The first few terms of the sequence, generated by D(n), are 1, 1, 5, 1, 10, 4, 12, 2, (sequence A333448 in OEIS). This will loop through every number from 1 to 600 and check if they are multiples of 5: Answering the question exactly as it's currently written, disregarding the title (which was edited). The result must be divisible by 13. The result must be divisible by 11. Add 3 times the last two digits to the rest. Next, and after eliminating the known multiple of 7, the result is. {\displaystyle [\sum _{k=1}^{n}(a_{2k}a_{2k-1})\times 10^{2k-2}]{\bmod {7}}}, n ), Subtract twice the last three digits from the rest. n 559+333=658. How can I shave a sheet of plywood into a wedge shim? But this makes it harder to handle the case when $value is not found among the given numbers without using some sort of flag: I would personally go for the case esac implementation. Please use our new forums! Quick! [6] Note that checking 3 and 12, or 2 and 18, would not be sufficient. For example, testing divisibility by 24 (24 = 83 = 233) is equivalent to testing divisibility by 8 (23) and 3 simultaneously, thus we need only show divisibility by 8 and by 3 to prove divisibility by 24. Subtract 11 times the last digit from the rest. To test the divisibility of a number by a power of 2 or a power of 5 (2n or 5n, in which n is a positive integer), one only need to look at the last n digits of that number. = 10a+3 7 Number must be divisible by 331 with the sum of all digits being divisible by 3. 3675: 75 is at the end and 3 + 6 + 7 + 5 = 21 = 37. Also, the result of the second test returns the same result as the original number divided by 6), Now convert the first digit (3) into the following digit in the sequence, Add the result in the previous step (2) to the second digit of the number, and substitute the result for both digits, leaving all remaining digits unmodified: 2+0=2. 140,625: 625 = 1255 and 1 + 4 + 0 + 6 + 2 + 5 = 18 = 63. Example: What is the remainder when 1036125837 is divided by 7? . 3 plus 8 is 11. Likewise, since 10 (28) = 280 = 1 mod 31 also, we obtain a complementary rule y+28x of the same kind - our choice of addition or subtraction being dictated by arithmetic convenience of the smaller value. According to this, our 7 becomes, Now, change 1 into the following digit in the sequence (3), add it to the second digit, and write the result instead of both: 3+1=, Repeat the procedure, since the number is greater than 7. 3,300: The result of sum the digits is 6, and the last two digits are zeroes. Also sometimes I'm getting the error message O. utput argument "valid" (and maybe others) not assigned during call to "valid_date". 2 Remainder = 17 mod 13 = 9, Example: What is the remainder when 1234567 is divided by 13? Since nine is a factor of eighty-one. i knew there had to be a simpler way but wasn't having any luck. Direct link to Peter Collingridge's post It will only work with nu, Posted 11 years ago. Direct link to CamdenMounts's post To test divisibility by 2, Posted 11 years ago. Let us understand this with the help of the following example. 2 You didn't answer my question. A divisibility rule is a shorthand and useful way of determining whether a given integer is divisible by a fixed divisor without performing the division, usually by examining its digits. Add the last three digits to three times the rest. automaton. If the number is divisible by six, take the original number (246) and divide it by two (246 2 = 123). Fifth rightmost digit = 2 3 = 6 But theres even something more interesting. {\displaystyle \sum _{k=1}^{n}(a_{2k}a_{2k-1}{\bmod {7}})\times (10^{2k-2}{\bmod {7}})}, Remainder Test The same reason applies for all the remaining conversions: First method example a 34,400: The third last digit is 4, and the last two digits are zeroes. Example 1 : Just in the interest of syntax neutrality and mending the overt infix notation bias around these parts, I've modified nagul's solution to use dc. 10 rev2023.6.2.43474. I have done it in a different way. After entering the number and divisor I get: test.sh: 7: test.sh: dc: not found Remainder: test.sh: 10: [: unexpected operator Your number is not divisible by 2 Do you have any idea why? if( i%3 == 0 ) Example: What is the remainder when 321 is divided by 13? 3 Because the resulting 182 is less than six digits, we add zero's to the right side until it is six digits. a Divisibility properties of numbers can be determined in two ways, depending on the type of the divisor. @Kusalananda in which case something like. Fourth rightmost digit = 5 1 = 5 Grey, 3 studs long, with two pins and an axle hole. However, I think since. Any tips on how I should fix it would be much appreciated! Plus 9, it's 20. k That. {\displaystyle 10\equiv -1{\pmod {11}}} Add the digits in blocks of two from right to left. 9 Add the last digit to the hundreds place (add 10 times the last digit to the rest). The number formed by the last eight digits must be divisible by 256. Asking for help, clarification, or responding to other answers. ) Step A: Rationale for sending manned mission to another star? automata. Direct link to junaid's post Why doesn't this work wit, You're just walking down the street and someone comes up to you and says "Quick! The number formed by the last two digits is divisible by 20. How does the number of CMB photons vary with time? ), Add 2 times the last digit to 3 times the rest. So to check if a number is divisible by 3, you need to determine if dividing the number by three has a remainder of zero. a Is this divisible by 3? If that number is an even natural number, the original number is divisible by 4. (1 <= day) will be either false (treated as 0) or true (treated as 1) both of which are less than or equal to 29. This follows from Pascal's criterion. But it doesn't work with 4 because when you get to 12, you subtract 9, which isn't a multiple of 4. Direct link to Shlomo Fingerer's post Sure! The result must be divisible by 9. Here is my code; I want to compare $COUNTER to various multiple times. Take the digits in blocks of five from right to left and add each block. 10,0000,000 mod 7 = 2^4 = 16; 16 mod 7 = 2, 1,000,0000,000 mod 7 = 2^5 = 32; 32 mod 7 = 4. To test for divisibility by D, where D ends in 1, 3, 7, or 9, the following method can be used. 10 The number must be divisible by 11 and 13. , The modulous operator gives just the remainder part as its output. + Language links are at the top of the page across from the title. (Works because 98 is divisible by 7.). We know how hard it can be working out some crossword answers, but weve got you covered with the clues and answers for the Number that is not divisible by 2 crossword clue right here. Why doesnt SpaceX sell Raptor engines commercially? Example. {\displaystyle D(n)\equiv {\begin{cases}9a+1,&{\mbox{if }}n{\mbox{ = 10a+1}}\\3a+1,&{\mbox{if }}n{\mbox{ = 10a+3}}\\7a+5,&{\mbox{if }}n{\mbox{ = 10a+7}}\\a+1,&{\mbox{if }}n{\mbox{ = 10a+9}}\end{cases}}\ }. Sum the ones digit, 4 times the 10 digit, 4 times the 100s digit, 4 times the 1000s digit, etc. rev2023.6.2.43474. The piece wise form of D(n) and the sequence generated by it were first published by Bulgarian mathematician Ivan Stoykov in March 2020.[13]. Remainder = 5. This method is especially suitable for large numbers. 1050 105 0=105 10 10 = 0. Web 3 comments ( 210 votes) Upvote Flag Peter Collingridge 11 years ago It will only work with numbers that are factors of 9 (i.e. Using the example above: 16,499,205,854,376 has four of the digits 1, 4 and 7 and four of the digits 2, 5 and 8; Since 4 4 = 0 is a multiple of 3, the number 16,499,205,854,376 is divisible by 3. Why you can add the digits to see if something is divisible by 3. Recurring numbers: 1, 3, 2, 6, 4, 5 1 the if [$number] statement is what I don't know how to set up. Making statements based on opinion; back them up with references or personal experience. Take any number, say 8,675,309. Is there a faster algorithm for max(ctz(x), ctz(y))? A more complicated algorithm for testing divisibility by 7 uses the fact that 1001, 1013, 1022, 1036, 1044, 1055, 1061, (mod7). Does this divisibility apply to negative numbers? Or, the number formed by the last four digits is divisible by 625. where in this case a is any integer, and b can range from 0 to 99. For example: 7%3 == 1 because 7 is divisible by 3 two times, with 1 left over. 15 is divisible by 3 at which point we can stop. partially. If it is not 0, then the number is not divisible by 3. It makes the expression look like finding non-divisible numbers Because it's examining the remainder of counter divided by 5. Well, you employ the same tactic You say, what's 3 plus 8 plus 6 plus 8 plus 0 plus 2? Martin Gardner explained and popularized these rules in his September 1962 "Mathematical Games" column in Scientific American.[1]. Direct link to rchakrjr's post Is there a divisibility r, Posted 11 years ago. While reading from a file, choose specific fields and compare them with another file in Unix bash? 3 Enter a number: 15 The number 15 is divisible by 3. Subtract 59 times the last digit from the rest. if( i % 3 == 0 ){ 1 Another digit pair method of divisibility by 7. We say that a natural number n is divisible by another natural number k if dividing n / k leaves no remainder (i.e., the remainder is equal to zero). If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked. So it works with 3, because when you get to 12, the sum of the digits is 12-9 or 3 (which is divisible by 3). Note: The reason why this works is that if we have: a+b=c and b is a multiple of any given number n, then a and c will necessarily produce the same remainder when divided by n. In other words, in 2+7=9, 7 is divisible by7. Then, break the integer into a smaller number that can be solved using Step B. D 4,675: 467 3 + 5 2 = 1,411; 238: 23 3 + 8 2 = 85. Divisibility Rules of 3 - check if a number is divisible by 3 Number must be divisible by 19 ending in 0 or 5. I'm trying to write a function that determines if the input is a valid date, and for that I need a function that will tell me if an input number is divisible by 4 and at the same time not divisible by 100 (this has to do with leap years but I'm not supposed to use the special function for that). 77,925,613,961: 7 + 79,256 + 13,961 = 93,224 = 271344. The result must be divisible by 11. Direct link to Nikki Ranjit's post Sal confused me when he s, Posted 4 years ago. Does substituting electrons with muons change the atomic shell configuration? 30,855: 3085 25 = 3060 = 17180. 1,986 = 6 331. Use the modulo operator. if(i % 3 == 0) condition is always true, because it is interpreted as (1 <= day) <= 29. 1 It works for all numbers. 2 Number must be divisible by 17 ending in 0 or 5. mod Can you please wrap the code in your question in code tags (or use the code button on the editor)? By writing a number as the sum of each digit times a power of 10 each digit's power can be manipulated individually. 345: 3 + 4 + 5 = 12 = 3 4, and 34 + 5 9 = 69 = 3 23. Site design / logo 2023 Stack Exchange Inc; user contributions licensed under CC BY-SA. Alternatively, one can just add half of the last digit to the penultimate digit (or the remaining number). Eighth rightmost digit = 3 3 = 9 In some cases the process can be iterated until the divisibility is obvious; for others (such as examining the last n digits) the result must be examined by other means. mod Only the last n digits need to be checked. The same for all the higher powers of 10: The result is the same as the result of 40 divided by 5(40/5 = 8). Any integer is divisible by 1. Direct link to Hanato Jingo's post partially. Multiply the remainders with the appropriate multiplier from the sequence 1, 2, 4, 1, 2, 4, : the remainder from the digit pair consisting of ones place and tens place should be multiplied by 1, hundreds and thousands by 2, ten thousands and hundred thousands by 4, million and ten million again by 1 and so on. Subtract 29 times the last digit from the rest. [12] Find any multiple of D ending in 9. 2 how to compare all files inside ONE directory in bash? The original number is divisible by 7 if and only if the number obtained using this procedure is divisible by 7 (hence 371 is divisible by 7 since 28 is).[10]. n 576), AI/ML Tool examples part 3 - Title-Drafting Assistant, We are graduating the updated button styling for vote arrows, How to write a script to accept any number of command line arguments? Add the digits in blocks of three from right to left. 3 and 9). Linux is a registered trademark of Linus Torvalds. 8,937: 8 + 7 = 15. To compare an integer in a variable to a number of other integer values, where the other values are determined ahead of time (it is unclear what "dynamically" actually means in the question): This may also be done in a loop, of course. Subtract 27 times the last digit from the rest. The result must be divisible by 11. 7 Grey, 3 studs long, with two pins and an axle hole. Ans: 1 1 + 2 3 + 3 4 = 17 The last three digits are divisible by 8. {\displaystyle {\overline {a_{2n}a_{2n-1}a_{2}a_{1}}}\mod 7}, [ WebRepeat the process for larger numbers. 6013->601 then subration 601-3x2(6)= 595. I works for al, Posted 10 years ago. {\displaystyle {\overline {a_{2n}a_{2n-1}a_{2}a_{1}}}} The last digit is even (0, 2, 4, 6, or 8). When the number is larger than six digits, then repeat the cycle on the next six digit group and then add the results. Add the last two digits to 6 times the rest. Can I change my terminal environment to bash shell using a script and then execute other commands after changing the shell in the same script? Check if it works for you. And the number ends in 5. A table of prime factors may be useful. Subtract 18 times the last digit from the rest. bash scripting has always been a bit of a black art to me. 6,507: 65 8 7 = 520 7 = 513 = 27 19. Subtract 12 times the last digit from the rest. Subtract the last two digits from four times the rest. The number formed by the last three digits must be divisible by 125. :), Alternatively, you could use expr instead of bc: remainder=, @Dan Yes it should suffice for the OP. In Portrait of the Artist as a Young Man, how can the reader intuit the meaning of "champagne" in the first chapter? 1 (Works because 51 is divisible by 17. Subtract 5 times the last digit from the rest. A composite divisor may also have a rule formed using the same procedure as for a prime divisor, given below, with the caveat that the manipulations involved may not introduce any factor which is present in the divisor. Is there a legal reason that organizations often refuse to comment on an issue citing "ongoing litigation"? Similarly a number of the form 10x+y is divisible by 7 if and only if x+5y is divisible by 7. 1 Well, what you could do (it might be a bit faster; it is faster on my machine) is: boolean canBeDevidedBy3 = ((int) (i * 0x55555556L >> 30) & 3 The = is an assignment operator in MATLAB. The result 42 is divisible by seven, thus the original number 157514 is divisible by seven. 2 For example, if you add the digits of 81 together (8 + 1) you get 9, which is clearly divisible by 3. Recurring numbers: 1, 3, 2, 1, 3, 2 Add three times the last digit to the rest. Example: Check number divisibility . Quick! For the higher powers of 10, they are congruent to 1 for even powers and congruent to 1 for odd powers: Like the previous case, we can substitute powers of 10 with congruent values: which is also the difference between the sum of digits at odd positions and the sum of digits at even positions. , Vedic method of divisibility by osculation Positive sequence, Multiply the right most digit by the left most digit in the sequence and multiply the second right most digit by the second left most digit in the sequence and so on and so for. Sum = 33 Add the last two digits to 3 times the rest. ), Subtracting 2 times the last digit from the rest gives a multiple of 7. 54778+533=54943. 6 2 = 3 (Check to see if the last digit is divisible by 2), 376 2 = 188 (If the last digit is divisible by 2, then the whole number is divisible by 2), 4 + 9 + 2 = 15 (Add each individual digit together). Subtract 47 times the last digit from the rest. n ) Subtract the last two digits from 3 times the rest. And that is actually the same as subtracting 710n (clearly a multiple of 7) from 1010n. (If D ends respectively in 1, 3, 7, or 9, then multiply by 9, 3, 7, or 1.) This method could be useful in a mathematics competition such as MATHCOUNTS, where time is a factor to determine the solution without a calculator in the Sprint Round. if Explore free or paid courses in topics that interest you. It works also for 6, 12, 18 etc because you just have to do the test for the non-three power number then for three. Was the breaking of bread in Acts 20:7 a recurring activity that the disciples did every first day and was this a church service? The result 77 is divisible by seven, thus the original number 15751537186 is divisible by seven. if [11][unreliable source?]. 10 Subtract 7 times the last digit from the rest. Unable to complete the action because of changes made to the page. =6013 you repete the process untill you get to a 2digte or 1 digte number. And the number ends in 5. 3 This can be done by subtracting the first three digits from the last three digits. Divisibility by seven can be tested by multiplication by the Ekhdika. However it gives me a few errors and I honestly don't know why. Although there are divisibility tests for numbers in any radix, or base, and they are all different, this article presents rules and examples only for decimal, or base 10, numbers. If the last digit in the number is 5, then the result will be the remaining digits multiplied by two, plus one. The result must be divisible by 3. Welcome, Roger. mod The PohlmanMass method provides a quick solution that can determine if most integers are divisible by seven in three steps or less. Notice that leading zeros are permitted to form a 6-digit pattern. Multiply by 5, add the product to the next digit to the left. Here is the Multiplication of the rightmost digit = 1 7 = 7 Step 3: As per the divisibility rule of 3, if the sum of the digits of a number is k {\displaystyle 10=-3{\bmod {1}}3} For example: The fact that 999,999 is a multiple of 7 can be used for determining divisibility of integers larger than one million by reducing the integer to a 6-digit number that can be determined using Step B. Sum the digits. Start here for a quick overview of the site, Detailed answers to any questions you might have, Discuss the workings and policies of this site. 2,651,272: 72 - 11 x 2 + 51 - 11 x 6 + 2 = 37. 925 = 3725. Subtract twice the last two digits from the rest. n (Works because 91 is divisible by 7. 1 So now I came up with this and it works with some dates but not with most; For example, I tried it with the input (2001,1,32) and it returns "1" although the input is not a correct date. Browse other questions tagged. There is an operator called modulus that tells you what the remainder is after doing division with two numbers. If we add 1 and 5 it will become 6. + ) n 2,651,272: 2 + 651 + 272 = 925. In other words, you will find the remainder of dividing the number by 7. For example, to determine divisibility by 36, check divisibility by 4 and by 9. . a I'm a student studying DFAs looking for a DFA that could find if a decimal number is divisible by 6. i tried the traditionel (mod) method and it doesn't work. ), Subtract the last two digits from 8 times the rest. k This implies that a number is divisible by 13 iff removing the first digit and subtracting 3 times that digit from the new first digit yields a number divisible by 13. Connect and share knowledge within a single location that is structured and easy to search. And 34 + 5 2 = 1,411 ; 238: 23 3 + +! Note that checking 3 and 12, or responding to other answers. ) we add zero to. 0, then the number by 7 're behind a web filter, make... Only work with nu, Posted 11 years ago, What 's 3 8. Inc ; user contributions licensed under CC BY-SA, Posted 11 years ago 0 plus?! Be determined in two ways, depending on the next digit to the rest 4 5... Clearly a multiple of 7 ) from 1010n quick solution that can be determined in two,... 4 times the rest 3 two times, with 1 left over } the. 1 ( Works because 91 is divisible by 19 ending in 9 division with two pins an! 2023 Stack Exchange Inc ; user contributions licensed under CC BY-SA and easy to search that the disciples every! By 5 substituting electrons with muons change the atomic shell configuration another file in bash... = 27 19 by writing a number: 15 the number must be divisible seven. 15 the number of the divisor: Rationale for sending manned mission to another star from a file, specific... 13., the result is determined in check if number is divisible by 3 ways, depending on the next digit the. How does the number formed by the Ekhdika if the last eight digits must divisible. From right to left Find any multiple of 7. ): What the! ) from 1010n 3 studs long, with 1 left over long, with two numbers 1255 and 1 2! 1 another digit pair method of divisibility by seven in three steps or less from four times the digit... = 2 3 + 5 = 18 = 63 and add each block a number! 8 plus 0 plus 2 5 it will only work with nu Posted. Number 15 is divisible by 4 divisibility r, Posted 11 years ago 5 9 = 69 3... 6 plus 8 plus 6 plus 8 plus 0 plus 2 to be checked on opinion ; back up. Is 6, and after eliminating the known multiple of 7 ) from 1010n direct link rchakrjr., and after eliminating the known multiple check if number is divisible by 3 7. ) 10 each digit power! Digit 's power can be done by subtracting the first three digits to 6 times the last digit from rest! Works for al, Posted 10 years ago in the number formed the. = 925 4 times the 10 digit, 4 times the last digit from the rest gives a of... Connect and share knowledge within a single location that is structured and easy to search it! Is there a divisibility properties of numbers can be solved using step...., one can just add half of the divisor with two pins and an axle hole an! 10 the number must be divisible by seven, thus the original number 15751537186 is divisible 7! The 10 digit, etc and an axle hole by 9. in three steps or less here is code! And 5 it will become 6 digits multiplied by two, plus one this with the sum of all being... Check divisibility by 2, 1, 3, 2, 1 3! 345: 3 + 4 + 5 2 check if number is divisible by 3 1,411 ; 238 23... Or responding to other answers. ) Grey, 3 studs long with! Any tips on how I should fix it would be much appreciated *... 19 ending in 0 or 5 method of divisibility by seven in three steps or.. Subtracting 2 times the last digit from the rest 10\equiv -1 { \pmod { 11 } } } add... Step a: Rationale for sending manned mission to another star add each block 8 plus 6 8! Recurring activity that the domains *.kastatic.org and *.kasandbox.org are unblocked subtract 27 the. Is actually the same tactic you say, What 's 3 plus 8 plus 0 2!, check divisibility by seven to a 2digte or 1 digte number 34 + 5 = 18 63. In Scientific American. [ 1 ] 77 is divisible by 4 to left to.. 3 and 12, or 2 and 18, would not be sufficient on the type of form. Or the remaining digits multiplied by two, plus one sending manned mission to another star the... Of sum the digits is divisible by 3 7 number must be divisible by 7 + n! Step B 520 7 = 520 7 = 513 = 27 19 making statements based on ;! Digits to three times the rest and the last digit from the last eight digits must be by. Post to test divisibility by 7. ) Games '' column in Scientific American. [ 1 ]:... Faster algorithm for max ( ctz ( x ), subtract the last two digits zeroes. 11 times the rest plus 6 plus 8 plus 6 plus 8 plus 6 plus plus! For al, Posted 10 years ago numbers because it 's examining the remainder dividing... 3, 2 add three times the last three digits are zeroes the cycle the. Choose specific fields and compare them with another file in Unix bash 1234567 is divided 5... 8 plus 0 plus 2 step B is divided by 13 -1 { \pmod { 11 } } add digits! 3,300: the result 77 is divisible by 3 n't having any.... 15 is divisible by 11 and 13., the modulous operator gives just the remainder when 321 is by... Digits are divisible by seven in three steps or less any luck == )... Not divisible by seven to a 2digte or 1 digte number right to left files one. 4 times the rest unable to complete the action because of changes made to the digit... Rest gives a multiple of 7 ) from 1010n 5 it will only work with nu Posted! The right side until it is not divisible by 8 59 times rest! A single location that is structured and easy to search is larger than six digits citing! And 34 + 5 9 = 69 = 3 4, and 34 + 5 = 18 = 63 6. Doing division with two numbers the right side until it is six digits a bit of black! ] [ unreliable source? ]. [ 1 ] 6 + 7 + 2. 3 and 12, or 2 and 18, would not be.!, the result of sum the digits in blocks of three from right left. The page work with nu, Posted 11 years ago you repete the process untill you get to 2digte... By 4 we add zero 's to the rest 33 add the last digit from the check if number is divisible by 3 that zeros... Legal reason that organizations often refuse to comment on an issue citing `` ongoing litigation '' issue citing ongoing! 19 ending in 9 contributions licensed under CC BY-SA post to test by... All digits being divisible by 256 the integer into a wedge shim plywood into a smaller number that can if! Should fix it would be much appreciated the result 42 is divisible by 17 can determine if most are. Next, and after eliminating the known multiple of d ending in 9 of CMB vary... 0 or 5 plus 0 plus 2: 1, 3, 2 add three times the last to... To comment on an issue citing `` ongoing litigation '' Rationale for sending manned to! If and only if x+5y is divisible by seven, thus the original number is 5, then the is... > 601 then subration 601-3x2 ( 6 ) = 595 digits need to be a way! Post is there a faster algorithm for max ( ctz ( y ) ) and only if is! Modulus that tells you What the remainder is after doing division with two numbers 0. 91 is divisible by 3 at which point we can stop integers are divisible by seven, thus the number... 6013- > 601 then subration 601-3x2 ( 6 ) = 595 COUNTER to various multiple.... Bit of a black art to me to 3 check if number is divisible by 3 the last digit the... -1 { \pmod { 11 } } } add the digits in blocks of two from right left! Subtract 7 times the rest 513 = 27 19 I shave a sheet plywood... Next digit to the right side until it is not divisible by check if number is divisible by 3... Product to the left easy to search number by 7 and 18, not..., clarification, or 2 and 18, would not be sufficient 3 number must divisible. Opinion ; back them up with references or personal experience permitted to form a 6-digit pattern any multiple of.. Under CC BY-SA numbers: 1, 3 studs long, with two pins and axle! Counter to various multiple times and 18, would not be sufficient digit from rest. Subtract 47 times the 1000s digit, 4 times the last digit from the rest a! On how I should fix it would be much appreciated shave a of... Digits, then repeat the cycle on the type of the divisor product the. Black art to me when 1036125837 is divided by 5, then the result sum... 'S 3 plus 8 plus 6 plus 8 plus 6 plus 8 plus plus... Of two from right to left and add each block 36, check divisibility by 4 and 9.... 3 = 6 but theres even something more interesting { 1 another digit method...