Hands-on math!

Washington State Math Olympiad
Hints and Solutions
2016 Grade 8 Number Sense
Problem
Solution
1) Rewrite 0.72 as a fraction.

 0.72 is a repeating decimal.
  1. The number (we'll call it x) is 0.72727272 . . .
  2. Multiply it times 100 to get the repeated part to the left of the decimal:
    100x = 72.72727272 ....
  3. Subtract the original number from this number:
        72.727272 ....
       - 0.727272 ...
        72
  4. This value is 100x - x = 99x
  5. 99x = 72
        x = 72/99 = 8/11
2) A doctor makes a box-and-whisker plot(see below) to show the number of patients she sees each day. Identify upper and lower quartiles.
  1. The upper quartile is 27
  2. The lower quartile is 18
3) How many numbers between 100 and 1,000 have exactly 3 factors?
  1. All numbers have 2 factors: 1 and the number itself. Prime numbers have these factors only.
  2. The only numbers that have exactly 1 more factor are squares. For example the factors of 25 are 1, 5 and 25. But only numbers whose square roots are primes, otherwise there will be additional factors.
  3. Numbers that have prime square roots:
    11x11 = 121
    13x13 = 169
    17x17 = 289
    19x19 = 361
    		       23x23 = 529
    29x29 = 841
    31x31 = 961
  4. 7 numbers between 100 and 1000 that have 3 factors.

Problem
Solution
4) How many dots would there be in the 20th step of the sequence below?
This problem involves the sum of an arithmetic sequence
  1. Starting with 2 dots at step 1.
  2. 2 dots are added to the first 2 steps ( = 4 and 6)
  3. 3 dots are added to steps 3 and 4 (= 9 and 12)
  4. The number of dots added is incremented every other step
  5. The sum of an arithmetic sequence is
    (A1 + An) n / 2 where
    A1 is the first element, An is the last element and n is the number of elements in the sequence.
  6. Over 20 steps ((2 + 11) x 10/2) x 2 = 130 dots are added, total.
  7. Add the starting 2 dots and you get
    130 + 2 = 132 dots in the 20th step.
  8. This is complicated, so below is a step by step explanation:
    n=  1:   2    + 2 = 4
    n=  2:   dots in step n-1 + 2 = 6
    n=  3:   dots in step n-1 + 3 = 9
    n=  4:   dots in step n-1 + 3 = 12
    n=  5:   dots in step n-1 + 4 = 16
    n=  6:   dots in step n-1 + 4 = 20
    n=  7:   dots in step n-1 + 5 = 25
    n=  8:   dots in step n-1 + 5 = 30
    n=  9:   dots in step n-1 + 6 = 36
    n=10:   dots in step n-1 + 6 = 42
    n=11:   dots in step n-1 + 7 = 49
    n=12:   dots in step n-1 + 7 = 56
    n=13:   dots in step n-1 + 8 = 64
    n=14:   dots in step n-1 + 8 = 72
    n=15:   dots in step n-1 + 9 = 81
    n=16:   dots in step n-1 + 9 = 90
    n=17:   dots in step n-1 + 10 = 100
    n=18:   dots in step n-1 + 10 = 110
    n=19:   dots in step n-1 + 11 = 121
    n=20:   dots in step n-1 + 11 = 132

Problem
Solution
5) What is the one's digit of 3 to the power of 99?

 This kind of problem relies on the fact that some numbers have repeating one's place digits when taken to increasing powers. Keep taking powers of 3 (2,3,4, ...) until you notice a pattern. Divide the power you want by the number of repeating digits. The remainder will be the number of the digit in the repeating sequence.

    31 = 3
    32 = 9
    33 = 27
    34 = 81
    35 = 243
    36 = 729
    37 = 2187

The one's digits are 3, 9, 7, 1 and then they repeat. It's a cycle of 4 digits.
99/4 = 24, remainder 3, so the one's digit for the 99th power is 7 (the 3rd digit of the sequence)

Interesting note: Any number whose 1's digit squared results in the same 1's digit (for example 52 = 25) will have that 1's digit in any power! So a problem like "what is the one's digit of 6 to the power of 4545" will be easy! It's a 6 because 6x6 = 36 x 6 = 216 x 6 = 1296 ...
The 1's digit that behave this way are 1,5 and 6.