Question
What minimum value should be assigned to 'x' so that 2346x45 is exactly divisible by 9?
This question was previously asked in
RRB NTPC 2019 CBT 1 Question Paper (8-Mar-2021) (Shift 2)
The correct answer is
3
Finding Minimum Value for Divisibility by 9
The question asks for the minimum digit 'x' that makes the number 2346x45 divisible by 9.
Divisibility Rule for 9
A number is exactly divisible by 9 if the sum of its digits is divisible by 9.
Applying the Rule
Let the number be $N = 2346x45$.
The sum of the digits of N is:
$ S = 2 + 3 + 4 + 6 + x + 4 + 5 $Calculate the sum of the known digits:
$ S = (2 + 3 + 4 + 6 + 4 + 5) + x $ $ S = 24 + x $For N to be divisible by 9, the sum S must be a multiple of 9.
Determining the Minimum Value of 'x'
We need to find the smallest single digit (0-9) for 'x' such that $24 + x$ is a multiple of 9.
- List multiples of 9: 9, 18, 27, 36, ...
- Test values for $24 + x$:
- If $24 + x = 27$, then $x = 27 - 24 = 3$.
Since $x=3$ is a single digit and is the smallest value that satisfies the condition, it is the minimum value required.
Therefore, the minimum value for 'x' is 3.
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