Wednesday, 1 February 2012

Divisibility test for 59

I had asked readers to attempt a  test for divisibility by 59 of 2 numbers based  on the procedure shown by me for testing divisibility by 17 in an earlier blog.The same procedure is applicable for both because the product of 17 and 59 is 1003 and the procedure was based on this fact.
I had suggested that a minor alteration is needed.
This alteration is to add a multiple of 59( instead of a multiple of 17) for evenly dividing by 3.I am giving below the steps for one of the numbers viz 50364080202 indicated by me as this involves a new fact not specifically explained by me.
Number under test-50364080202
Step 1--Groups  A-202...B-080...C-364...D-50
Step 2--A multiple of 59 has to be added to 202 as this is not divisible by 3.We add 59 altering 202 as 261.Division by 3 gives us 87.
Step 3-As 87 or 087 is more than 080 in Group B we borrow 1 from Group C .Thus 080 becomes 1080 and subtraction of 087 will give us 993.The number in Group C is now 363 as 1 has been borrowed from 364.
Step 4-Division of 993 by 3 gives us 331 and subtraction of 331 from 363 will give us 032
Step 5-As 032 is not evenly divisible by 3 we add 118( a multiple of 59) to it making it 150 .Division by 3 will give us 050.
Step 6-This is now  to be deducted from the number in Group D and the balance tested for divisibility by 59.But there is no balance as the subtraction of 050 from the number 050 in Group D gives us zero.I had not specifically indicated earlier what should be done in such cases.I have to state that in such cases the final figure should be treated as evenly divisible by the divisor viz 59.-as zero times 59 is zero.
The final conclusion is that the test number viz 50364080202 is divisible by 59.
The actual quotient is 853628478
Readers may now test the second number mentioned by me viz 5036402781   . 

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