Rule :
If a number is divisible by both 2 and 3 then it is divisible by 6.
Problem 1 :
Which of these numbers are divisible by 6?
a) 162 b) 381 c) 1602 d) 2156 e) 5364
Solution :
If a number is divisible by both 2 and 3, then it is divisible by 6.
We already know that all even numbers are divisible by 2
So, all even numbers which are divisible by 3 are divisible by 6.
a) 162
The given number 162 ends with 2.
So, it is even number and divisible by 2.
Check whether the number 162 is divisible by 3.
Add all the digits.
1 + 6 + 2 = 9
The sum of the digits in the given number 162 is 9 which is a multiple of 3.
Therefore, the given number 162 is divisible by both 2 and 3.
So, the given number 162 is divisible by 6.
b) 381
The given number 381 ends with 1.
So, it is not an even number and it is not divisible by 2.
Check whether the number 381 is divisible by 3.
So, the given number 381is not divisible by 6.
c) 1602
The given number 1602 ends with 2.
So, it is even number and divisible by 2.
Check whether the number 1602 is divisible by 3.
Add all the digits.
1 + 6 + 0 + 2 = 9
The sum of the digits in the given number 1602 is 9 which is a multiple of 3.
Therefore, the given number 1602 is divisible by both 2 and 3.
So, the given number 1602 is divisible by 6.
d) 2156
The given number 2156 ends with 6.
So, it is even number and divisible by 2.
Check whether the number 2156 is divisible by 3.
Add all the digits.
2 + 1 + 5 + 6 = 14
The sum of the digits in the given number 2156 is 14 which is a multiple of 3.
Therefore, the given number 2156 is divisible by 2, but not by 3.
So, the given number 2156 is not divisible by 6.
e) 5364
The given number 5364 ends with 4.
So, it is even number and divisible by 2.
Check whether the number 5364 is divisible by 3.
Add all the digits.
5 + 3 + 6 + 4 = 18
The sum of the digits in the given number 5364 is 18 which is a multiple of 3.
Therefore, the given number 5364 is divisible by both 2 and 3.
So, the given number 5364 is divisible by 6.
Problem 3 :
What smallest number should be added to 4456 so that sum of the number is completely divisible by 6 ?
a) 4 b) 3 c) 2 d) 1
Solution :
If the number is divisible by 6, then the number must be divisible by 2 and 3.
Checking for divisible by 2 :
If the number is divisible by 2, then it must be a even number.
4456 + 4 ==> 4460 (even)
4456 + 2 ==> 4458 (even)
Checking for divisible by 3 :
Sum of the number 4456 is
= 4 + 4 + 5 + 6
= 19
19 + 4 ==> 23 (not divisible by 3)
19 + 2 ==> 21 (divisible by 3)
So, 2 is the smallest number to be added to make it as divisible by 6.
Problem 3 :
Paul believes that the number forms alongside are always divisible by 6 :
a) Check that the first four of them are divisible by 6.
(i) 2^{3} – 1^{3} – 1
(ii) 3^{3} – 2^{3} – 1
(iii) 4^{3} – 3^{3} – 1
(iv) 5^{3} – 4^{3} – 1
b) Check that 10^{3} – 9^{3} – 1 is divisible by 6.
Solution :
i) To check :
2^{3} – 1^{3} – 1
= 8 – 1 – 1
= 6
So, 2^{3} – 1^{3} – 1 is divisible by 6.
ii) 3^{3} – 2^{3} – 1
= 27 – 8 – 1
= 18
18 is even number and it is multiple of 3. So, it is divisible by 6.
iii) 4^{3} – 3^{3} – 1
= 64 – 27 – 1
= 36
36 is even number and it is multiple of 3. So, it is divisible by 6.
So, it is even number and divisible by 2.
iv) 5^{3} – 4^{3} – 1
= 125 – 64 – 1
= 60
60 is even number and it is multiple of 3. So, it is divisible by 6.
b) To check :
= 10^{3} – 9^{3} – 1
While observing the above questions, this also follows the same pattern. So, 10^{3} – 9^{3} – 1 it is also divisible by 6.
May 21, 24 08:51 PM
May 21, 24 08:51 AM
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