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IRISSAK [1]
3 years ago
11

How many integers from 1 through 100,000 contain the digit 6 exactly once?

Mathematics
1 answer:
wariber [46]3 years ago
4 0
Consider the set of all (not-all-zero) decimal strings of length 6. This is the set of strings


000001
000002
...
099998
099999
100000

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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The 6th term is 22 and the common difference is 6 what is the fiftieth term?
SOVA2 [1]

Answer:

the fiftieth term is 76

Step-by-step explanation:

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Answer: 76

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Hope this helps!

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Step-by-step explanation:

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What is the equation of a line, in general form, that passes though points (-1,2) and (5,2)?
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A

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