April 17, 2017
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Now that our national competitors have been determined, teams are meeting to practice for next month’s competition. On the team round, working together is key. Grab yourself some fellow Mathletes® and see if you can tackle a few team round problems from some of the 2017 competitions!

Raquel uses six different digits to fill in the blanks below, writing one digit in each blank, so that the resulting addition statement is correct. What is the least possible sum of the six digits?

­__ __ + __ __ = __ __

[School, Team #5]         

The smallest sum of six different digits is 0 + 1 + 2 + 3 + 4 + 5 = 15. Let’s see if we can fit these digits into the blanks. The digit 0 has two possible placements. It could be the units digits of the sum or in the tens place of one of the numbers being added. Each of these placements require two digits that sum to 10 or greater, which no two digits 1 through 5 do. So, let’s look at the next smallest sum of six different digits, 0 + 1 + 2 + 3 + 4 + 6 = 16. The same criteria are required for the placement of the 0. Only the digits 4 and 6 sum to 10. If we place a 4 and 6 in the units digits of the numbers being added and the 0 in the units digit of the sum, we are left with 1, 2 and 3. Because of the 1 carried over from summing the units digits, these numbers will not fit in the tens digits. So, let’s try the next smallest sum of six, 0 + 1 + 2 + 3 + 5 + 6 = 17. The same criteria apply to the placement of the 0. We know that 5 + 6 = 11, so we can place these two in the two units places and the number 1 in the units place of the sum. This leaves us with 0, 2 and 3. Placing 0 and 2 in the tens place gives us 05 + 26 = 36. So the smallest possible sum of the digits is 17.

 

What is the greatest number that evenly divides the sum of any six consecutive whole numbers?

 

[Chapter, Team #6]         

Six consecutive whole numbers have the form n, n + 1, n + 2, n + 3, n + 4, and n + 5, the sum of which is 6n + 15 = 3(2n + 5), where n is a non-negative integer, so 3 explicitly divides this value for all non-negative integers n. Now, how about 2n + 5? Well, 2n + 5 is 5 for n = 0 and 7 for n = 1, which have only 1 as a common factor. Therefore, the greatest positive integer guaranteed to divide 6n + 15 is 3.

 

Lorie wants to create a coffee blend that has regular coffee mixed with premium coffee in a ratio of 4:1. She starts with 100 kg of blend in which the ratio of regular to premium coffee is 3:2. How many kilograms of regular coffee does she need to add to get her desired 4:1 ratio?

[State, Team #2]         

The starting amounts are 3/(3 + 2) × 100 kg = 60 kg of regular and 100 kg ­– 60 kg = 40 kg of premium. We want to keep the same amount of premium and add regular until there is 4/1 = 4 times as much regular as there is premium. Therefore, a total of 4 × 40 kg = 160 kg of regular is needed, so 160 kg – 60 kg = 100 kg of regular needs to be added.

 

Page 1 of the linked PDF contains PROBLEMS & SOLUTIONS.
Page 2
contains ONLY PROBLEMS.