If 2 is added to each odd digit and 1 is subtracted from each even digit in the number 8354671, then how many digits will appear more than once in the new number thus formed?

Correct Answer: Three

Step-by-step Explanation:

Step 1: Identify the given number and its digits

The original number given in the problem is 8354671.

Let's list all the individual digits in this number:

8, 3, 5, 4, 6, 7, 1

Step 2: Apply the given rules to each digit

According to the problem instructions:

1. If a digit is odd, add 2 to it (Digit + 2).

2. If a digit is even, subtract 1 from it (Digit - 1).

Let's perform the operations for each digit from left to right:

• 8 is an even digit: $8-1=7$

• 3 is an odd digit: $3+2=5$

• 5 is an odd digit: $5+2=7$

• 4 is an even digit: $4-1=3$

• 6 is an even digit: $6-1=5$

• 7 is an odd digit: $7+2=9$

• 1 is an odd digit: $1+2=3$

Step 3: Form the new number and check digit frequencies

The new sequence of digits formed is 7, 5, 7, 3, 5, 9, 3 (or 7573593 as a whole number).

Now, let's count the frequency of each digit present in the new number:

• Digit 7 appears 2 times (appears more than once)

• Digit 5 appears 2 times (appears more than once)

• Digit 3 appears 2 times (appears more than once)

• Digit 9 appears 1 time

Conclusion:

The digits that appear more than once are 7, 5, and 3.

Therefore, exactly Three digits appear more than once in the newly formed number.