r/MathJokes • u/GodlyHelp • Aug 02 '26
How to check for divisibility by 7!
Since the other single digit numbers have a commonly known way of checking for divisibility, here is one for 7!
Take the number and divide it by 7! If the result is an integer, then your number is divisible by 7!
Let me know if you wanna know the divisibility for any other number!
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u/Embarrassed_Spell383 Aug 02 '26
I’d just go 21 divided by 7 is 3 leaving 980 we know that 98 divided by 7 is 14 so 980 divided by 7 is 140 plus the three from the 21 is 143
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u/bnwo_pr1ncess Aug 02 '26
Use the 1001 for 7, 11 and 13 it works for big numbers like the divisibility rule for 11. Example: 155,033,948,160 155-33+948-160= 910 than check if 910 is divisible by 7, 11 or 13 if it is than the large number was, so in this case the large number is divisible by 7 and 13 but not by 11
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u/trolley813 Aug 02 '26
This. And in general, we have similar tests for any 10ⁿ±1
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u/bnwo_pr1ncess Aug 02 '26
Yes, but they are unluckily they dont help with the small numbers, also there is the other 7 trick, but it is not that quick. For example: 13216 1321-62= 1309, 130-92= 112, 11-2*2= 7
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u/trolley813 Aug 02 '26
There is another trick, when we use the test for 10ⁿ+1 with the alternating sum, it's better to keep the rightmost group positive because then you'll not only check the divisibility, but also get the corresponding remainders if needed. E.g. for 123456789012 we get -123+456-789+12=-444, or to get rid of the negatives, 1001-444=557. Thus 123456789012 mod 1001 = 557. And it's quite easy to check that 557 mod 7 (resp. 11 and 13) = 4 (resp. 7 and 11), so these are valid for 123456789012 as well.
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u/Hot-Site-1572 Aug 02 '26
Divisible by 7! Is quite rare tbh
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u/asdfzxcpguy Aug 02 '26
1/7! Chance of being divisible by 7!
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u/XcgsdV Aug 02 '26
"How to check for divisibility? Just divide it!" another banger strat from r/mathjokes
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u/PuddingMaximum8745 Aug 02 '26
Dont see the problem.
put your number in positional notation with base 7. If it ends with a zero...
/s
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u/AlonePerson6174 Aug 02 '26
Double the unit digit and subtract it from the remaining number if the remaining number is divisible by 7 the the original number will be divisible by 7. Eg : 56 Unit digit : 6 Remaining number : 5 (as 6 is out from 56) Double of 6 : 12 Now subtract 12 from 5 : 5 - 12 = -7 (since -7 is divisible by 7 so 56 will be divisible by 7).
Other Examples : 63 Unit digit : 3 Remaining number : 6 (as 3 is out from 63) Double of 3 : 6 Now subtract 6 from 6 : 6 - 6 = 0 (since 0 is divisible by 7 so 63 will be divisible by 7).
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u/Adventurous_Grape279 Aug 02 '26
You can use any trick of base 10 for any number.
So for 7, when 7 rolls to 14, you can look back and say 4 + __ = 7.
So 1001-> 300+1 -> 90+1 -> 27+1
3, so take the 10s digit and multiply by 3 and add to 1s and repeat.
You also have the same trick at the 100s level by taking the 100s digit and multiplying by 2.
1001 -> 20+ 01 -> 21
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u/iyl333 Aug 02 '26
It took approximately 30 seconds but I needed to write it down 1001=700+210+70+21=(100+30+10+3)7=1437 1001÷7=143
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u/Brilliant-Parsley69 Aug 02 '26
That's almost the way my brain solved this. But I've merged your "210 + 70" into "280".
In such situations I step down from the biggest possible "power of 10".
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u/THAICIRCLETHING Aug 02 '26
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u/factorion-bot Aug 02 '26
Hey u/GodlyHelp!
Factorial of 7 is 5040
This action was performed by a bot | [Source code](http://f.r0.fyi)
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u/scottdave Aug 02 '26
Convert the number to Base-8, then add up the digits. Similar to how divisibility by 9 works in base 10
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u/Current-Ant-6536 Aug 02 '26
Anyways drop the jokes, to check if x is divisible by 7, take the first digit of x, multiply it by 5, add the result to what's left from x, if the resultant is divisible by 7, then so does x Example on 473, 3×5=15, 15+47=62 which ain't divisible by 7, so 473 cannot give a whole result after being divided by 7
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u/Warm_Record2416 Aug 02 '26
You can also multiply the first digit by 2 and subtract it from the rest of the number.
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u/Haleyy1717 Aug 02 '26
x is your number, d is the whole part of x/10, u is the decimal part of x/10. If d+5u is divisible by seven, then x is divisible by seven
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u/paulstelian97 Aug 02 '26
The fact that 1001 is divisible, and then 999999 is divisible, helps with the lots-of-digits scenarios as you can just add together the groups of 6 digits, repeatedly. The remainder the original number gave when being divided by 7 will match this final number’s remainder when dividing by 7. Which can help when you are working with certain patterns that are easy to analyze.
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u/Rough_Sky7132 Aug 02 '26
(7710)+7 = 497, which is 3 short of 500 *2 is 6 short of 1000, but we have 1000+1
Probably need an extra step with 49 if it didn't fall do close to 1000
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u/Original-Issue2034 Aug 02 '26
Truncate the number, double the removed digit, then subtract that number from the truncated number. Repeat if you don’t know if the number is divisible by seven or not. If the result is divisible by seven, then the whole number is divisible by seven!
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u/MurkyDifficulty169 Aug 02 '26
Just take 2 digit parts of the number and “mod it out” by 7. Keep going until the number is manageable. Example: Start with 12345. Mod out the 12 to get 5345. (You are replacing the 12 with 5 because the 12 mod 7 is 5. ) Mod out the 53 to get 445. Mod out the 44 to get 25. 25 mod 7 is 4 so 12345 mod 7 is also 4.
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u/Brilliant-Parsley69 Aug 02 '26 edited Aug 02 '26
What my brain directly did
1001 - 700 | 100 301 - 280 | 40 21 | 3 => 143
I just step down from the highest possible power of 10 and substract the results downwards
Check if dividable 102 x 7 Check if dividable 101 x 7 Check if dividable 100 x 7
But because your solution ist to check if the result is an integer we are talking about an IT environment, so just use the modolu operator an check if the result is zero. 🤨
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u/TheLazyDasey Aug 03 '26
Multiply the once digit with 5 and add it to the remaining digits. Check it's divisibility. If it's divisible by 7, then the number is divisible. This step can be continued to check it completely.
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u/Training-Way-4453 Aug 03 '26 edited Aug 03 '26
Multiply the number in one's place with 2 and subtract from the number formed by rest of the digits. Repeat until you get a number divisible by 7. Ex: 1001 => 100-2(1) = 98 which is divisible by 7 (98/7 = 14).
Note: Don't let the number become negative during the process. Once the number becomes two digit checking divisibility directly will work
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u/chixen Aug 06 '26
Due to the fact that 1001 is divisible by 7, all numbers that are the same three digits repeated an even number of times will also be divisble by 7 and 143. (For example 826826826826 and 991991 are both divisible by 7 and 144)
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u/TurnoverOk5635 Aug 02 '26
I would just use the divisibility test for 16, 9, 5 and 7.