If 1 is added to each odd digit and 2 is subtracted from each even digit in the number 7436518, how many digits will appear more than once in the new number thus formed?

Given: 7436518

Add 1 to each odd digit and subtract 2 from each even digit.

7 odd → 7 + 1 = 8

4 even → 4 − 2 = 2

3 odd → 3 + 1 = 4

6 even → 6 − 2 = 4

5 odd → 5 + 1 = 6

1 odd → 1 + 1 = 2

8 even → 8 − 2 = 6

So, the new number formed is: 8, 2, 4, 4, 6, 2, 6

2, 4 and 6 are appears twice in new number.

3 digit will appear more than once in the new number thus formed.

Thus, correct option is (c).