62 lines
5.0 KiB
Markdown
62 lines
5.0 KiB
Markdown
---
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title: "Divisibility rule"
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chunk: 3/8
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source: "https://en.wikipedia.org/wiki/Divisibility_rule"
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category: "reference"
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tags: "science, encyclopedia"
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date_saved: "2026-05-05T08:13:56.367542+00:00"
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instance: "kb-cron"
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---
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Take for instance the number 371
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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.
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Now convert the first digit (3) into the following digit in the sequence 13264513... In our example, 3 becomes 2.
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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.
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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.
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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.
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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:
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First, change the 8 into a 1: 116.
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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.
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Repeat the procedure, since the number is greater than 7. Now, 4 becomes 5, which must be added to 6. That is 11.
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Repeat the procedure one more time: 1 becomes 3, which is added to the second digit (1): 3 + 1 = 4.
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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.
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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.
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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.
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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.
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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:
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20×10n − 6×10n=14×10n
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60×10n − 4×10n=56×10n
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40×10n − 5×10n=35×10n
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50×10n − 1×10n=49×10n
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First method example
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1050 → 105 − 0=105 → 10 − 10 = 0. ANSWER: 1050 is divisible by 7.
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Second method example
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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.
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Pohlman–Mass method of divisibility by 7
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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.
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Step A:
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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:
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112 -> 11 − (2×2) = 11 − 4 = 7 YES
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98 -> 9 − (8×2) = 9 − 16 = −7 YES
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634 -> 63 − (4×2) = 63 − 8 = 55 NO
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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:
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001 001 = 1,001 / 7 = 143
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010 010 = 10,010 / 7 = 1,430
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011 011 = 11,011 / 7 = 1,573
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100 100 = 100,100 / 7 = 14,300
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101 101 = 101,101 / 7 = 14,443
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110 110 = 110,110 / 7 = 15,730
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01 01 01 = 10,101 / 7 = 1,443
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10 10 10 = 101,010 / 7 = 14,430
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111,111 / 7 = 15,873
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222,222 / 7 = 31,746
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999,999 / 7 = 142,857
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576,576 / 7 = 82,368 |