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How would you calculate 3001 plus 1999 in two different ways and 3001 minus 1999 in two different ways but working out all four calculations entirely in Roman numerals with explanations?
Wiki User
∙ 9y ago
The modern way of converting 1999 into Roman numerals is now considered to be MCMXCIX but the ancient Romans would have worked out the equivalent of 1999 on an abacus counting device as MDCCCCLXXXXVIIII and then probably abridged it to IMM thus facilitating the speed and ease of the required calculations as follows:-
MMMI+IMM = (V) => 3001+(2000-1) = 5000
MMMI+MDCCCCLXXXXVIIII = (V) => 3001+1999 = 5000
MMMI-IMM = MII => 3001-(2000-1) = 1002
MMMI-MDCCCCLXXXXVIIII = MII => 3001-1999 =1002
Note that in mathematics -(2000-1) changes to 1-2000 and that the Roman numeral of (V) means 1000*5 = 5000
QED
Wiki User
∙ 9y ago
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How would you actually add together 1666 and 1999 in two different ways entirely in Roman numerals throughout both calculations with explanations?
See answer to question: ' How do you add together 1666 and 1999 in two different ways using Roman numerals'
How would you work out 1776 plus 549 in two different ways and 1776 minus 549 in two different ways but working out all four calculations entirely in Roman numerals with explanations?
Roman numerals are entirely inappropriate for doing such calculations. I believe the people in Roman times did such calculations on an abacus or something similar - which is basically similar to converting them to the Arabic numbers we use. If you really want to do it in Roman numerals - which is basically NOT a good idea - you would have to keep the thousands, hundreds, etc. separate, and handle carry (for addition) and borrowing (for subtraction).
How would you work out 1776 plus 9 in two different ways and 1776 minus 9 in two different ways but working out all four calculations entirely in Roman numerals with explanations?
When 9 is converted into Roman numerals it is IX which is an abridged version of VIIII and so the required calculations are as follows:-MDCCLXXVI+IX = MDCCLXXXV => 1776+(10-1) = 1785MDCCLXXVI+VIIII = MDCCLXXXV => 1776+9 = 1785MDCCLXXVI-IX = MDCCLXVII => 1776-(10-1) = 1767MDCCLXXVI-VIIII = MDCCLXVII => 1776-9 = 1767Note that in mathematics -(10-1) changes to 1-10QED
How would you correctly calculate 52 plus 49 and 52 minus 49 but working out both calculations entirely in Roman numerals with explanations?
The modern way of expressing 49 into Roman numerals is now XLIX but the ancient Romans would have probably worked out the equivalent of 49 on an abacus counting frame as XXXXVIIII and then wrote it out as IL thus expediently working out the required calculations as follows:-LII+IL = CI => 52+(50-1) = 101LII-IL = III => 52-(50-1) = 3Note that in mathematics -(50-1) becomes -50+1 and that if we were to use the longer version of 49 in the above calculations the results would be exactly the same.QED
What is the sum of 1999 and 2014 added together in two different ways entirely in Roman numerals with explanations?
MIM + MMXIV = MMMXIII or MMCXCIX + MMXIII = MMMXIII There is only one way to write the solution (3013)
How would you actually add together 1666 and 1999 in two different ways entirely in Roman numerals throughout both calculations with explanations?
See answer to question: ' How do you add together 1666 and 1999 in two different ways using Roman numerals'
How do you calculate 66013.42 to get a number 66087?
You cannot.66013.42 and 66087 are two entirely different numbers. You cannot "calculate" one to the other.You cannot.66013.42 and 66087 are two entirely different numbers. You cannot "calculate" one to the other.You cannot.66013.42 and 66087 are two entirely different numbers. You cannot "calculate" one to the other.You cannot.66013.42 and 66087 are two entirely different numbers. You cannot "calculate" one to the other.
How would you work out 1776 plus 549 in two different ways and 1776 minus 549 in two different ways but working out all four calculations entirely in Roman numerals with explanations?
Roman numerals are entirely inappropriate for doing such calculations. I believe the people in Roman times did such calculations on an abacus or something similar - which is basically similar to converting them to the Arabic numbers we use. If you really want to do it in Roman numerals - which is basically NOT a good idea - you would have to keep the thousands, hundreds, etc. separate, and handle carry (for addition) and borrowing (for subtraction).
How would you work out 1776 plus 9 in two different ways and 1776 minus 9 in two different ways but working out all four calculations entirely in Roman numerals with explanations?
When 9 is converted into Roman numerals it is IX which is an abridged version of VIIII and so the required calculations are as follows:-MDCCLXXVI+IX = MDCCLXXXV => 1776+(10-1) = 1785MDCCLXXVI+VIIII = MDCCLXXXV => 1776+9 = 1785MDCCLXXVI-IX = MDCCLXVII => 1776-(10-1) = 1767MDCCLXXVI-VIIII = MDCCLXVII => 1776-9 = 1767Note that in mathematics -(10-1) changes to 1-10QED
How would you correctly calculate 52 plus 49 and 52 minus 49 but working out both calculations entirely in Roman numerals with explanations?
The modern way of expressing 49 into Roman numerals is now XLIX but the ancient Romans would have probably worked out the equivalent of 49 on an abacus counting frame as XXXXVIIII and then wrote it out as IL thus expediently working out the required calculations as follows:-LII+IL = CI => 52+(50-1) = 101LII-IL = III => 52-(50-1) = 3Note that in mathematics -(50-1) becomes -50+1 and that if we were to use the longer version of 49 in the above calculations the results would be exactly the same.QED
What is the sum of 1999 and 2014 added together in two different ways entirely in Roman numerals with explanations?
MIM + MMXIV = MMMXIII or MMCXCIX + MMXIII = MMMXIII There is only one way to write the solution (3013)
What is 1776 plus 249 added in two different ways and 1776 minus 249 subtracted in two different ways but working out all four calculations entirely in Roman numerals with explanations?
Under today's modern rules now governing the Roman numeral system the equivalent of 249 when converted into Roman numerals is now considered to be CCXLIX which does not lend itself quite easily to arithmetical operations but there exist credible evidence to suggest that the ancient Romans would have carried out the requested calculations as follows:-MDCCLXXVI+ICCL = MMXXV => 1776+(250-1) = 2025MDCCLXXVI+CCXXXXVIIII = MMXXV => 1776+249 = 2025MDCCLXXVI-ICCL = MDXXVII => 1776-(250-1) = 1527MDCCLXXVI-CCXXXXVIIII = MDXXVII => 1776-249 = 1527Note that in mathematics -(250-1) becomes -250+1 or as 1-250The above calculations were fairly simple and straight forward to work out but for more complicated calculations the Romans would make use of an abacus calculating device.QED
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 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 calculate 1776 plus 549 and 1776 minus 549 but working out both calculations entirely in Roman numerals with explanations?
The modern way of expressing 549 in Roman numerals is now DXLIX but the ancient Romans would have probably worked it out on an abacus calculating device as DXXXXVIIII and then abridged it to IDL in wrtten format thus facilitating the speed and ease of calculations as follows:-MDCCLXXVI+IDL = MMCCCXXV => 1776+(550-1) = 2325MDCCLXXVI-IDL = MCCXXVII => 17776-(550-1) = 1227Note that in mathematics -(550-1) becomes +1-550 and that if we were to use the elongated version of 549 instead of the abridged version the results would be exactly the same in both calculations.QED
What is 1776 plus 89.5 added in two different ways and 1776 minus 89.5 subtracted in two different ways entirely in Roman numerals with explanations?
The rules now governing the Roman numeral system as we know them today had nothing to do with the Romans because they were introduced during the Middle ages but there exist credible evidence to show that the ancient Romans would have worked out all four calculations in the following formats:-MDCCLXXVI+SXC = MDCCCLXVS => 1776+(100-10.5) = 1865.5MDCCLXXVI+LXXXVIIIIS = MDCCCLXVS => 1776+89.5 = 1865.5MDCCLXXVI-SXC = MDCLXXXVIS => 1776-(100-10.5) = 1686.5MDCCLXXVI-LXXXVIIIIS = MDCLXXXVIS => 1776-89.5 = 1686.5Note that in mathematics -(100-10.5) becomes 10.5-100 and that the above calculations were fairly simple and straightforward to work out but for more complicated calculations the Romans would have used an abacus calculating device.QED
What is 1218 plus 558 added in two different ways and 1218 minus 558 subtracted in two different ways but working out all four calculations entirely in Roman numerals with explanations?
The following answers are self explanatory:-IIMCCXX+IIDLX = MDCCLXXVI => (1220-2)+(560-2) = 1776MCCXVIII+DLVIII = MDCCLXXVI => 1218+558 = 1776IIMCCXX-IIDLX = DCLX => (1220-2)-(560-2) = 660MCCXVIII-DLVIII = DCLX => 1218-558 = 660Note that the rules as we now know them today governing the Roman numeral system had little or nothing to do with the ancient Romans whatsoever because they were introduced during the Middle Ages and so therefore the Romans would have probably worked out the given simple calculations as shown above but for more complicated calculations the Romans would have made usage of an abacus calculating device.QED
