Answer to Question #350060 in Algebra for Phumza

Question

5 Test the following numbers for divisibility by 6, 9 and 11. (Do not divide or factorise.)

a) 6 798 340

b) 54 786 978

Answer 1

Solution:

Divisibility rule of 6: Numbers which are divisible by both 2 and 3 are divisible by 6. That is, if the last digit of the given number is even and the sum of its digits is a multiple of 3, then the given number is also divisible by 6;

Divisibility rule of 9: if the sum of digits of the number is divisible by 9, then the number itself is divisible by 9;

Divisibility rule of 11: if difference between the sum of digits at even places and the sum of the digits at odd places is either 0 or divisible by 11, then that number is divisible by 11. 

a) 6 798 340; Sum of its digits=$6+7+9+8+3+4+0=37;$

The last digit of given number 0, even number, but sum of the its digits is not multiple of 3. So, 6 798 340 is not divisible by 6.

Sum of the its digits is not divisible by 9. So, 6 798 340 is not divisible by 9.

For 11: let's find difference of sum even and odd places of its digits;

$(7+8+4)-(6+9+3+0)=1;$ 1 is not divisible by 11. So, So, 6 798 340 is not divisible by 11.

b) 54 786 978; Sum of its digits= $5+4+7+8+6+9+7+8=54;$

The last digit of given number is even number (8) and sum of the its digits is multiple of 3. So, 54 786 978 is divisible by 6.

Sum of the its digits is divisible by 9. So, 54 786 978 is divisible by 9.

For 11: let's find difference of sum even and odd places of its digits;

$(4+8+9+8)-(5+7+6+7)=-4;$ -4 is not divisible by 11. So, So, 54 786 978 is not divisible by 11.