If \(7254*98\) is divisible by 22, the digit \(*)\) is
1
2
6
0
Key idea: A number divisible by 22 must be divisible by both 2 and 11.
Step 1: Check divisibility by 2
Last digit is 8, so it is divisible by 2.
Step 2: Check divisibility by 11
Use the alternating-sum rule: (sum of digits in odd places) − (sum of digits in even places) must be 0 or a multiple of 11.
Number: \(7\;2\;5\;4\;*\;9\;8\)
Odd places (1st, 3rd, 5th, 7th): \(7,\;5,\;*,\;8\)
Even places (2nd, 4th, 6th): \(2,\;4,\;9\)
Sum(odd) = \(7 + 5 + * + 8 = 20 + *\)
Sum(even) = \(2 + 4 + 9 = 15\)
Difference = \((20 + *) - 15 = * + 5\)
For divisibility by 11: \(* + 5\) must be \(0, 11, 22,\ldots\)
Since \(*)\) is a single digit (0–9), \(* + 5\) can be from 5 to 14.
The only value from \(\{5,6,\ldots,14\}\) that is a multiple of 11 is \(11\).
So, \(* + 5 = 11 \Rightarrow * = 6\).
Answer: \(6\) (Option C).