Hands-on math!

Washington State Math Olympiad
Hints and Solutions
2017 Grade 7 Number Sense
Problem
Solution
1) Find the largest whole number x, so x 8 > 8x
  1. Rearrange the expression as:
        x8   >   1
        8x
  2. Since
        88 = 1
        88
    x must be less than 8 because as x increases beyond 8, the ratio gets less and less:
        98 = 43,046,721 = 0.32 (approx)
        89   134,217,728
  3. Therefore x = 7
2) Place groups of parentheses to make the statement true
    6 - 2 x 3 + 2 ÷ 5 - 3 - 43 = 5
  1. First note that 43 = 64. There is no way the rest of the expression can be close to 64, therefore a set of parentheses must enclose the 4 with at least the 3 and possibly other numbers in this expression.
  2. Try just enclosing the 3 and 4:
        6 - 2 x 3 + 2 ÷ 5 - (3 - 4)3 = 5
        6 - 2 x 3 + 2 ÷ 5 - (-1)3 = 5
        6 - 2 x 3 + 2 ÷ 5 + 1 = 5
  3. All of the expressions to the left of +1 must reduce to 4 for the result to sum to 5. With a divide by 5, the first 4 terms must reduce to 20. The only way to do that is to group the (3+2) and the (6-2) to make 4x5:
        (6 - 2) x (3 + 2) ÷ 5 + 1 = 5
  4. The answer is
        (6 - 2) x (3 + 2) ÷ 5 - (3-4)3 = 5
Note: we solved this by working from right to left in the expression.

Problem
Solution

3) The same relationship is used in each square. What is the missing number?
 Call the 4 numbers in the square a (top left), b (top right), c (bottom left) and d (bottom right)
  1. The first square shows that there must be a multiplication involved in order to result in d = -34, but which ones?
  2. Try different combinations of
    • axb ± c
    • axc ± b
    • bxc ± a
  3. The combination that works is bxc -a: 9x(-3)-7=-34
  4. Therefore ? = 5x2-11 = -1
Please note: You don't need the other 3 squares!
4) Simplify each expression. Find what the 11th term in the sequence would be. Express your answer as a fraction in lowest terms.
  1. The first 3 terms (simplified) are:
        3⁄4   -1⁄3   4
  2. The next term is 1 - 1⁄4 = 3⁄4 which repeats the first term
  3. They repeat in groups of 3, so the 11th term is 11/3 = 3, remainder 2, so the 11th term repeats the second which is -1⁄3

Problem
Solution
5) A box of red pens holds 13 pens. A box of black pens holds 21 pens. Chuck needs to order exactly 1,550 pens. For his order, he wants the number of red pens to be as close to the number of black pens as possible. How many boxes of black pens should he order? Let R = # boxes of red pens and B = # boxes of black pens. To restate the problem, we have 2 equations with 2 unknowns:
  • 13R + 21B = 1550 (exactly)
  • 13R = 21B (approximately)
Since we need exactly 1550 pens, we must have integer numbers for R and B that result in exactly 1550 pens.
  • Solving the above 2 equations for B we get:
        21B + 21B = 1550, so B is approximately 36.9
  • The easiest way is to create a table and increase B until we find an integer value for R, like this:
    B R =
    (1550-21B)/13
     37  55.6
     38  57.8
     39  56.23
     40  54.6
     41  53 (exactly!)
So, the answer is 41 boxes of black pens.