## Elementary Math (K-6th) Tutorial

#### Intro

The rules for divisibility helps to easily determines when a number divides another number exactly without any remainder.

#### Sample Problem

The rules for the divisors of whole numbers is given below :
Any whole number is exactly divisible by
(1.) 2 if the last digit is even or 0 .
(2.) 3 if the sum of its digit is divisible by 3 .
(3.) 4 if the last two digit form a number divisible by 4 .
(4.) 5 if the last digit is 5 or 0.
(5.) 6 if the last digit is even and the sum of the digits is divisible by 3
(6.) 8 if the last three digits form a number divisible by 8 .
(7.) 9 if the sum of its digit is divisible by 9
(8.) 10 if the last digit is 0.
NOTE :There is no easy rule for division by 7

#### Solution

EXAMPLE : Test this numbers to which are exactly divisible by 9
(a) 51060 (b) 9039 (c ) 48681
SOLUTION
(a) 51060 = 5 + 1 + 0 + 6 + 0 = 18
18 is divisible by 9 .Thus , 51060 is divisible by 9

(b) 9039 = 9 + 0 + 3 + 9 =21
21 is not divisible by 9 . Thus 9039 is not divisible by 9

(c) 48681 = 4 + 8 + 6 + 8 + 1 = 27
27 is divisible by 9 . Thus 48681 is divisible by 9

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 Fasesin J Math Expert My name is Fasesin James Opeyemi.I am a christain from nigeria based in lagos .i finished from YABA COLLEGE OF TECHNOLOGY ,YABA,LAGOS,NIGERIA.There i study electrical/electronics engineering.I started my teaching career in the year 2000 with MERIT COMPREHENSIVE COLLEGE,JAKANDE ESTATE,LAGOS.I Joined ...
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