Hands-on math!

Washington State Math Olympiad
Hints and Solutions
2017 Grade 8 Number Sense
Problem
Solution
1) Insert parentheses to make the following equation true.
    60 + 12 ÷ 2 - 3 x 22 = 0
  1. To sum to 0, the expressions to either side of that negative sign must be equal.
  2. Try putting parentheses around the first half of that expression:
        (60 + 12 ÷ 2) - 3 x 22 = 0     (66) - 3 x 22 = 0
    No way this can work!
  3. On the right side, try putting parentheses around everthing but the square:
        (3 x 2)2 = 36
  4. See if you can make the left expression equal to 36:
    (60 + 12) ÷ 2) = 36 This works!
  5. The answer is
        (60 + 12) ÷ 2 - (3 x 2)2 = 0

2) Rewrite 0.132 as a fraction in lowest terms.
 The explanation for repeating decimals shows you how to do this.
  1. Call the repeating decimal X
  2. Multiply the number by a power of 10 to get the repeating part to the left of the decimal:
        0.1 32 x 1000 = 132.32 = 1000X
  3. If there are leading non-repeating decimals, get them also to the left of the decimal point by multiplying the number by a power of 10 to get the repeating part to the right of the decimal:
        0.132 x 10 = 1.32 = 10X
  4. Subtract this second value from the larger value:
    132.323232 ...
    -   1.323232 ...
    131
  5. 1000X - 10X = 131
    990X = 131
    X = 131/990

Problem
Solution
3) What is the one's place of 4355?

 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 43 (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.
    431 = 43
    432 = 1849
    433 = 79507
    434 = 3418801
    435 = 147008443
    436 = 5321363049
    437 = 271818611107 		The one's digits are 3, 9, 7, 1 and then they repeat. It's a cycle of 4 digits.
55/4 = 13, remainder 3, so the one's digit for the 55th 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.
4) What numbers between 1 and 100 have exactly 5 factors?
  1. If a number has exactly 5 factors, then there are only 3 factors that are not 1 and the number itself.
    (1 and the number itself are considered factors).
  2. Square numbers have an ODD number of factors. This is because factors are always in pairs.
    Think of 8: the factors of 8 are 1,2,4 and 8. 4 factors.
    But in a square (like 25), one of the factors is multiplied by itself and it is only counted once. (the square root). Therefore the numbers that have 3 factors will be squares. (factors of 25: 1,5,25).
  3. The numbers that have 5 factors will be "squares of squares" or 4th powers of some basal number.
  4. The only numbers which have 3 factors, other than 1 and the number itself that are less than 100 are 16 and 81 which are 24 and 34:
    factors of 16: 1,2,4,8,16
    factors of 81: 1,3,9,27,81
The 2 numbers between 1 and 100 that have 5 factors are 16 and 81.
5) Given the following sequence, find what the 10th term in the sequence is. Express your answer in a simplified form.
Solution: Starting at the first term, perform the addition, leaving the result in the form of an improper fraction. Keep substituting this value in the subsequent terms and simplifying. You should notice a pattern.

  1. 1 + 2⁄3 = 5⁄3
  2. 1 + 1/ 5⁄3 = 1 + 3⁄5 = 8⁄5
  3. 1 + 1/ 8⁄5 = 1 + 5⁄8 = 13⁄8
  4. 1 + 1/ 13⁄8 = 1 + 8⁄13 = 21⁄13
Calling the fraction A / B, each successive element is
(A + B) / A of the previous element, so, extending this to the 10th term:
Term=5678910
  34⁄21 55⁄34 89⁄55 144⁄89 233⁄144 377⁄233