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How would you work out 1776 plus 999 in two different ways and 1776 minus 999 in two different ways showing all four calculations from start to finish entirely in Roman numerals?
Wiki User
∙ 9y ago
The modern way of expressing 999 as Roman numerals is CMXCIX but there exist historical evidence to suggest that the ancient Romans would have probably calculated the equivalent of the given numbers as follows:-
MDCCLXXVI+IM = MMDCCLXXV => 1776+(1000-1) = 2775
MDCCLXXVI+DCCCCLXXXXVIIII = MMDCCLXXV => 1776+999 = 2775
MDCCLXXVI-IM = DCCLXXVII => 1776-(1000-1) = 777
MDCCLXXVI-DCCCCLXXXXVIIII = DCCLXXVII => 1776-999 = 777
The above calculations are fairly simple and straightforward to work out but for more complicated calculations the Romans would have used an abacus counting device.
QED
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∙ 9y ago
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How would you add together 31 plus 19 in two different ways and 31 minus 19 in two different ways showing all four calculations entirely in Roman numerals?
The modern way of expressing 19 into Roman numerals is now XIX but the ancient Romans would have probably worked out the above calculations as follows: A: XXXI+IXX = L => 31+(20-1) = 50. B: XXXI+XVIIII = L => 31+19 = 50. C: XXXI-IXX = XII => 31-(20-1) = 12. D: XXXI-XVIIII = XII => 31-19 = 12.
How would you add together 1999 plus 199 plus 19 in two different ways but showing all calculations and answers entirely in Roman numerals with explanations?
With great difficulty because under today's modern rules now governing the Roman numeral system the equivalent of 1999, 199 and 19 when converted into Roman numerals are now considered to be MCMXCIX, CXCIX and XIX respectively which does not lend themselves quite easily conducive to logical mathematical operations amongst themselves but notwithstanding the aforementioned inasmuch that there exist credible historical evidence to suggest that the ancient Romans would have calculated the required calculations as follows:-IMM+ICC+IXX = MMCCXVII => (2000-1)+(200-1)+(20-1) = 2217MDCCCCLXXXXVIIII+CLXXXXVIIII+XVIIII = MMCCXVII => 1999+199+19 = 2217Note that the above calculations are fairly simple and straightforward but for more complicated calculations the ancient Romans would have made use of an abacus calculating device.QED
How would you add together 1776 plus 99 plus 1907 in two different ways showing both completed calculations entirely in Roman numerals with reasons why?
When in Rome do as the Romans do because the ancient Romans would have probably added together the given numbers in either of the following configurations:- As: MDCCLXXVI+IC = MDCCCLXXV => 1776+(100-1) = 1875. MDCCCLXXV+CMMVII = MMMDCCLXXXII => 1875+(2007-100) = 3782. Or as: MDCCLXXVI+LXXXXVIIII = MDCCCLXXV => 1776+99 = 1875. MDCCCLXXV+MDCCCCVII = MMMDCCLXXXII => 1875+1907 = 3782. Note that in todays modern configuration of Roman numerals 99 and 1907 are now considered to be XCIX and MCMVII respectively. The above calculations are fairly simple and straightforward but for more complicated calulations the Romans would have used an abacus counting device.
What is 1700.75 plus 75.25 and 1700.75 minus 75.25 but showing all work and answers entirely in Roman numerals?
They are worked out as in the following formats:- MDCCS:. + LXXV:. = MDCCLXXVI => 1700.75 + 75.25 = 1776 MDCCS:. - LXXV:. = MDCXXVS => 1700.75 - 75.25 = 1625.5
How would you add together 1999 plus 19 plus 199 in two different ways but showing both calculations entirely in Roman numerals with explanations?
Because of the changes made to the rules now governing the Roman numeral system introduced during the Middle Ages nowadays 1999, 19 and 199 when converted into Roman numerals are now configured as MCMXCIX, XIX and CXCIX respectively which restricts some sort of sensible arithmetical interaction amongst them.Notwithstanding the aforementioned inasmuch that there exist historical evidence to suggest that the ancient Romans would have added together the given numbers in either of the following formats:-A: IMM+IXX+ICC = MMCCXVII => (2000-1)+(20-1)+(200-1) = 2217.B: MDCCCCLXXXXVIIII+XVIIII = MMXVIII => 1999+19 = 2018.MMXVIII+CLXXXXVIIII = MMCCXVII => 2018+199 = 2217.Note that the above calculations are fairly simple and straightforward to work out but for more complicated calculations the Romans would have used an abacus calculating device.QED
How would you add together 1776 and 49 in two different ways but showing both calculations entirely in Roman numerals from start to finish?
In todays modern configuration of Roman numerals 49 is now considered to be XLIX but the ancient Romans would have probably added the given numbers in either of the following formats:-A: MDCCLXXVI+IL = MDCCCXXV => 1776+(50-1) = 1825.B: MDCCLXXVI+XXXXVIIII = MDCCCXXV => 1776+49 = 1825.For more complicated calculations the Romans would have used an abacus calculating device.QED
How would you add together 31 plus 19 in two different ways and 31 minus 19 in two different ways showing all four calculations entirely in Roman numerals?
The modern way of expressing 19 into Roman numerals is now XIX but the ancient Romans would have probably worked out the above calculations as follows: A: XXXI+IXX = L => 31+(20-1) = 50. B: XXXI+XVIIII = L => 31+19 = 50. C: XXXI-IXX = XII => 31-(20-1) = 12. D: XXXI-XVIIII = XII => 31-19 = 12.
How would you add together 51 plus 49 in two different ways and 51 minus 49 in two different ways but showing all calculations entirely in Roman numerals with explanations?
Doing arithmetic with Roman numerals is exasperating, and imho a pointless waste of time, except to demonstrate the obvious superiority of our "normal numbers," which use base-10 radix / positional notation that includes a zero digit as a placeholder. I'd venture to say science & technology -- commerce, too -- could never have developed in recent centuries if we still used Roman numerals for calculations. However, this web site explains some methods: http://turner.faculty.swau.edu/mathematics/materialslibrary/roman/
How would to add together 1999 and 3001 in two different ways showing both calculations entirely in Roman numerals?
When in Rome do as the Roman do and the ancient Romans would have probably added together 1999 and 3001 in either of the following ways:- IMM+MMMI = (V) => (2000-1)+3001 = 1000*5 = 5000 MDCCCCLXXXXVIIII+MMMI = (V) => 1999+30001 = 5000 Note that in todays modern configuration of Roman numerals 1999 is now considered to be MCMXCIX
What is the sum of 999 plus 14 plus 1987 added together in two different ways but showing all work and answers entirely in Roman numerals?
Cmxcix + xiv + mcmlxxxvii = mmm. It should be all in capital letters, but the system changed them.
How would you evaluate 49 plus 19 in two different ways and 49 minus 19 in two different ways showing all four calculations entirely in Roman numerals with explanations?
In today's modern conversion of Roman numerals 49 and 19 are now considered to be XLIX and XIX respectively but the ancient Romans would have probably worked out the above as follows:- IL+IXX = LXVIII => (50-1)+(20-1) = 68 XXXXVIIII+XVIIII = LXVIII => 49+19 = 68 IL-IXX = XXX => (50-1)-(20-1) = 30 XXXXVIIII-XVIIII => 49-19 = 30 For more complicated calculations the Romans would use an abacus calculating device.
How would you calculate 1776 plus 1449 and 1776 minus 1449 but showing both calculations worked out entirely in Roman numerals with explanations?
Today's modern way of expressing 1449 as Roman numerals is now MCDXLIX which prohibits sensible interaction with other numerals but the ancient Romans would have worked out the equivalent of 1449 on an abacus counting device as MCCCCXXXXVIIII and probably abridged it to ILMD thus facilitating the speed and ease of calculations as follows:-MDCCLXXVI+ILMD = MMMCCXXV => 1776+(1500-51) = 3225MDCCLXXVI-ILMD = CCCXXVII => 1776-(1500-51) = 327Note that the results would be exactly the same if we were to use the longer version of the equivalent of 1449.QED
How would you add together 3001 plus 1999 in two different ways and 3001 minus 1999 in two different ways but showing all four calculations entirely in Roman numerals?
In today's modern configuration of Roman numerals 1999 is now considered to be MCMXCIX but the ancient Romans would have probably worked out the above as in the following formats:- A: MMMI+IMM = (V) => 3001+[2000-1] = 5000. B: MMMI+MDCCCCLXXXXVIIII = (V) => 3001+1999 = 5000. C: MMMI-IMM = MII => 3001-[2000-1] = 1002. D: MMMI-MDCCCCLXXXXVIIII = MII => 3001-1999 = 1002.
How would you add together 1999 plus 199 plus 19 in two different ways but showing all calculations and answers entirely in Roman numerals with explanations?
With great difficulty because under today's modern rules now governing the Roman numeral system the equivalent of 1999, 199 and 19 when converted into Roman numerals are now considered to be MCMXCIX, CXCIX and XIX respectively which does not lend themselves quite easily conducive to logical mathematical operations amongst themselves but notwithstanding the aforementioned inasmuch that there exist credible historical evidence to suggest that the ancient Romans would have calculated the required calculations as follows:-IMM+ICC+IXX = MMCCXVII => (2000-1)+(200-1)+(20-1) = 2217MDCCCCLXXXXVIIII+CLXXXXVIIII+XVIIII = MMCCXVII => 1999+199+19 = 2217Note that the above calculations are fairly simple and straightforward but for more complicated calculations the ancient Romans would have made use of an abacus calculating device.QED
How would you work out 1776 plus 49 in two different ways and 1776 minus 49 in two different ways but showing all four calculations entirely in Roman numerals from start to finish?
In today's modern world it's generally accepted that 49 when converted into Roman numerals is now XLIX but in the ancient world of the Romans it probably once was IL which is an abridged version of XXXXVIIII and so the above numbers would be worked out as follows:- MDCCLXXVI+IL = MDCCCXXV => 1776+(50-1) = 1825 MDCCLXXVI+XXXXVIIII = MDCCCXXV => 1776+49 = 1825 MDCCLXXVI-IL = MDCCXXVII => 1776-(50-1) = 1727 MDCCLXXVI-XXXXVIIII = MDCCXXVII => 1776+49 = 1727 For more complicated calculations the ancient Romans would have used an abacus calculating device.
What is a mathematical scroll?
old documents showing calculations of the Mayan Calendar
How would you work out 1776 plus 499 in two different ways and 1776 minus 499 in two different ways but showing all four calculations entirely in Roman numerals with explanations?
Nowadays the modern way of converting 499 into Roman numerals is now considered to be CDXCIX but there exist credible evidence to suggest that the ancient Romans would have probably worked out the equivalent of 499 on an abacus counting device as CCCCLXXXXVIIII and then logically abridged it to ID in written format thus facilitating the speed and ease of the aforesaid required calculations as follows:-MDCCLXXVI+ID = MMCCLXXV => 1776+(500-1) = 2275MDCCLXXVI+CCCCLXXXXVIIII = MMCCLXXV => 1776+499 = 2275MDCCLXXVI-ID = MCCLXXVII => 1776-(500-1) = 1277MDCCLXXVI-CCCCLXXXXVIIII = MCCLXXVII => 1776-499 = 1277Note that in mathematics -(500-1) changes to 1-500QED
