Question

write the smallest digit and greatest possible digit in the blank space of each of the following numbers so that the numbers formed are divisible by 3​

Answer 1

Answer:

Thus, The smallest digits to be placed in blank space = 0. Then, sum = 24 + 0 = 24, which is divisible by 3. The greatest digit to be placed in blank space = 9. Then, the sum = 24 + 9 = 33, which is divisible by 3.

Step-by-step explanation:

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Answer 2

Answer:

Write the smallest digit and the greatest digit in the blank space of each of the following number so that the number formed is divisible by 3

(a) ____ 6724 (b) 4765 ____ 2

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Solution:

We will be using the concepts of divisibility by 3 to solve this.

A number is divisible by 3 if the sum of all digits in the number is also divisible by 3.

answer

(a) ___ 6724

Sum of the digits = 4 + 2 + 7 + 6 = 19

Thus, The smallest digit to be placed is blank space = 2.

Then the sum = 19 + 2 = 21, which is divisible by 3.

The greatest digit to be placed in blank space = 8.

Then, the sum = 19 + 8 = 27, which is divisible by 3

Therefore, the required digits are 2 and 8.

(b) 4765 ____ 2.

Sum of digits = 2 + 5 + 6 + 7 + 4 = 24

Thus, The smallest digits to be placed in blank space = 0.

Then, sum = 24 + 0 = 24, which is divisible by 3.

The greatest digit to be placed in blank space = 9.

Then, the sum = 24 + 9 = 33, which is divisible by 3.

Therefore, the required digits are 0 and 9.

Step-by-step explanation:

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