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Patterns, Functions, and Algebra
Session 9 Part A Part B Part C Part D Part E Homework
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Session 9 Materials:



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.

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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.

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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.

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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.

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Problem C6


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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.

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Problem C9

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

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Problem C10

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

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Problem C11

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

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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.

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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.

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Problem C14

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

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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.

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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.

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Problem C17

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

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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.

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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.

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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.

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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.

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