Sunday, 15 January 2012

Divisibility test for 17

Under my blog account Nadamadum numbers I had given details about a divisibility test for 7,11 and13.This was based on the fact that the product of 7,11 and 13 is 1001.As the product of 17 and 59 is 1003,I have discovered a test for divisibility for 17 on a similar basis.Details are shown below-
Three examples were taken for the test-numbers 159327944,..65482725...45327029331.
I will give full step by step details for the first example 159327944.
Step 1-The number is divided into groups of 3 digits starting from right.They are A-944,B-327,C-159
Step2-You have to divide the number in group A by 3 and subtract the result from group B.If the number in group A is not evenly divisible by 3,you can add a small multiple of 17 to it before division.Thus we add 34 to 944 in group A,making it 978 and the result of division by 3 is thus 326.Subtracting from 327 in group B,we have the result 001.
Step3-We have now to divide the new number in group B by 3 and similarly subtract the result from
Group C..We add 17 to 001 making it 018 and the result of division by 3 being 006,we subtract 006 from 159 in group C making in 153.
Step4-We now test the final number in group C for divisibility by 17.If it is so divisible the original start number would be divisible by 17.In fact 153 is divisible by 17 and hence the test number 159327944 is divisible by 17.The actual quotient is 9372232.
Example 2-Number 65482725.
In this case the number in group B(the penultimate group) will come as 229.If we add 17 to it for making it divisible by 3,the number to be subtracted fromC(the final group) comes as 82 but the number in C is less viz 65.We therefore have to change the procedure and instead of subtracting one-third of B from C and testing the same,we subtract 3 times the number in C from B and test the same.Thus we subtract 3*65=195 from 229 giving us the result 34.This result is divisible by 17 and hence the start number 65482725 will be divisible by 17.The quotient is actually  3851925.
Example 3-Test number 45327029331,-Here we will have a fourth group D=45.
The changed procedure shown in Example 2 is adoptable only when the number to be subtracted from the next group is the penultimate group.Otherwise like in a normal subtraction sum we can borrow 1 from the next group before subtraction..Thus for subtracting (331+17)/3=116 from 029 in group B we borrow 1 from group C and subtract.The new number in group B will be 1029 and the subtraction will give us the result 913 in group B while the number in group C will now be 326 after the borrowal of 1 from 327.The rest of the procedure is the same.The final figure in group D for testing comes as 45-11=34 and as this is divisible by 17 the test/start number viz 45327029331 would be divisible by 17.In fact the actual quotient is 2666295843.
You have to read the steps carefully and make the calculations.Read the steps a couple of times if neceesary..

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