Teacher resources and professional development across the curriculum

Teacher professional development and classroom resources across the curriculum

Monthly Update sign up
Mailing List signup
Search
Follow The Annenberg Learner on LinkedIn Follow The Annenberg Learner on Facebook Follow Annenberg Learner on Twitter
MENU
Learning Math Home
Patterns, Functions, and Algebra
 
Session 9 Part A Part B Part C Part D Part E Homework
 
Glossary
Algebra Site Map
Session 9 Materials:
Notes
Solutions
 

A B C D
Homework

Video

Solutions for Session 9, Part C

See solutions for Problems: C1 | C2 | C3 | C4 | C5 | C6| C7| C8| C9| C10| C11
C12| C13| C14| C15| C16| C17| C18| C19 | C20| C21| C22


Problem C1

Don't forget about the order of operations. The units digit will be identical to that of (4 * 6) + (3 * 4), or 4 + 2 = 6.

<< back to Problem C1


 

Problem C2

The units digit of the second number is the same as that of (2 * 2) + (6 * 2), or 4 + 2 = 6. Therefore, it's true.

<< back to Problem C2


 

Problem C3

The units digit is the same as that of (2 * 3 + 7 * 7) * (8 + 5 * 8 * 8 * 8), or (6 + 9) * (8 + 0), or 5 * 8 = 0.

<< back to Problem C3


 

Problem C4

Yes, this is true. At each step, we reduce our calculation to the units digit -- for example, 7 * 7 = 49, so we use 9. The end result is the units digit of the sum, or product, of all the units digits used throughout. We can do this only because things that occur in the tens digit, or any other higher digit, cannot affect the result in the units digit.

<< back to Problem C4


 

Problem C5

When dividing by 10, the quotient will always be the number, beginning with the tens digit. For example, 7,536 divides into 10 a total of 753 times. The remainder is, therefore, the units digit only.

<< back to Problem C5


 

Problem C6

 

<< back to Problem C6


 

Problem C7

No, order doesn't count. This can be seen from the symmetry in the table -- each side of the main diagonal is identical.

<< back to Problem C7


 

Problem C8

Because addition on numbers is commutative already (a + b = b + a), we would expect the remainder of two identical numbers to still be identical. So our table, made of remainders, must be commutative.

<< back to Problem C8


 

Problem C9

No. Again, both sides of the main diagonal are identical.

<< back to Problem C9


 

Problem C10

Multiplication on numbers is commutative (ab = ba), so our table of remainders, based on a full multiplication table, must also be commutative.

<< back to Problem C10


 

Problem C11

Sure, adding 0 to an original number doesn't change it, so it cannot change the remainder.

<< back to Problem C11


 

Problem C12

Yes. It can be read from the table: we want to find the second number that makes the sum 0. In each row and column of our table, 0 occurs exactly once, so there is a single opposite number for each given number.

<< back to Problem C12


 

Problem C13

If our number is n, the opposite is (10 - n). If our number is 0, the opposite is 0, since 10 is not in the system.

<< back to Problem C13


 

Problem C14

Sure, multiplying by 1 won't change an original number, so it can't change the remainder.

<< back to Problem C14


 

Problem C15

Only 1, 3, 7, and 9 have reciprocals, since they each have exactly one 1 in their rows and columns of the table. No other number has any 1 in its row or column.

<< back to Problem C15


 

Problem C16

The numbers 1, 3, 7, and 9 are all the numbers that do not share a common factor with 10. This allows the 1 to appear as a units digit. For example, multiples of 2 will all be even, so there cannot be a multiple of 2 that ends with a 1. The same is true of any number that shares a common factor with 10, so these do not have reciprocals. It is more difficult to guarantee (prove!) that the others must have a single 1 in their rows and columns, but a useful observation is that each number of 0 through 9 occurs exactly once in these numbers' rows and columns.

<< back to Problem C16


 

Problem C17

For addition and multiplication, both systems have commutativity, associativity, and identity. In both systems, every number has an opposite.

<< back to Problem C17


 

Problem C18

No, it isn't true -- for example, 5 * 4 = 0. One explanation is that there is now more than one number that equals 0: 10, 20, 30, 40 ... . If we can find two numbers that multiply to make any multiple of 10, this will be a 0 in our table. Also, there are numbers in our table that share factors with 10, so it is possible to find a pair that creates a multiple of 10. This is not possible in ordinary arithmetic, because no number has 0 as a factor.

There is a lot more here; you might try making the same table with other bases. Low numbers are good starting points, because their tables are easy to produce. Some tables do have the property that if two numbers multiply to 0, one must be 0 -- maybe you can find which ones.

<< back to Problem C18


 

Problem C19

There are tons of patterns. One is that two numbers which are opposites (3 and 7, for example) will have rows and columns which reverse each other. Another is that 0 appears in a multiplication row twice for multiples of 2, five times for multiples of 5, and ten times for multiples of 10. Another is that 1, 3, 7, and 9 occur only in their own rows and columns in the multiplication table.

<< back to Problem C19


 

Problem C20

All are closed under addition, except for the odd integers. All five sets are closed under multiplication. The only set which is closed under division is the positive real numbers.

<< back to Problem C20


 

Problem C21

 

The first set does not have identity or inverses. The identity should be 0, but 0 is not in the set.

 

The second set has identity, but does not have inverses. The identity, 1, is in the set, but most numbers do not have inverses.

 

The third set has identity and inverses. The identity, 0, is in the set, and every number has an inverse (its opposite).

 

The fourth set has identity, but does not have inverses. The identity, 1, is in the set, but most numbers do not have inverses.

 

The fifth set has neither identity nor inverses. There is no identity element for division; many people think 1 is the identity, since 7 / 1 = 7 works for any number. However, 1 / 7 = 7 would also have to be true, and it is not.

<< back to Problem C21


 

Problem C22

 

The first set is not a field, since there is no identity for addition (0 is not in the set).

 

The second is a field.

 

The third is not a field, since there is no identity for multiplication (1 is not in the set).

 

The fourth is not a field, since one element (2) does not have an inverse under multiplication.

 

The fifth is a field.

<< back to Problem C22

 

Learning Math Home | Algebra Home | Glossary | Map | ©

Session 9: Index | Notes | Solutions | Video

© Annenberg Foundation 2014. All rights reserved. Legal Policy