GMAT 700+ Quant Question #32

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?

1. 55,500
2. 1300
3. 44,400
4. 247,275
5. 54,400

Video Explanation

Explanatory Answer | GMAT Counting Methods

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}

Step 1: Compute the total possible numbers

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

Step 2: Compute the value of sum of units and tens digits

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

Step 3: Compute the value of sum of hundreds digits

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

Step 4: Compute the sum of such 3-digit numbers

Required Sum = Sum of units place + Sum of tens place + Sum of hundreds place
= 400 + 4000 + 50000 = 54,400

Choice E is the correct answer.