write the smallest digits and greatest digits in the black space each of the following numbers so that the number formula is divisible by 3 b)4765 2
Answers
Answer:
Hint: -Any number is divisible by 3, if the sum of its digits is divisible by 3.
Given number:
4765....2.
It is given that the number is divisible by 3.
By the divisibility rule of 3, any number is divisible by 3, if the sum of its digits is divisible by 3.
Let the digit on the blank space be b.
4+7+6+5+b+2⇒b+24
As we see that 24 is divisible by 3 = 243=8 times.
Therefore in order to make the required number divisible by 3, the least value of b=0 and the maximum value of b=9.
If b=0
, then0+24=24
. This is divisible by 3.
So, the number becomes 476502
If b=9
, then9+24=33
. This is divisible by 3.
So, the number becomes 476592
Therefore the smallest digit is 0 and the greatest digit is 9.
Note: - In such types of questions the key concept we have to remember is that always remember the divisibility rule of 3 which is stated above, then according to this rule calculate the minimum and maximum value in the black space so that number is exactly divisible by 3.
Step-by-step explanation:
4765....2.
It is given that the number is divisible by 3.
By the divisibility rule of 3, any number is divisible by 3, if the sum of its digits is divisible by 3.
Let the digit on the blank space be b.
4+7+6+5+b+2
⇒b+24
As we see that 24 is divisible by 3 = 243=8 times.
Therefore in order to make the required number divisible by 3, the least value of b=0 and the maximum value of b=9.
If b=0, then
0+24=24. This is divisible by 3.
So, the number becomes 476502
If b=9, then
9+24=33. This is divisible by 3.
So, the number becomes 476592
Therefore the smallest digit is 0 and the greatest digit is 9.