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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
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Problem C1 | |
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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.
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Problem C2 | |
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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 | |
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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 | |
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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 | |
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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 C7 | |
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No, order doesn't count. This can be seen from the symmetry in the table -- each side of the main diagonal is identical.
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Problem C8 | |
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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 | |
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No. Again, both sides of the main diagonal are identical.
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Problem C10 | |
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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 | |
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Sure, adding 0 to an original number doesn't change it, so it cannot change the remainder.
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Problem C12 | |
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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 | |
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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 | |
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Sure, multiplying by 1 won't change an original number, so it can't change the remainder.
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Problem C15 | |
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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 | |
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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 | |
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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 | |
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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.
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Problem C19 | |
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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 | |
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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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