Download 2013 State Convention – Speed Math Solutions 1. For each square

2013 State Convention – Speed Math Solutions
1. For each square root, factor out the largest perfect square:
 48  27  363  16  3  9  3  121 3
A square root of a product is the same as the product of two square roots:
 16 3  9 3  121 3
Simplifying:
 4 3  3 3  11 3  12 3
2. This can be done quickly by considering that both 32 and 28 are 2 away from 30.
 28  30   30  2  30  2 
This means the product is like a difference of squares, so applying the concept that
  x  a  x  a   x2  a 2
We can rewrite our product in this way. These squares can be calculated easily, leaving with a simple
subtraction.
 302  22  900  4  896
3. First, rewrite each number as a fraction (repeated portions are over as many 9s as there are repeated
numbers and if a 0 precedes the repeated portion, it is 1/10 of the number without it).



 0.54 0.04 
54 4
6 2
62
22
4
   


99 90 11 45 45 11 15 11 165
4. First, break down the sum in the exponent to a product of two terms, then factor.
 3x  3x  32  270  3x 1  9  270
Then, solving:
 3x 10  270  3x  27  x  3
5. First, isolate the term involving the exponent:
 3y2  5  x  3y2  x  5
Next, remove the exponent by taking a log of both sides:
 log3 3y 2  log3  x  5  y  2  log3  x  5  y  log3  x  5  2
6. The possible roots of a polynomial can be determined by finding all combinations of quotients with
the numerator being a factor of the constant and the denominator being a factor of the leading
coefficient. However, we don’t need to consider all factors, just the largest and smallest. The largest
possibility will come from the largest numerator and smallest denominator and the smallest will come
from the smallest numerator and largest denominator. The largest will always be the number itself and
the smallest will always be 1.
y  4 x5  3x3  5x 2  7 x  12
12
 The largest possible root:
1
The smallest possible root:
1
4
The difference:

12 1 48 1 47
 
 
1 4 4 4
4
7. Rewrite each logarithm using the change of base, and write each argument as a lowest base possible
to a power:
  log16 25 log9 8 log5 243 
log 52 log 23 log 35


log 24 log 32 log 5
Use the power rule for logarithms:

2 log 5 3log 2 5log 3


4 log 2 2 log 3 log 5
Cancel common logs:

2 log 5 3 log 2 5 log 3


4 log 2 2 log 3 log 5
This leaves:

2 3 5 15
  
or 3.75
4 2 1
4
8. This difference of sums can be easiest to calculate by placing one atop the other:
  2  4  6  ...  2014 
 1  3  5  ...  2013
Here, it can be seen that grouped together, each difference is 1. Therefore, the sum is the same as the
total number of pairs. Noticing that the first series is arithmetic:
 an  a1  d (n  1)  2014  2  2(n  1)  2012  2(n  1)  1006  n 1  n  1007
9. To quickly do this calculation, rewrite the difference of squares:
 522  482   52  48 52  48   4 100   400
10. Do each determinant, then solve algebraically:
 5 2  3 5  14  5(4)  2(1)  3x  5(2)  14  18  3x 10  14  3x  6  x  2
1 4
2 x
11. To determine the units digit of the number, it is necessary only to know how the original units digit
changes as the number is raised to subsequent powers (only the units digit will ever affect the units
digit):
 31  3 , 32  9 , 33  _ 7 , 34  _1
Therefore, we can see that beyond the 4th power, this pattern will repeat. Thus, we can set aside all
groups of 4 from the exponent and find the remainder:

 20132013  20134
  2013
503
1
Because the units digit of the first part will be 1, the remaining part dictates the final digit:
 3
12. Consider that when writing out the factors of a number that is not a perfect square, the numbers
can always be grouped into pairs that will multiply to get the original number, such as 1x200 and 2x100
(this is true for perfect squares as well, except the square root will remain by itself). Therefore, rather
than actually finding all the factors and multiplying them out, it is easier to determine the total number
of factors to determine the number of pairs that will multiply to 200. First find the prime factorization:
 200  23  52
The number of total factors comes from considering that any factor can be built up from the constituent
prime factors, each including anywhere from 0 of the factor to the maximum power of that factor (there
are 4 options for the power on 2 and 3 options for the power on 5).
Total number of factors = 4  3  12
Thus, the total number of pairs is 12/2 = 6.

 2006  2 102

6
 26 1012  64 1012  6.4 1013
13. Set the expression equal to a variable:
1
 x  2
2
1
...
In order to solve, it’s necessary to determine what the repeated portion is equal to so a recursive
approach can be used.
1
 x2  2 
2
1
...
The infinitely repeated expression is equal to x2, so the second repeated portion can be replaced with
that quantity.
 x2  2 
1
x2
Now solve algebraically:


 x 4  2 x 2  1  x 4  2 x 2  1  0  x2 1
2
 0   x  1  x  1  0  x  1
2
2
However, because the original expression is a square root, it must be equal to a positive value.
Therefore: x  1
14. Use the internal angle equation and solve algebraically:
180  n  2 
 156  180  n  2  156n  180n  360  156n  180n 156n  360
n
 24n  360  n  15

15. Factor the numerator and the denominator:
 y
 x  3 x  2 
x2  x  6
 y
2
x  8 x  15
 x  3 x  5
The removable discontinuity, or “hole”, occurs at the x-value that would cause the fraction to be 0/0,
which is x = -3. Cancel the common factor and substitute that value for x:
 y
 x  3  x  2  x  2

 x  3  x  5  x  5
 y
3  2
5

3  5
2
Therefore, the point is:


5
2
  3,  
16. First convert the base 7 number to base 10:
 2457  5  7 0  4  71  2  7 2  5 1  4  7  2  49  5  28  98  13110
Next, convert to base 5.
 131  125  5  1  1 53  0  52  1 51  1 50  1011 5
17. Given that a7  14 and a10  5 , the common difference is added 3 times to get from the 7th to the
10th term.
 5 14  3d  d  3
a30 is 20 terms away from a10 , so the common difference is added 20 times to that term:
 a30  5  20(3)  5  60  55
18. Because the cube’s side length is decreased by 80%, it is 20% of the normal side ( 0.2s 
2
s ). The
10
volume is the side length cubed:
3
8 3
 2 
s 
s  0.008s 3
 10  1000
 V 
Because the new volume is 0.008 times the original (or 0.8% of it), then it has decreased by:
 100%  0.8%  99.2%
19. Factor the left-hand side and solve algebraically:


  2 x  3 3x  1   2 x  3  3 x  1  0   2 x  3 3x 1  3x 1   2 x  3  0
2
2
3
2
2 1
5 3
  2 x  3 3x 1 5x  2   0  x   ,  ,
So, the product:



3  2  1  1
    
2  5  3  5
20. In order to change 89  519 into scientific notation, consider that 8 is a power of 2 and rewrite the
expression as a power of 2 and a power of 5. This can then be regrouped into the product of something
and a power of 10:
   5
 89  519  23
9
19
 227  519  28  219  519  28   2  5   256 1019  2.56 1021
19
21. Pascal’s triangle can be used to determine the coefficients for the terms of a binomial raised to any
power. However, it can also be used to determine the number of terms of any sized polynomial raised
to any power. These can be found by looking at successive diagonals. The first diagonal (1, 1, 1, …) is
the number of terms for any monomial raised to powers. The next diagonal (1, 2, 3, 4, …), the counting
numbers, represents the number of terms of a binomial raised to a power. The third (1, 3, 6, 10, …), the
triangular numbers, represent trinomials. So, for this case, the fourth diagonal (1, 4, 10, 20, …), the
tetrahedral numbers, represents 4-term polynomials. Because the first term represents the 0th power,
the 5th term represents the 4th power. So the answer is 35 .
22.
15
C3 
15! 15 14 13 12! 15 14 13 5  7 13



 455
12!3!
3  2 112!
3 2
1
23. If the diagonal is a length of 10, the side length must be 10 / 2 . This is the same length as the
diameter of the inscribed circle, so the circumference is simply that times pi. So, with a rationalized
denominator:

10 2 10 2


 5 2
2
2
2
24. Because each remainder is one less than the divisor, the product must be one less than a common
multiple of the divisors. So, first find the LCM of 2, 3, 4, and 5:
 LCM  2,3, 4,5  3  4  5  60
Next, to find the lowest 3-digit number with the desired properties, find the lowest multiple of 60 that is
3-digits, which is 120. Therefore, the lowest 3-digit number that has remainders of 1 less than each of
the divisors would be one less than 120:
 119
25. To solve the given system of equations, multiply the first equation by 2 and the second equation by
3 and add:
4 x  6 y  34
  9 x  6 y  18 
13x  52
 x4
Next, substitute this value into either equation:
 2(4)  3 y  17  y  3
So, evaluating the desired expression:
 43  64