Using the digits 3, 4, 5, 6, 7, and 8 exactly once, what is the minimum possible sum of three two-digit natural numbers?
Solution:
To make the sum as small as possible, we should place the smallest digits in the tens places, because tens digits have a value 10 times greater than ones digits.
Tens digits: 3, 4, 5
Ones digits: 6, 7, 8
So the minimum sum is:
30 + 40 + 50 + 6 + 7 + 8
= 141