Divisibility tests for various numbers
A simple test for large numbers for divisibility by 7, 11 and 13
Take groups of three digits of the candidate number, starting from the last three, and alternately subtract and add each group of digits (the last one might have less than three). If the end result is zero, or divisible by any of the three numbers 7, 11 and 13 then so is the whole candidate number.
For example: 1008049343 is divisible by 7, because 343-49+8-1=287, which is divisible by 7.
The commonest divisibility test for 11
Alternately subtract and add the digits of the candidate number, starting from the last one. If the result is zero or divisible by 11, then the candidate number is divisible by 11.
For example, 161051 is divisible by 11, because 1-5+0-1+6-1=0. Or, you could have used 51-161=-110 as above. -110 is divisible by 11, so 161051 is divisible by 11.
A relatively simple divisibility test for 7
Starting from the last digit of the candidate number, multiply it by 2, and alternately subtract and add the next digit. Multiply the result by 2, and continue this way to the first digit. Remember to multiply your result by 2 each time, or else the 'trick' won't work.
For example: 1008049343 as above. 3*2-4=2. 2*2+3=7. 7*2-9=5. 5*2+4=14. 14*2-0=28. 28*2+8=64. 64*2-0=128. 128*2+0=256. 256*2-1=511=7*73.
Base change
A test for both divisibility by 7 and 49 consists of taking chunks of two digits from the end of the candidate number. Multiply the second pair by 2, the next by 4, the next by 8 and so on, and add the results. If the sum total is divisible by 7, the original number is divisible by 7, if it is divisible by 49, the original number is divisible by 49. In this view, there is also the test for divisibility by 17, whereby, the multiplied chunks are alternately subtracted and added, starting from the end, just as the single digits in the divisibility by 11 test, and the end result is checked for divisibility by 17. Essentially, we change from base 10 to base 50 and check for divisibility by that base's equivalents of 9 and 11.
A simple test for divisibility by 37
This works quite simply. You take groups of three digits as in the first test, and always add!
For example 456790086 is divisible by 37, because 86+790+456=1332=332+1=333=9*37.
Okay, I'll give you a clue. If you do exactly the same thing except with groups of two digits, you have the simplest test for divisibility by 11.
And, last but not least...
A test for divisibility by 19
Starting from the last digit of the candidate, multiply it by 2, and add the next digit. Multiply the result by 2, and continue this way to the first digit. Repeat that procedure for the total sum, if its divisibility by 19 is not immediately obvious.
For example, 257876892 is divisible by 19, as 2*2=4, 4+9=13, 13*2=26, 26+8=34, 34*2=68, 68+6=74, 74*2=148, 148+7=155, 155*2=310, 310+8=318, 318*2=636, 636+7=643, 643*2=1286, 1286+5=1291, 1291*2=2582, 2582+2=2584. 4*2=8, 8+8=16. Okay, I'll let you in on a little secret. We can multiply 16*4=64. Then, taking the other two digits, 5*2=10, 10+2=12. 64+12=76=19*4, and so this formidable number is divisible by 19.
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