What is the smallest digit and the greatest digit that can be inserted as the ten’s digit in the number 47652, so that the number formed is divisible by 3?

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1, 9

0, 9

2, 8

3, 9

answer is B.

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Detailed Solution

It is given that by inserting a smallest digit and the greatest digit as the ten’s digit in the number 47652, the number formed will be divisible by 3.
When the sum of the digits in a number are divisible by 3, then the number is also divisible by 3.
Sum of the given digits is,
= 4 + 7 + 6 + 5 + 2
=24.
If we add 0, then the digit is divisible by 3.
i.e., 24 + 0 = 24.
Hence, 0 is the smallest digit.
If we add 9, then the digit is divisible by 3.
i.e., 24 + 9 = 33.
Hence, 9 is the largest digit.
The smallest digit is 0 and can be written as 476502 and the greatest digit is 9 and can be written as 476592, where 476502 and 476592 are divisible by 3.
Therefore, option 2 is correct.

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