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Hint: The smallest number of n-digits = 10….0(n-1 times). Find the smallest 4 digit number in the decimal system. Find the representative numeral for that number in the Roman Number system.
Complete step-by-step answer:
Roman Number system: In the Roman system, numerals are used to represent a number. The system works almost similarly in the way coins of various denominations are used to represent any sum of money. In the Roman system, I =1, V = 5, X = 10, L = 50, C = 100, D = 500 , M = 1,000, etc.
We know that the smallest number of n-digits = 10….0(n-1 times). Hence the smallest number of 4 digits is 1000.
Hence the numeral for the smallest four-digit number in the Roman system is M.
Hence option [b] is correct
Note: The smallest number of n-digits is 1000…0 (n-1 times).
Proof: Consider an arbitrary n -digit number $a={{a}_{1}}{{a}_{2}}...{{a}_{n}},a\ne 100...00$
Since a is an n-digit number we have ${{a}_{1}}\ge 1$
If ${{a}_{1}}>1$, then clearly a>100…00 and we are done.
If ${{a}_{1}}=1$, then since a is not equal to 100…00, they differ by at least one decimal place.
Let i=k be the smallest value of i such that $i>1,{{a}_{i}}\ne 0$.
So, we have $a=100...{{a}_{k}}{{a}_{k+1}}\ldots {{a}_{n}}$
Since $0\le {{a}_{k}}\le 9$we have ${{a}_{k}}>0$
So a>100…00.
Hence proved.
Complete step-by-step answer:
Roman Number system: In the Roman system, numerals are used to represent a number. The system works almost similarly in the way coins of various denominations are used to represent any sum of money. In the Roman system, I =1, V = 5, X = 10, L = 50, C = 100, D = 500 , M = 1,000, etc.
We know that the smallest number of n-digits = 10….0(n-1 times). Hence the smallest number of 4 digits is 1000.
Hence the numeral for the smallest four-digit number in the Roman system is M.
Hence option [b] is correct
Note: The smallest number of n-digits is 1000…0 (n-1 times).
Proof: Consider an arbitrary n -digit number $a={{a}_{1}}{{a}_{2}}...{{a}_{n}},a\ne 100...00$
Since a is an n-digit number we have ${{a}_{1}}\ge 1$
If ${{a}_{1}}>1$, then clearly a>100…00 and we are done.
If ${{a}_{1}}=1$, then since a is not equal to 100…00, they differ by at least one decimal place.
Let i=k be the smallest value of i such that $i>1,{{a}_{i}}\ne 0$.
So, we have $a=100...{{a}_{k}}{{a}_{k+1}}\ldots {{a}_{n}}$
Since $0\le {{a}_{k}}\le 9$we have ${{a}_{k}}>0$
So a>100…00.
Hence proved.
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