Octal divisibility

Number Theory Level 4

In base $10$ , there are simple divisibility rules formulated for division by $2,3,4,5,6,8,9$ and $11$ based on last digit, sum of digits or difference of alternate digits,(i.e., all numbers less than or equal to $(10+1)$ . The number $1$ is of course ignored in this context.

Which is the only number less than or equal to $9_{10}=11_{8}$ , for which a divisibility rule cannot be formulated in base $8$ , based on the either last digit(s), sum of digits or difference of alternate digit sums. The answer is to be given in base $10$ ?.

4 9 5 There is no such number 3 7 There are more than one such numbers 6

2 solutions

Abhay Tiwari
Jun 16, 2016

When looking in base 8

Divisibility for 2 , will be the same as in base 10.
Divisibility for 3 , will be equivalent to the divisibility for 11 in base 10. Reason being in base n, n+1 follows the alternate digit addition rule, similarly in base 8, n+1 $={{(11)}_{8}}={{(9)}_{10}}$ , which is a factor of 9, and so this rule follows for 3 also in base-8.
For 4 , any number ending in 4 or 0 will be divisible by 4.
For 5 , Since we are looking in base 8, which is not a divisible by 5, and neither , $n-1$ (so that sum of digit rule can be applied) or $n+1$ (so that alternate digit sum rule can be applied) is divisible by 5. So no rule for 5.
For 6 , Any number divisible by 2 and 3 will be divisible by 6.
For 7 , sum of digit rule, same as that we follow for 9 in base 10.
For ${{11}_{8}}$ , same as that we use for 11 in base 10. For divisibility by 8 is any number ending in zero, same as that we use for 10 in base-10.

Keanu Ac
May 8, 2017

Converting base numbers in base 10 with divisibility rules, note that 5 has no divisibility rules as it serves the same function as 7 in base 10.

Octal divisibility

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