Sam has forgotten his friend's seven-digit telephone number. He remembers the following: the first three digits are either 635 or 674; the number is odd, and the number nine appear... Sam has forgotten his friend's seven-digit telephone number. He remembers the following: the first three digits are either 635 or 674; the number is odd, and the number nine appears once. If Sam were to use a trial-and-error process to reach his friend, what is the minimum number of trials he has to make before he can be certain to succeed? Select the appropriate answer from the options given below: (a) 1000 (b) 2430 (c) 3402 (d) 3006 Read more

Steps to Solve

1. Identify the Possible Starting Digits

   Sam's friend's number starts with either 635 or 674, giving us 2 possible options for the first three digits.

2. Determine the Characteristics of the Remaining Digits

   The remaining four digits must form a valid 7-digit telephone number that is odd. Thus, the last digit (7th digit) must be an odd number: 1, 3, 5, 7, or 9.

3. Define the Placement of the Digit '9'

   Since the digit '9' appears once in the number and is odd, the last digit could be '9', or the '9' can be placed in any of the first six positions (except the 7th). We will analyze both scenarios:

   • If '9' is the last digit.
   • If '9' is one of the first six digits.

4. Count the Combinations with '9' as the Last Digit

   If '9' is the last digit, we then have three remaining positions for which we need to determine valid digits (0-9, excluding '9'). The first digit must also be odd (1, 3, 5, 7), but the second and third digits can each be any digit:

   • Possible odd digits for the first digit: 1, 3, 5, 7 = 4 options.
   • Possible choices for the second digit: 0-9 (but excluding what has been chosen so far) = 9 options.
   • Possible choices for the third digit: 0-9 (but excluding what has been chosen so far) = 8 options.

   Therefore: $$ \text{Total combinations} = 4 \times 9 \times 8 = 288 $$

5. Count the Combinations with '9' not as the Last Digit

   If '9' is in one of the first six positions, assume it could be the first, second, third, fourth, fifth, or sixth digit:

   • Total positions for '9': 6 (any of the first six).
   • First digit must be 1, 3, 5, 7 (still 4 options).
   • The digit '9' can go into any of the first five positions (5 possible choices remaining).

   The remaining digits can still be chosen from 0-9 excluding '9' and any previously chosen digit:

   • Possible choices for the second digit (0-9, excluding): 9 options.
   • Possible choices for the third digit: 8 options.
   • The remaining digit choices for the last position (0-9, excluding): This must also be odd.

   Therefore, if we consider it with '9' in, say, the 2nd position or 5th:

   • Combinations per case will yield variations of: $$ 5 \times 4 \times 8 \times 7 $$ for positions that '9' can occupy but being cautious on overlapping odd choices.

6. Calculate Valid Combinations

   Add the results from both cases:

   Total for when '9' is last + total for '9' not one last $$ = 288 + (other combinations) $$ to finish summing and finalize how many trials Sam will need.

7. Final Calculation

   Conclude with the final number of combinations available for a successful dialing.