There is one for 7, although it is not very useful. Take the last digit, truncate it and double it. Then subtract it from the rest. If the resulting number is divisible by 7, so was the original. For 8, if the last 3 digits are divisible by, so is the number in question. For example, 72248 is divisible by 8 because 248 is. And 259 is divisible by 7 because 9 times 2 is 18, and 25 minus 18 is 7... Voila!
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A number is a multiple of 312 if it's a multiple of 3, 8 and 13 at the same time
Those for 1, 2, 4, 5 and 8.
Any multiple of two must end in 0, 2, 4, 6 or 8.
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It is: 7
If the last digit doubled subtracted from the rest is a multiple of 7, the whole number is divisible by 7.
3+7=10
While it is possible to come up with divisibility rules for any number, some divisibility rules are more complicated than others, and in some cases it becomes easier to just carry out the division. A rule that works for 8 is: If the last three digits of a number are divisible by 8, then the entire number is divisible by 8. For 7: Separate a number's last three digits, and divide from the remaining number. If the result is divisible by 7, the original number is divisible by 7. For example, in 31969, separate as 31; 969. Subtracting the second part from the first will give you a negative number, but you can also subtract the other way: 969 - 31. Since the result is divisible by 7, the original number also is. This SAME rule also works for 11 and 13, and is based on the fact that 7, 11 and 13 are factors of 1001.
The divisibility rules for a prime number is if it is ONLY divisible by 1, and itself.
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If the last 3 digits are divisible by 8 and the sum of the digits are divisible by 9.
Oh honey, divisibility rules have been around longer than your grandma's secret meatloaf recipe. But if you want a name to drop at your next trivia night, credit goes to good ol' Euclid. He's the OG mathematician who laid down the law on how numbers can play nice and divide evenly.
A number is a multiple of 312 if it's a multiple of 3, 8 and 13 at the same time
Those for 1, 2, 4, 5 and 8.
Any multiple of two must end in 0, 2, 4, 6 or 8.
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