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siniylev [52]
3 years ago
14

Plz help out with this!!!!

Mathematics
1 answer:
liberstina [14]3 years ago
4 0
1. 6 times 6 = 36 cm
2. 9 times 11 = 99 yd

3. 1/2(13)(8+3)
6.5(11)
71.5 ft

4. 12 times 4 = 48 m
5. 5 times 3 = 15 in
6. 8 times 4 = 32 mm

7. 1/2(8)(10+2)
4(12)
48 yd

8. 4 times 4 = 16 cm
9. 12 times 15 = 180 in
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How many integers from 1 through 100,000 contain the digit 6 exactly once?
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Consider the set of all (not-all-zero) decimal strings of length 6. This is the set of strings


000001
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099999
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There are obviously 100,000 strings in this set, so we have a one-to-one correspondence to the integers between 1 and 100,000. Think of any string starting with 0s as the number with the leading 0s chopped off.

There are two choices for the first digit, either 0 or 1, but a number can only contain a 6 if the first digit is 0; otherwise, the number would exceed 100,000. For every digits place afterward, if a given digits place contains a 6, then the remaining four places have 9 possible choices each, choosing from 0-9 excluding 6. If we fix the 6 in, say, the second digits place, then the number of integers between 1 and 100,000 containing exactly one 6 is


1\cdot1\cdot9^4=6561


where the first 1 refers to the only choice of 0 in the first digits place, the second 1 refers to the unique 6 in the next place, and the remaining four places are filled with one of 9 possible choices.


Now, notice that we can permute the digits of such a number in 5 possible ways. That is, there are 5 choices for the placement of the 6 in the number, so we multiply this count by 5.

5(1\cdot1\cdot9^4)=32,805
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