The given question is a challenging GMAT 700+ level quant problem solving question combining concepts in permutation combination (counting methods) and number properties.

Question 32: What is the sum of all 3-digit positive integers such that all the digits of each of the number is even?

- 55,500
- 1300
- 44,400
- 247,275
- 54,400

@ INR

**Constraint**: All the digits of the number are even. Therefore, the digits of each of the numbers are from the set {0, 2, 4, 6, 8}

**Hundreds place**: 4 possible values {2, 4, 6, 8}. Hundreds place cannot be zero.

**Tens place and units place**: All 5 values possible.

Therefore, total possible 3-digit positive integers such that all the digits are even = 4 × 5 × 5 = 100

Each of the 5 digits is equally likely to appear in the units place.

Therefore \\frac{100}{5}) = 20 is the number of times each digit will appear in the units place.

Therefore, value of sum of digits in units place = 20 × (0 + 2 + 4 + 6 + 8 ) = 400.

For the same reason, sum of tens digits = 400

Hence, value of the sum of digits in tens place = 400 × 10 = 4000

Each of the 4 digits is equally likely to appear in the hundreds place.

Therefore \\frac{100}{4}) = 25 numbers will begin with each of {2, 4, 6, and 8}

Therefore sum of digits in hundreds place = 25(2 + 4 + 6 + 8) = 500

Value of the sum of digits in 100s place = 500 × 100 = 50,000

Required Sum = Sum of units place + Sum of tens place + Sum of hundreds place

= 400 + 4000 + 50000 = **54,400**

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