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Learning Math Home
Patterns, Functions, and Algebra
 
Session 8 Part A Part B Part C Homework
 
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Session 8 Materials:
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A B C 
Homework

Video

Solutions for Session 8, Part H

See solutions for Problems: H1 | H2 | H3 | H4 | H5 | H6| H7 | H8


Problem H1

The output will be 2. Five goes into 3 once, and the remainder is 2.

<< back to Problem H1


 

Problem H2

The output will be 0. Twelve goes into 3 four times, with no remainder.

<< back to Problem H2


 

Problem H3

The output will be 2. Two doesn't go into 3 at all, so the remainder is still 2.

<< back to Problem H3


 

Problem H4

a. 

There are an infinite number of answers, beginning with 1. The next is 4, then 7, then 10. Any number which is one more than a multiple of 3 produces an output of 1.

b. 

The "pullback" is all numbers which are two more than a multiple of 3: 2, 5, 8, 11, 14, ... .

c. 

The numbers producing an output of zero will be all numbers which are zero more than a multiple of 3, or in other words, the multiples of 3: 0, 3, 6, 9, 12, ... .

d. 

The only possible outputs are 0, 1, and 2. You might have noticed in H4(a)-H4(c) that every whole number falls into one of the three categories. Another explanation is that the only possible remainders when dividing by 3 are 0, 1, and 2.

<< back to Problem H4


 

Problem H5

It is a cyclic function. Here is a table:

x

y

0

0

1

1

2

2

3

0

4

1

5

2

6

0

7

1

8

2

9

0

10

1

<< back to Problem H5


 

Problem H6

This can be done by looking through the table of numbers which are one more than a multiple of 3 and looking for a pattern of those numbers which are also three more than a multiple of 5. The list consists of: 13, 28, 43, 58, 73, 88, .... The first of these numbers occurs with 13, and they repeat every 15 numbers. One rule to find all the numbers would be that they are all 13 more than a multiple of 15, or all numbers whose output is 13 in the "mod 15" operation.

<< back to Problem H6


 

Problem H7

This is harder to find, but the first two conditions mean that the age is two more than a multiple of 15 -- those numbers are 2, 17, 32, 47, 62, 77, ... . Among those numbers, the only one which is five more than a multiple of 7 is 47. There are other possible answers, but the others are greater than 100 (the first of these is 152). Overall, the condition is that the age must be equal to 47 (mod 105).

<< back to Problem H7


 

Problem H8

Here is one possibility for a completed table:

answers

Of course, there are lots of other possibilities!

<< back to Problem H8


 

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