Question

The sum of all possible numbers that can be formed by using the digits 2,3,5,7 without repetition of digits is

• A. \( 17 \times \frac{10^3-1}{9} \)
  % This looks like sum of GP series related
• B. \( 33 \times 34 \times 101 \)
  % Product of numbers
• C. \( 6 \times \frac{10^3-1}{9} \)
• D. \( 33 \times 35 \times 1001 \)

Hint:

To find the sum of all n-digit numbers formed by n distinct non-zero digits: 1. Calculate the sum of the given digits (S). 2. Each digit appears in each position \((n-1)!\) times. 3. The sum is \( S \times (n-1)! \times (11\dots1 \text{ [n times]}) \).
(11...1 [n times]) can be written as \(\frac{10^n-1}{9}\).

Answer 1

The Correct Option is B

We have 4 distinct digits: 2, 3, 5, 7. Numbers can be 1-digit, 2-digit, 3-digit, or 4-digit. The question implies "sum of all possible numbers" meaning numbers using all 4 digits without repetition (4-digit numbers). If it means sum of all numbers of any length, the problem is different. Let's assume it means sum of all 4-digit numbers formed by these digits. Sum of digits = \(2+3+5+7 = 17\). Number of digits \(n=4\). The number of permutations (4-digit numbers) is \(4! = 24\). In these 24 numbers, each digit (2, 3, 5, 7) will appear in each position (units, tens, hundreds, thousands) an equal number of times. Number of times each digit appears in a specific position = \((n-1)! = (4-1)! = 3! = 6\) times. Sum of digits in the units place = \(6 \times (2+3+5+7) = 6 \times 17 = 102\). Sum of digits in the tens place = \(6 \times (2+3+5+7) = 102\). Contribution = \(102 \times 10\). Sum of digits in the hundreds place = \(6 \times (2+3+5+7) = 102\). Contribution = \(102 \times 100\). Sum of digits in the thousands place = \(6 \times (2+3+5+7) = 102\). Contribution = \(102 \times 1000\). Total sum = \(102 \times 1 + 102 \times 10 + 102 \times 100 + 102 \times 1000\) \( = 102 \times (1 + 10 + 100 + 1000) \) \( = 102 \times (1111) \). \( 102 \times 1111 = 113322 \). Let's check the options. % Option (a) \(17 \times \frac{10^3-1}{9} = 17 \times \frac{999}{9} = 17 \times 111 = 1887\). (Incorrect for 4-digit sum) If n digits, sum is (sum of digits) \(\times (n-1)! \times (11\dots1 \text{ n times})\). Here, sum of digits = 17. \((n-1)! = 3! = 6\). String of ones = 1111. Sum = \(17 \times 6 \times 1111 = 102 \times 1111 = 113322\). Now evaluate option (b): \(33 \times 34 \times 101\). \(33 \times 34 = (30+3)(30+4) = 900 + 120 + 90 + 12 = 900 + 210 + 12 = 1110 + 12 = 1122\). \(1122 \times 101 = 1122 \times (100+1) = 112200 + 1122 = 113322\). This matches my calculated sum. So option (b) is correct. The formula in options (a) and (c) \(\frac{10^k-1}{9}\) gives a string of k ones (e.g., \(111\dots1\)). Option (a) \(17 \times (111)\) is not right. The general formula for sum of all n-digit numbers formed by n distinct non-zero digits is: Sum = (Sum of the n digits) \(\times\) ((n-1)!) \(\times\) (111...1, n times). Here, Sum = \(17 \times 3! \times 1111 = 17 \times 6 \times 1111 = 102 \times 1111 = 113322\). Option (b) evaluates to 113322. \[ \boxed{33 \times 34 \times 101} \]