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marshall27 [118]
3 years ago
8

Which is the largest three-digit number of the form 9k + 1, where k is any positive integer?​

Mathematics
1 answer:
Contact [7]3 years ago
5 0

Answer:

991

Step-by-step explanation:

We are looking for a number that is one more than a multiple of 9(denoted by the 9k) and is a three digit number. We can start by looking for 3-digit number divisible by 9 which are close to 1,000, since that is the next number larger than the largest three-digit number. We can tell that 999 is divisible by 9 because when divided it does not leave a remainder(you can also figure this out with divisibility tricks). We add one to get 1,000. This is not a three-digit number,  so we need to look for a smaller multiple of 9. Subtracting 9 from 999, we get the next largest multiple of 9. We can add 1, and this time, the number, 991, is a three digit number, and the largest that can be in the form 9k + 1.

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\qquad \qquad \textit{direct proportional variation} \\\\ \textit{\underline{y} varies directly with \underline{x}}\qquad \qquad \stackrel{\textit{constant of variation}}{y=\stackrel{\downarrow }{k}x~\hfill } \\\\ \textit{\underline{x} varies directly with }\underline{z^5}\qquad \qquad \stackrel{\textit{constant of variation}}{x=\stackrel{\downarrow }{k}z^5~\hfill } \\\\[-0.35em] \rule{34em}{0.25pt}

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3 years ago
Read 2 more answers
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