1 General Mental and Psychological Abilities · 1.2.1 Number system and simplification

1.2.1.14 Place value, Roman numerals and counting digits in a range

Read every digit by its place, write Roman numerals from the seven symbols and six subtractive pairs, and count digits place by place or block by block, never number by number.

  • APPSC Group 1
  • Advanced level
  • 7 min read
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Core facts

  • Indian system: ones, tens, hundreds, thousands, ten thousands, lakhs, ten lakhs, crores, ten crores. Commas: 4,52,46,317.
  • International system: thousands, millions, billions. Commas every three digits: 45,246,317.
  • 1 lakh = 100 thousand; 10 lakh = 1 million; 1 crore = 10 million; 100 crore = 1 billion.
  • Place value = digit × its place; face value = the digit. Place value of 0 is 0.
  • Roman numerals: I = 1, V = 5, X = 10, L = 50, C = 100, D = 500, M = 1,000.
    • Subtractive pairs only: IV (4), IX (9), XL (40), XC (90), CD (400), CM (900).
    • I, X, C, M may repeat up to three times; V, L, D never repeat and are never subtracted.
    • There is no Roman symbol for zero.
  • Digits used in writing numbers: 1–9 use 9; 10–99 use 180; 100–999 use 2,700. So 1–99 use 189 and 1–999 use 2,889.

Example 1. Place value of 8 in 3,87,45,612. Solution. 8 is in the ten-lakh place: 8 × 10,00,000. Answer: 80,00,000 (8 million)

Example 2. Write MCMXCIV in numerals. Solution. M = 1,000; CM = 900; XC = 90; IV = 4. Answer: 1,994

Example 3. How many digits are used to write the numbers 1 to 60? Solution. 9 single-digit numbers, then 51 two-digit numbers (10–60): 9 + 102. Answer: 111

Explained

Type 1: place values of repeated digits.

Example 4. Difference between the place values of the two 4s in 4,52,46,317. (a) 3,99,60,000 (b) 4,00,00,000 (c) 3,99,96,000 (d) 39,960 Solution. The first 4 is in the crore place: 4,00,00,000. The second is in the ten-thousand place: 40,000. Difference = 3,99,60,000. Answer: (a) 3,99,60,000

Type 2: number names in both systems.

Example 5. Write "seven crore four lakh thirty thousand nine" in figures, and read it in the international system. Solution. 7,04,30,009 = 70,430,009, read "seventy million four hundred thirty thousand nine". Answer: 7,04,30,009

Type 3: writing Roman numerals. Break the number into thousands, hundreds, tens and units, and write each part.

Example 6. Write 2,749 in Roman numerals. Solution. 2,000 = MM; 700 = DCC; 40 = XL; 9 = IX. Answer: MMDCCXLIX

Type 4: pages from digits used.

Example 7. 1,002 digits are used to number the pages of a book from page 1. How many pages does it have? Solution. Pages 1–99 use 189 digits. The remaining 813 digits give 813 ÷ 3 = 271 three-digit pages (100 to 370). Answer: 370

Example 8. How many digits are used to write 1 to 1,000? Solution. 9 + 180 + 2,700 + 4. Answer: 2,893

Type 5: how many times a digit is written. Count each place separately.

Example 9. How many times is the digit 5 written from 1 to 500? How many numbers contain at least one 5? Solution. Units place: 5, 15, …, 495 → 50. Tens place: 50–59 in each of the 5 hundreds → 50. Hundreds place: only 500 → 1. Total 101. For numbers containing a 5, count 0–499 with no 5: 5 × 9 × 9 = 405, which includes 0, so 404 in 1–499. That leaves 499 − 404 = 95 containing a 5, plus 500 itself. Answer: 101 times; 96 numbers

Type 6: complement counting. "Contains at least one" = total − "contains none".

Example 10. How many numbers from 1 to 400 contain the digit 4 at least once? Solution. From 0 to 399, numbers with no 4: hundreds 0–3 (4 choices) × 9 × 9 = 324, which includes 0, so 323 in 1–399. 400 contains a 4. So 1–400 has 400 − 323 = 77 containing a 4. Answer: 77

Speed tips.

  • "How many times is the digit written" counts 55 twice; "how many numbers contain it" counts 55 once.
  • For page problems, subtract 189 (pages 1–99) before dividing by 3.

Deeper

Example 11. A 99-digit number is formed by writing 1, 2, 3, … in a row. Find (i) the last number written and (ii) the remainder when the 99-digit number is divided by 8. (a) 54, 2 (b) 54, 4 (c) 49, 2 (d) 55, 6 Solution. (i) 9 digits for 1–9 leave 90 digits = 45 two-digit numbers, 10 to 54. (ii) The number ends …5354; its last three digits 354 leave 2 on 8. Answer: (a) 54, 2

  • Its digit sum leaves the same remainder on 9 as 1 + 2 + … + 54 = 1,485, which is divisible by 9, so the number is divisible by 9.

Example 12. In the string 123456789101112…, which digit is in the 200th place? Solution. 1–9 use 9 places; 10–99 use 180 (189 in all). 11 places remain: 100, 101, 102 use 9 (198), so places 199 and 200 are the first two digits of 103: "1" and "0". Answer: 0

Example 13. Find CDXLIV + LVI, in Roman numerals. Solution. 444 + 56 = 500 = D. Answer: D

Example 14. How many three-digit numbers have all three digits different? Solution. Hundreds: 9 choices (1–9). Tens: 9 (0–9 except the first). Units: 8. 9 × 9 × 8. Answer: 648

Example 15. Find the sum of all digits used in writing 1 to 999. Solution. Treat the numbers as 000 to 999: each of the 3 places shows every digit 0–9 exactly 100 times. Sum = 3 × 100 × 45. Answer: 13,500

  • Similarly, 1 to 99 gives 2 × 10 × 45 = 900.
  • Zeros are special: from 1 to 1,000 the digit 0 is written 192 times (units place 100, tens place 91, hundreds place 1), and 181 numbers contain a 0. Count place by place and do not pad with leading zeros.

Common traps

  • ✗ "1 million = 1 crore." ✓ 1 million = 10 lakh; 1 crore = 10 million.
  • ✗ "1,999 = MIM." ✓ Subtract only within the six pairs: MCMXCIX.
  • ✗ "4 = IIII and 40 = XXXX." ✓ No symbol repeats four times: IV and XL.
  • ✗ "1,002 digits ÷ 3 = 334 pages." ✓ Pages 1–99 use only 189 digits: 370 pages.
  • ✗ "Digit 5 appears 96 times from 1 to 500." ✓ 96 is how many numbers contain 5; the digit is written 101 times.
  • ✗ "Numbers with no 4 from 1 to 399 = 4 × 9 × 9 = 324." ✓ That counts 0 too: 323.

Quick revision

  • 1 lakh = 100,000; 10 lakh = 1 million; 1 crore = 10 million; 100 crore = 1 billion.
  • Place value = digit × place; face value = digit.
  • Roman: I 1, V 5, X 10, L 50, C 100, D 500, M 1,000.
  • Subtractive pairs: IV, IX, XL, XC, CD, CM only.
  • V, L, D never repeat; I, X, C, M at most three times.
  • Digits: 9 (1–9), 180 (10–99), 2,700 (100–999).
  • Pages from digits: (digits − 189)/3 + 99.
  • Count a digit place by place; numbers containing it by complement.
  • Sum of digits 1–99 = 900; 1–999 = 13,500.

Sources: NCERT Mathematics textbook, Class 6 (knowing our numbers: Indian and international systems, Roman numerals); Standard counting results (place-wise digit counting, complement counting)

Last checked 2026-10-17.

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