Q1.
Represent the following numbers in the Roman system. (i) 1222 (ii) 2999 (iii) 302 (iv) 715
Answer
Use the book’s own rule: take as many 1000s as possible, then as many 500s, then 100s, 50s, 10s, 5s and finally 1s.
(i) 1222 = 1000 + 100 + 100 + 10 + 10 + 1 + 1
= MCCXXII
= MCCXXII
(ii) 2999 = 1000 + 1000 + 900 + 90 + 9
= MM + CM + XC + IX = MMCMXCIX
= MM + CM + XC + IX = MMCMXCIX
(iii) 302 = 100 + 100 + 100 + 1 + 1
= CCCII
= CCCII
(iv) 715 = 500 + 100 + 100 + 10 + 5
= DCCXV
= DCCXV
Check by reading back: MCCXXII = 1000 + 200 + 20 + 2 = 1222 ✓ MMCMXCIX = 2000 + 900 + 90 + 9 = 2999 ✓ CCCII = 300 + 2 = 302 ✓ DCCXV = 500 + 200 + 15 = 715 ✓
Why it happens: 2999 is the interesting one. The plain grouping rule would give 900 as DCCCC and 90 as LXXXX and 9 as VIIII, producing MMDCCCCLXXXXVIIII — 17 symbols. Using the “one less than” shortcut (IV for 4, XL for 40) gives CM, XC, IX and only 8 symbols. The book itself notes that people using this system were not always consistent about it, so both forms occur in old inscriptions; the short form is the one in use today.
Tip: a symbol placed before a larger one is subtracted (IX = 10 − 1), and after a larger one is added (XI = 10 + 1). Read a Roman numeral left to right and watch for a small letter sitting in front of a big one.