places that [math]0[/math] at the beginning, which represents the same number as the reversal of [math]n[/math].Now for the last two relations. Take any natural number [math]n[/math] represented in binary. Attach [math]01[/math]at the end; this is now [math]4n+1[/math]. Reverse; you get [math]g(4n+1)=2^k+g(n)[/math], where the term [math]2^k[/math] comes from moving the rightmost [math]1[/math] to the very left, and we don’t bother to calculate [math]k[/math]: we won’t need to.Indeed, [math]g(2n+1)=2^{k-1}+g(n)[/math], where the added power of [math]2[/math] is clearly one less than [math]k[/math]. So,[math]2g(2n+1)=2^k+2g(n)=g(4n+1)+g(n)[/math]which is exactly the defining relation of [math]f(4n+1)[/math].Finally, [math]4n+3[/math] is represented by the binary representation of [math]n[/math] with [math]11[/math] appended in the end, so[math]g(4n+3)=2^k+2^{k-1}+g(n)[/math]while[math]3g(2n+1)=3(2^{k-1}+g(n))=(2+1)2^{k-1}+3g(n)[/math]or[math]3g(2n+1)=2^k+2^{k-1}+3g(n)=g(4n+3)+2g(n)[/math]which confirms the last relation.This concludes the verification that [math]g(n)[/math], the “binary reversal” function, satisfies the same recursive relations as [math]f[/math][math][/math], so the two functions are one and the same.The only thing left to do is count how many palindromic numbers exist between [math]1[/math] and [math]1988[/math]. The number of such nonzero numbers with [math]d[/math] binary digits is:[math]2^{e-1}[/math] if [math]d=2e[/math] is even (you must start with [math]1[/math], and then freely choose [math]e-1[/math] digits)[math]2^e[/math] if [math]d=2e+1[/math] is odd (start with [math]1[/math], choose [math]e-1[/math] digits, and choose the middle digit).This gives [math]1+1+2+2+4+4+8+8+16+16[/math] palindromes of up to [math]10[/math] bits. With [math]11[/math] bits we should have had [math]32[/math] palindromes, but two of them start with five [math]1[/math]’s and are therefore larger than [math]1988[/math] (notice that [math]1988[/math] is very close to [math]2^{11}=2048[/math]). So we only get [math]30[/math] palindromic numbers of [math]11[/math] bits which do not exceed [math]1988[/math], and so the final answer is [math]92[/math].

Try this:65537^(10101....0101-1)%10101....0101If it equals to 1, it's likely prime.Repeat for 37, 7, 5, 3, 11 (known primes).This can be solved in log2(n) time or approximately 4033 operations.7^12 mod 131 72 7*7 % 13 = 104 10*10 % 13 = 98 9*9 % 13 = 3Find binary of 12 which is 8,4((F(8)*F(4))%133*9 % 13 = 27 %13 = 1Same concept except bigger numbers.

When we type some letters or words, the computer translates them in numbers as computers can understand only numbers. A computer can understand the positional number system where there are only a few symbols called digits and these symbols represent different values depending on the position they occupy in the number.The value of each digit in a number can be determined using −The digitThe position of the digit in the numberThe base of the number system (where the base is defined as the total number of digits available in the number system)Decimal Number SystemThe number system that we use in our day-to-day life is the decimal number system. Decimal number system has base 10 as it uses 10 digits from 0 to 9. In decimal number system, the successive positions to the left of the decimal point represent units, tens, hundreds, thousands, and so on.Each position represents a specific power of the base (10). For example, the decimal number 1234 consists of the digit 4 in the units position, 3 in the tens position, 2 in the hundreds position, and 1 in the thousands position. Its value can be written as(1 x 1000)+ (2 x 100)+ (3 x 10)+ (4 x l) (1 x 103)+ (2 x 102)+ (3 x 101)+ (4 x l00) 1000 + 200 + 30 + 4 1234 As a computer programmer or an IT professional, you should understand the following number systems which are frequently used in computers.Binary Number SystemCharacteristics of the binary number system are as follows −Uses two digits, 0 and 1Also called as base 2 number systemEach position in a binary number represents a 0 power of the base (2). Example 20Last position in a binary number represents a x power of the base (2). Example 2x where x represents the last position - 1.ExampleBinary Number: 101012Calculating Decimal Equivalent −Step Binary Number Decimal NumberStep 1 (10101)2((1 x 24) + (0 x 23) + (1 x 22) + (0 x 21) + (1 x 20))10Step 2 (10101)2 (16 + 0 + 4 + 0 + 1) 10Step 3 (10101)2 (21)10Note − (10101)2 is normally written as 10101.Octal Number SystemCharacteristics of the octal number system are as follows −Uses eight digits, 0,1,2,3,4,5,6,7Also called as base 8 number systemEach position in an octal number represents a 0 power of the base (8). Example 80Last position in an octal number represents a x power of the base (8). Example 8x where x represents the last position - 1ExampleOctal Number: (12570)8Calculating Decimal Equivalent −Step Octal Number Decimal NumberStep 1 (12570)8 ((1 x 84) + (2 x 83) + (5 x 82) + (7 x 81) + (0 x 80))10Step 2 (12570)8 (5496)10Note − (12570)8 is normally written as 12570.Hexadecimal Number SystemCharacteristics of hexadecimal number system are as follows −Uses 10 digits and 6 letters, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, A, B, C, D, E, FLetters represent the numbers starting from 10. A = 10. B = 11, C = 12, D = 13, E = 14, F = 15Also called as base 16 number systemEach position in a hexadecimal number represents a 0 power of the base (16). Example, 160Last position in a hexadecimal number represents a x power of the base (16). Example 16x where x represents the last position - 1ExampleHexadecimal Number: (19FDE)16Calculating Decimal Equivalent −Step Hexadecimal Number Decimal NumberStep 1 (19FDE)16 ((1 x (16)4) + (9 x (16)3) + (F x (16)2) + (D x (16)1) + (E x (16)0))10Step 2 (19FDE)16 (106462)10Note − (19FDE)16 is normally written as 19FDE