General Divisibility Rule for any Number Ending 1, 3, 7 or 9 : GMAT Quantitative Section
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# General Divisibility Rule for any Number Ending 1, 3, 7 or 9

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General Divisibility Rule for any Number Ending 1, 3, 7 or 9 [#permalink]

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22 Apr 2014, 07:17
This was discovered by me while I was studying divisibility rules. This is true for any number ending with 1, 3, 7 or 9.

We have a number P which is divided in XY where Y is unit’s digit and X is rest of the no. Hence P=10*X+Y..e.g if we have a no P=1643 then X=164 and Y=3.

Based on the above assumption, consider the following rules:
A no P=(10*X+Y) is divisible by M if N is divisible by M.

M: 3 13 23 33 43 53 63 73 83 93
N: X+Y X+4Y X+7Y X+10Y X+13Y X+16Y X+19Y X+22Y X+25Y X+28Y

M: 7 17 27 37 47 57 67 77 87 97
N: X-2Y X-5Y X-8Y X-11Y X-14Y X-17Y X-20Y X-23Y X-26Y X-29Y

M: 9 19 29 39 49 59 69 79 89
N: X+Y X+2Y X+3Y X+4Y X+5Y X+6Y X+7Y X+8Y X+9Y

M: 11 21 31 41 51 61 71 81 91
N: X-Y X-2Y X-3Y X-4Y X-5Y X-6Y X-7Y X-8Y X-9Y

If we observe all the Divisibility rules follow a known pattern.

In general the following hypothesis is true. To find the divisibility of any number whose end values are 1, 3, 7 or 9 we have the following test:

Number XY (10*X+Y) is divisible by A1 where 1 is unit place and A is Ten's place Divisibility Test will be: X - (A * Y) is divisible by A1
Number XY (10*X+Y) is divisible by A3 where 3 is unit place and A is Ten's place Divisibility Test will be: X + [(3A+1)*Y] is divisible by A3
Number XY (10*X+Y) is divisible by A7 where 7 is unit place and A is Ten's place Divisibility Test will be: X - [(3A+2)*Y] is divisible by A7
Number XY (10*X+Y) is divisible by A9 where 9 is unit place and A is Ten's place Divisibility Test will be: X + [(A+1) * Y] is divisible by A9

This is the General Divisibility Rule for any number ending with 1. 3, 7 or 9.

Regards
Souvik Jana..
General Divisibility Rule for any Number Ending 1, 3, 7 or 9   [#permalink] 22 Apr 2014, 07:17
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# General Divisibility Rule for any Number Ending 1, 3, 7 or 9

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