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course Mth 151
7/19 1
Question: `qquery 4.3.6 number following base-six 555
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Your solution:
555{base 6} = 5 * 6^2 + 5 * 6^1 + 5 * 6^0. If you add 1 you get
5 * 6^2 + 5 * 6^1 + 5 * 6^0 + 1, which is equal to
5 * 6^2 + 5 * 6^1 + 6 * 6^0. But 6 * 6^0 is 6^1, so now you have
5 * 6^2 + 6 * 6^1 + 0 * 6^0. But 6 * 6^1 is 6^2 so you have
6 * 6^2 + 0 * 6^1 + 0 * 6^0. But 6 * 6^2 is 6^3 so the number is
6 * 6^3 + 0 * 6^2 + 0 * 6^1 + 0 * 6^0.
So the number following 555{base 6} is 1000{base 6}.
The idea is that we can use a number greater than 5 in base 6, so after 555 we go to 1000, just like in base 10 we go from 999 to 1000
3
Ok
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Question: `qquery 4.3.20 34432 base five
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Your solution:
34432 base five means 3 * 5^4 + 4 * 5^3 + 4 * 5^2 + 3 * 5^1 + 2 * 5^0.
5^2 = 625, 5^3 = 125, 5^2 = 25, 5^1 = 5 and 5^0 = 1 so
3 * 5^4 + 4 * 5^3 + 4 * 5^2 + 3 * 5^1 + 2 * 5^0= 3 * 625 + 4 * 125 + 4 * 25 + 3 * 5 + 2 * 1 = 1875 + 500 + 100 + 15 + 2 = 2492. **
3
Ok
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Question: `qExplain how you use the calculator shortcut to get the given number.
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Your solution:
2 + 5 (3 + 5(4 + 5(4 + 3 * 5))) multiplies out to the given number because, for example, the 3 * 5 at the end will be multiplied by the three 5’s to its left to give 3 * 5 * 5 * 5 * 5 = 3 * 5^4, which matches the 3 * 5^4 in the sum 3 * 5^4 + 4 * 5^3 + 4 * 5^2 + 3 * 5^1 + 2 * 5^0.
So we do 3 + 5, then 4 + the answer, then 5 * the answer, then 4 + the answer, then 5 * the answer, then 3 + the answer, then 5 * the answer, then 2 + the answer. **
3
Ok
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Question: `qquery 4.3.40 11028 decimal to base 4
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Your solution:
Thus our number is 2230110 base 4.
3
Ok
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Question: `qquery 4.3.51 DC in base 16 to binary
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Your solution:
C stands for decimal 12, which in binary is 1100.
D stands for decimal 13, which in binary is 1101.
Since our base 16 is the fourth power of 2, our number can be written 11011100, where the 1101 in the first four places stands for D and the 1100 in the last four places stands for C.
Note that this method works only when one base is a power of the other
3
Ok
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Question: `qIs a base-9 number always even, always odd, or sometimes even and sometimes odd if the number ends in an even number?
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Your solution:
You can investigate this question by trying a variety of examples.
For example, 82nine = 8 * 9^1 + 2 * 9^0 = 72 + 2 = 74 in decimal. This is even because you are adding two even multiples of the odd numbers 9^1 = 9 and 9^0 = 1.
You have to use an even multiple of 9^0 = 1 because the number ends with an even number. But you don't have to use an even multiple of 9^1 = 9.
So we try something like 72nine = 7 * 9^1 + 2 * 9^0 = 63 + 2 = 65 in decimal and we see that the result is odd.
The key is that in base nine, the powers of nine are always odd numbers.
So when converting you get from a base-9 number with two even digits a number which is even * odd + even = even, while from a base-9 number with an odd digit followed by an even digit you get odd * odd + even = odd; many other possible combinations show the same thing but this is one of the simpler examples that shows why we get odd for some base-9 numbers, and even for others.
3
Ok
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