kb/data/en.wikipedia.org/wiki/Divisibility_rule-2.md

title	chunk	source	category	tags	date_saved	instance
Divisibility rule	3/8	https://en.wikipedia.org/wiki/Divisibility_rule	reference	science, encyclopedia	2026-05-05T08:13:56.367542+00:00	kb-cron

Take for instance the number 371 Change all occurrences of 7, 8 or 9 into 0, 1 and 2, respectively. In this example, we get: 301. This second step may be skipped, except for the left most digit, but following it may facilitate calculations later on. Now convert the first digit (3) into the following digit in the sequence 13264513... In our example, 3 becomes 2. 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. So 301 becomes 21. Repeat the procedure until you have a recognizable multiple of 7, or to make sure, a number between 0 and 6. So, starting from 21 (which is a recognizable multiple of 7), take the first digit (2) and convert it into the following in the sequence above: 2 becomes 6. Then add this to the second digit: 6 + 1 = 7. If at any point the first digit is 8 or 9, these become 1 or 2, respectively. But if it is a 7 it should become 0, only if no other digits follow. Otherwise, it should simply be dropped. This is because that 7 would have become 0, and numbers with at least two digits before the decimal dot do not begin with 0, which is useless. According to this, our 7 becomes 0. If through this procedure you obtain a 0 or any recognizable multiple of 7, then the original number is a multiple of 7. If you obtain any number from 1 to 6, that will indicate how much you should subtract from the original number to get a multiple of 7. In other words, you will find the remainder of dividing the number by 7. For example, take the number 186:

First, change the 8 into a 1: 116. 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 = 4. So 116 becomes now 46. Repeat the procedure, since the number is greater than 7. Now, 4 becomes 5, which must be added to 6. That is 11. Repeat the procedure one more time: 1 becomes 3, which is added to the second digit (1): 3 + 1 = 4. Now we have a number smaller than 7, and this number (4) is the remainder of dividing 186/7. So 186 minus 4, which is 182, must be a multiple of 7. 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 by 7. So 2 and 9 must have the same remainder when divided by 7. The remainder is 2. Therefore, if a number n is a multiple of 7 (i.e.: the remainder of n/7 is 0), then adding (or subtracting) multiples of 7 cannot change that property. What this procedure does, as explained above for most divisibility rules, is simply subtract little by little multiples of 7 from the original number until reaching a number that is small enough for us to remember whether it is a multiple of 7. If 1 becomes a 3 in the following decimal position, that is just the same as converting 10×10n into a 3×10n. And that is actually the same as subtracting 7×10n (clearly a multiple of 7) from 10×10n. Similarly, when you turn a 3 into a 2 in the following decimal position, you are turning 30×10n into 2×10n, which is the same as subtracting 30×10n−28×10n, and this is again subtracting a multiple of 7. The same reason applies for all the remaining conversions:

20×10n − 6×10n=14×10n 60×10n − 4×10n=56×10n 40×10n − 5×10n=35×10n 50×10n − 1×10n=49×10n First method example 1050 → 105 − 0=105 → 10 − 10 = 0. ANSWER: 1050 is divisible by 7. Second method example 1050 → 0501 (reverse) → 0×1 + 5×3 + 0×2 + 1×6 = 0 + 15 + 0 + 6 = 21 (multiply and add). ANSWER: 1050 is divisible by 7. Pohlman–Mass method of divisibility by 7 The Pohlman–Mass method provides a quick solution that can determine if most integers are divisible by seven in three steps or less. 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. Step A: If the integer is 1000 or less, subtract twice the last digit from the number formed by the remaining digits. If the result is a multiple of seven, then so is the original number (and vice versa). For example:

112 -> 11 − (2×2) = 11 − 4 = 7 YES 98 -> 9 − (8×2) = 9 − 16 = −7 YES 634 -> 63 − (4×2) = 63 − 8 = 55 NO

Because 1001 is divisible by seven, an interesting pattern develops for repeating sets of 1, 2, or 3 digits that form 6-digit numbers (leading zeros are allowed) in that all such numbers are divisible by seven. For example:

001 001 = 1,001 / 7 = 143 010 010 = 10,010 / 7 = 1,430 011 011 = 11,011 / 7 = 1,573 100 100 = 100,100 / 7 = 14,300 101 101 = 101,101 / 7 = 14,443 110 110 = 110,110 / 7 = 15,730

01 01 01 = 10,101 / 7 = 1,443 10 10 10 = 101,010 / 7 = 14,430

111,111 / 7 = 15,873 222,222 / 7 = 31,746 999,999 / 7 = 142,857

576,576 / 7 = 82,368