Download Section 1.1 Solutions to Practice Problems For Exercises 1

Survey
yes no Was this document useful for you?
   Thank you for your participation!

* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project

Document related concepts

Large numbers wikipedia , lookup

Mathematics of radio engineering wikipedia , lookup

Georg Cantor's first set theory article wikipedia , lookup

Positional notation wikipedia , lookup

Arithmetic wikipedia , lookup

Series (mathematics) wikipedia , lookup

Addition wikipedia , lookup

Location arithmetic wikipedia , lookup

Proofs of Fermat's little theorem wikipedia , lookup

Sequence wikipedia , lookup

Collatz conjecture wikipedia , lookup

Elementary arithmetic wikipedia , lookup

Transcript
Section 1.1
Solutions to Practice Problems
For Exercises 1-4, use inductive reasoning to find a pattern, and then make a reasonable conjecture for the next number or
item in the sequence.
1) 1 2 4 7 11 16 22 29 ___
Solution:
The numbers in the sequence are obtained by adding the positive integers (in order) to the previous terms. In other
words, the sequence is:
1
1+1=2
2+2=4
4+3=7
7 + 4 = 11
11 + 5 = 16
16 + 6 = 22
22 + 7 = 29
Therefore, the next number in the sequence is 29 + 8 = 37.
2) 10 20 11 18 12 16 13 14 14 12 15 ___
Solution:
This sequence is comprised of two sequences whose terms alternate:
10 20 11 18 12 16 13 14 14 12 15
10 20 11 18 12 16 13 14 14 12 15 ___
The next term will be part of the second sequence in which each term is two less than the previous term. The next term
in the sequence is therefore the number 10.
3) 100 99 97 94 90 85 79 ___
Solution:
The numbers in the sequence are obtained by subtracting the positive integers (in order) from the previous terms. In
other words, the sequence is:
100
100 – 1 = 99
99 – 2 = 97
97 – 3 = 94
94 – 4 = 90
90 – 5 = 85
85 – 6 = 79
Therefore, the next number in the sequence is 79 – 7 = 72.
4)
Solution:
Each figure in the sequence is a rotation of the previous image 90° in a counterclockwise direction. Therefore, the next
figure in the sequence will be:
For Exercises 5 and 6, find a counterexample to show that each statement is false.
5) The sum of any three odd numbers is even.
Solution (Answers may vary):
The numbers 1, 3, and 5 are odd; their sum is 1 + 3 + 5 = 9, which is odd (not even).
6) When an odd number is squared and divided by 2, the result will be a whole number.
Solution (Answers may vary):
The number 3 is odd; 32 ÷ 2 = 9/2 = 4.5, which is not a whole number.
For Exercises 7 and 8, use inductive reasoning to make a conjecture about a rule that relates the number you selected to
the final answer.
7) Select a number:
2
Double it:
4
Subtract 20 from the answer:
-16
Divide by 2:
-8
Subtract the original number:
-10
Result: The result is always the number -10.
5
10
-10
-5
-10
100
200
180
90
-10
8) Select a number:
4
10
Add 3:
7
13
Multiply by 2:
14
26
Subtract 10:
4
16
Divide by 2:
2
8
Result: The result is always 2 less than the selected number.
200
203
406
396
198
For Exercises 9 and 10, use inductive reason to find a pattern for the answers. Then use the pattern to guess the result of
the final calculation, and perform the operation to see if your answer is correct.
9) 12,345,679 × 9 = 111,111,111
12,345,679 × 18 = 222,222,222
12,345,679 × 27 = 333,333,333
⁞
12,345,679 × 72 = ?
Solution:
The factors on the right are:
9×1
9×2
9×3
Because 72 = 9 × 8, the product should be:
12,345,679 × 72 = 888,888,888
10) 999,999 × 1 = 0,999,999
999,999 × 2 = 1,999,998
999,999 × 3 = 2,999,997
⁞
999,999 × 9 = ?
Solution:
In the product the first digit is one less than the second factor,
and the last digit is 10 minus the second factor. In other
words:
For 999,999 × 1, first digit: 1 – 1 = 0, last digit: 10 – 1 = 9
For 999,999 × 2, first digit: 2 – 1 = 1, last digit: 10 – 2 = 8
For 999,999 × 3, first digit: 3 – 1 = 2, last digit: 10 – 3 = 7
Following this pattern:
For 999,999 × 9, first digit: 9 – 1 = 8, last digit: 10 – 9 = 1
Therefore, 999,999 × 9 = 8,999,991.