Large Numbers Around Us · Lesson 8 of 9
Counting Digits and Exploring Number Names
“Count large writing patterns by blocks and explore the structure of English number names.”
• State consistent spelling and letter-counting conventions for number-name puzzles. • Explain a longest seven-digit number name by maximising its component words. • Investigate shared letters in consecutive names using both examples and carry cases. • Locate a requested digit in the endless string of consecutive positive integers. • Count occurrences of a chosen digit by place positions and complete blocks.
A counting convention comes first
Writing a number in words creates a new kind of counting problem. We are counting letters, not digits or the number’s value. Different spellings, optional “and”, and singular or plural group words can change the answer, so agree on a convention before comparing names.
In this lesson, ignore spaces and hyphens, omit “and”, and use the group words lakh, thousand, and hundred in the singular. Thus 10,30,285 is “ten lakh thirty thousand two hundred eighty-five”. Its letter count is 3 + 4 + 6 + 8 + 3 + 7 + 6 + 4 = 41. Using “lakhs” instead adds one letter, producing the chapter’s count of 42.
The agreed rules for which written characters count and how number names are spelled. A maximum-letter answer is meaningful only under a consistent convention.
A long name need not be the largest number
A seven-digit Indian numeral has a lakh group from 10 to 99, a thousand group from 0 to 99, and a final group from 0 to 999. Choosing 99 in every group gives the largest value, but the word “ninety” is shorter than “seventy”. We want to maximise letters in each named component, not maximise the digits.
Among two-digit names, seventy-three, seventy-seven, and seventy-eight each have twelve letters without the hyphen. The tens word seventy has seven letters; three, seven, and eight each have five. Other tens words have at most six letters, and other unit words have at most five. For the hundreds digit, three, seven, or eight gives five letters before the seven-letter word hundred.
Problem
Find a seven-digit number whose Indian name has the maximum letter count under our convention.
- 1.Choose 77 in the lakh group: “seventy-seven” contributes 12 letters, then “lakh” contributes 4.
- 2.Choose 77 in the thousand group: another 12 letters plus 8 for “thousand”.
- 3.Choose 777 in the last group: “seven” contributes 5, “hundred” contributes 7, and “seventy-seven” contributes 12.
- 4.The numeral is 77,77,777 and the total is 12 + 4 + 12 + 8 + 5 + 7 + 12 = 60 letters.
- 5.Every variable component has reached its maximum under these rules. Other choices using three, seven, or eight in the relevant unit positions can tie, so this is a maximum example, not a unique number.
- 6.With “lakhs” consistently used for the lakh group, the corresponding maximum becomes 61. Do not compare a singular-convention answer against a plural-convention answer without adjustment.
| Component of 77,77,777 | Letters counted |
|---|---|
| seventy-seven | 12 |
| lakh | 4 |
| seventy-seven | 12 |
| thousand | 8 |
| seven | 5 |
| hundred | 7 |
| seventy-seven | 12 |
| Total under the stated convention | 60 |
Do consecutive names always share a letter?
The chapter asks how far you must count to find consecutive number names with no English letter in common. Start by checking small pairs, but remember that examples alone cannot answer an unlimited question. The structure of number names lets us consider the places where a name changes most: the carry boundaries.
Zero and one share e and o; one and two share o; two and three share t. The adjacent single-digit pairs all share a letter. Ten and eleven share e and n, eleven and twelve share e and l, and twelve and thirteen share t and e. Most later teen pairs share letters in “teen”.
| Boundary pair | A shared letter or word |
|---|---|
| nine → ten | n and e |
| nineteen → twenty | n, e, or t |
| twenty-nine → thirty | t or i |
| thirty-nine → forty | r, t, or y |
| forty-nine → fifty | f, i, or y |
| fifty-nine → sixty | i, t, or y |
| sixty-nine → seventy | s, t, or n |
| seventy-nine → eighty | e, t, or y |
| eighty-nine → ninety | n, i, or t |
| ninety-nine → one hundred | n or e |
Away from a boundary, unchanged tens or hundreds words already provide shared letters. When a larger group changes but its scale word remains, both names share a word such as thousand, lakh, million, or crore. When an entirely new scale starts, the earlier name ends in nine and the new power-of-ten name begins with one, ten, or one hundred; n or e is shared. These cases explain why continuing through the usual decimal English naming patterns does not produce the requested letter-free consecutive pair.
Problem
Explain why 999 and 1,000 share letters even though their group words change.
- 1.The first is “nine hundred ninety-nine”; the second is “one thousand”.
- 2.The old final word nine and the new first word one share n and e.
- 3.For 1,999 and 2,000, the names both contain thousand, so there is a shared word even before examining the other letters.
- 4.Checking these boundary types is stronger than merely testing a long list of easy consecutive pairs.
Checking the first hundred pairs is useful evidence, but it does not by itself prove a statement about larger numbers. Explain both unchanged-prefix cases and carry boundaries, and keep the naming convention consistent.
Writing every positive integer without separators
Now imagine the string 1234567891011121314… formed by writing 1, 2, 3, and so on with no spaces or commas. A digit’s position in this string is different from the number containing it. The first nine numbers contribute nine digits, but each number from 10 onward initially contributes two.
There are 90 two-digit numbers, from 10 through 99 inclusive, so they contribute 90 × 2 = 180 digits. There are 900 three-digit numbers, each contributing three digits. Organising the writing into these blocks avoids counting thousands of characters one by one.
| Numbers in block | Number of numbers | Digits contributed | Total digits through block |
|---|---|---|---|
| 1–9 | 9 | 9 | 9 |
| 10–99 | 90 | 180 | 189 |
| 100–999 | 900 | 2,700 | 2,889 |
| 1,000–9,999 | 9,000 | 36,000 | 38,889 |
| 10,000–99,999 | 90,000 | 4,50,000 | 4,88,889 |
| 1,00,000–9,99,999 | 9,00,000 | 54,00,000 | 58,88,889 |
If k is a digit length, the count of k-digit positive integers is nine times 10 multiplied by itself k − 1 times. For k = 3 this means 9 × 100 = 900. Multiplying that count by k gives the number of written digits in the block; this distinction between numbers and digits is essential.
Problem
Find the 1,000th written digit and the number containing it.
- 1.The one- and two-digit blocks contribute 9 + 180 = 189 digits. The three-digit block extends to position 2,889, so position 1,000 lies in it.
- 2.The requested position within that block is 1,000 − 189 = 811.
- 3.270 complete three-digit numbers contribute 810 digits. Starting at 100, those numbers run through 369.
- 4.The next number is 370. Its first digit occupies the next position, so the 1,000th digit is 3 in 370.
- 5.Check the boundary: through 369, the string has 189 + 270 × 3 = 999 digits. The first digit of 370 is therefore exactly position 1,000.
Problem
Which number contains the 1,000,000th written digit?
- 1.Through all five-digit numbers, the total is 488,889 digits. The next block uses six-digit numbers beginning at 100,000.
- 2.The position within this block is 1,000,000 − 488,889 = 511,111.
- 3.85,185 complete six-digit numbers contribute 511,110 digits. They run from 100,000 through 185,184.
- 4.The next number is 185,185, and its first digit is at position 1,000,000.
- 5.Check: 488,889 + 85,185 × 6 = 999,999. Thus the millionth digit is in 185,185, and it is 1.
A safe general procedure is to find the block, remove the digits in all earlier blocks, and then remove complete numbers within the chosen block. If the remaining position is exactly a multiple of the block length, it is the last digit of a number, not the first digit of the next one. Testing the position immediately before your answer helps catch this boundary error.
An occurrence count is a different question
Finding the 5,000th occurrence of the digit 5 is not the same as finding position 5,000 in the string. We count only characters equal to 5, including more than one occurrence in a number such as 55. Complete blocks let us count by digit positions without writing the entire string.
Temporarily write the numbers 0000 through 9999 as four-position strings. In each position, 5 appears exactly 1,000 times because the other three positions have 1,000 combinations. This gives 4,000 occurrences. Removing leading zeros and dropping 0000 does not remove any 5, so the actual numbers 1 through 9,999 also contain 4,000 occurrences.
Problem
Find the number at which the digit 5 is written for the 5,000th time.
- 1.Through 9,999 there are 4,000 occurrences.
- 2.For each thousand-number block 10,000–10,999, 11,000–11,999, and 12,000–12,999, the last three positions contribute 300 fives. Their fixed first two digits contain no 5. Thus through 12,999 the count is 4,900.
- 3.In 13,000–13,399, the units position contributes 40 fives and the tens position contributes 40. The hundreds positions 0–3 add none. The running count is 4,980.
- 4.From 13,400 through 13,489, units ending in 5 occur nine times, and a tens digit of 5 occurs in 13,450–13,459 ten times. This adds 19, reaching 4,999 occurrences.
- 5.13,490 through 13,494 contain no 5. The next number, 13,495, ends in 5 and produces occurrence 5,000.
- 6.Therefore the required number is 13,495. A number such as 13,455 contributes two occurrences, because its two fives occupy different positions.
| Writing completed through | New occurrences in the described part | Running total of 5s |
|---|---|---|
| 9,999 | 4,000 | 4,000 |
| 12,999 | 900 | 4,900 |
| 13,399 | 80 | 4,980 |
| 13,489 | 19 | 4,999 |
| 13,494 | 0 | 4,999 |
| 13,495 | 1 | 5,000 |
The numeral 555 contributes three occurrences of 5, not one. In the padded-block method, removing leading zeros is harmless when counting 5, but would change a count of zeros. Use the method with that distinction in mind.
Quiz
Under the singular-lakh convention with no “and”, 10,30,285 has:
Under the same convention, a maximum-letter name for 77,77,777 has:
How many written digits do 10 through 99 contribute?
What is the 1,000th digit in 123456789101112…?
Which number contains the millionth written digit?
Which number produces the 5,000th occurrence of digit 5?
How many occurrences of 5 are in the number 5,055?
Practice Problems
- Count the letters in 10,30,285 and 77,77,777 under the stated convention. Repeat with “lakhs” and explain the one-letter change.
- Find another seven-digit number tying the maximum letter count. Explain why the largest seven-digit numeral need not have the longest name.
- Find shared letters in eight consecutive pairs near a tens or hundreds boundary. Explain why these observations alone are not the whole argument.
- Use unchanged scale words and new-scale carry boundaries to explain why consecutive names share letters in the naming patterns used here.
- Rebuild the digit-block table through the six-digit block. Distinguish the count of numbers from the count of characters.
- Find the digits at positions 189, 190, and 191 in the concatenated string. Explain the change between blocks.
- Locate the 1,000th and millionth written digits using a count of complete preceding numbers. Verify the immediately preceding position.
- Explain why counting 5s in padded four-position strings gives the same result as counting them in numbers 1–9,999, and why counting zeros would differ.
- Reconstruct the running occurrence count from 4,000 through 5,000 fives. Confirm that 13,494 is before and 13,495 is at the requested occurrence.
- Create a smaller digit-occurrence challenge that you can check by listing the numbers as well as by counting blocks.
Key Takeaways
• Number-name letter counts depend on spelling, plural, spacing, hyphen, and “and” conventions. • Maximise component word lengths rather than numeric value to find a long name. • Shared prefixes and carry-boundary cases explain patterns in consecutive names. • A digit position, a containing number, and an occurrence count are different quantities. • Complete digit-length blocks replace tedious character-by-character counting. • Check the preceding boundary to avoid moving one number too far. • Count every repeated occurrence of the chosen digit.