Question 6 - Mix Questions (Page 20 & 21) - Chapter 1 Class 7 - Large Numbers Around us (Ganita Prakash)

Transcript

Question 6 Suppose you write down all the numbers 1, 2, 3, 4, …, 9, 10, 11, ... The tenth digit you write is ‘1’ and the eleventh digit is ‘0’, as part of the number 10. (a) What would the 1000th digit be? At which number would it occur?Single-digit numbers (1-9): There are 9 numbers, each contributing 1 digit. Total digits: 9 × 1 = 9 digits. Remaining digits to reach 1000: 1000 − 9 = 991 digits. Two-digit numbers (10-99): There are 99 − 10 + 1 = 90 numbers. Each number contributes 2 digits. Total digits: 90 × 2 = 180 digits. Cumulative digits so far: 9 + 180 = 189 digits. Remaining digits: 991 − 180 = 811 digits (The 1000th digit is not in this range). Three-digit numbers (100-999): We need to find the 811th digit within the block of three-digit numbers. Each three-digit number contributes 3 digits. Number of three-digit numbers completely written: 811 ÷ 3 = 270 with a remainder of 1. This means we write 270 full three-digit numbers. The 1000th digit will be the 1st digit of the 271st three-digit number. The 270th three-digit number is 100 + 270 − 1 = 369. The total digits written up to 369 are 189 + (270 × 3) = 189 + 810 = 999 digits. The 999th digit is the last digit of 369, which is '9'. The 1000th digit will be the first digit of the next number, which is 370. The 1000th digit is '3', and it occurs as part of the number 370. Question 6 Suppose you write down all the numbers 1, 2, 3, 4, …, 9, 10, 11, ... The tenth digit you write is ‘1’ and the eleventh digit is ‘0’, as part of the number 10. (b) What number would contain the millionth digit? We keep counting the total number of digits as we write more numbers: 1-digit numbers (1-9): 9 digits. (Total so far: 9) 2-digit numbers (10-99): 90×2=180 digits. (Total so far: 9+180=189) 3-digit numbers (100-999): 900×3=2700 digits. (Total so far: 189+2700=2889) 4-digit numbers (1000-9999): 9000×4=36000 digits. (Total so far: 2889+36000=38889) 5-digit numbers (10000-99999): 90000×5=450000 digits. (Total so far: 38889+450000=488889) We are looking for the 1,000,000th digit. We've reached 488,889 digits by the time we finish writing 99,999. So, the millionth digit must be in the 6-digit numbers. Digits we still need to write: 1,000,000−488,889=511,111 digits. These digits come from 6-digit numbers, where each number has 6 digits. Number of full 6-digit numbers we need: 511,111÷6=85,185 with 1 left over. This means we'll write 85,185 complete 6-digit numbers. The first 6-digit number is 100,000. The 85,185th 6-digit number will be 100,000+85,185−1=185,184. By the time we finish writing 185,184, we would have written 488,889+(85,185×6)=488,889+511,110=999,999 digits. The 999,999th digit is the '4' from 185,184. The very next digit you write will be the 1,000,000th digit, and it will be the first digit of the next number, which is 185,185. The number that would contain the millionth digit is 185,185. Finding the remaining 1000 '5's: We need the 5000th '5', and we have 4000 so far. So we need 5000−4000=1000 more '5's. These '5's will be found in 5-digit numbers (10000 and onwards) Counting '5's in 5-digit numbers (10000-14999): Numbers like 10xxx, 11xxx, 12xxx, 13xxx, 14xxx. The first digit is never '5'. In each block of 1000 numbers (e.g., 10000-10999), the digit '5' appears 120 times (10 times in units place, 10 times in tens place, 100 times in hundreds place). Let's add these up: 10000-10999: +120 '5's (Total 4120) 11000-11999: +120 '5's (Total 4240) 12000-12999: +120 '5's (Total 4360) 13000-13999: +120 '5's (Total 4480) 14000-14999: +120 '5's (Total 4600) Finding the last 400 '5's (starting from 15000): We have written 4600 '5's up to 14999. We need 5000−4600=400 more '5's. Now we start writing numbers from 15000. All numbers from 15000 to 15999 have a '5' in the thousands place. This means each of these numbers contributes at least one '5'. The '5' in 15000 (thousands place) is the 4601st '5'. The '5' in 15001 (thousands place) is the 4602nd '5'. We need the 5000th '5'. This will be the (5000−4600)=400th '5' that appears in the 15xxx range. Let's count the occurrences of '5' in numbers like 15000, 15001, 15002, and so on, until we reach the 5000th overall '5'. Numbers 15000, 15001, 15002, 15003, 15004: Each has one '5' (in the thousands place). So far, 5 '5's from these numbers (4601st to 4605th). Number 15005: This number has two '5's (one in the thousands place and one in the units place). The '5' in the thousands place of 15005 would be the 4606th '5'. The '5' in the units place of 15005 would be the 4607th '5'. This continues until we find the 5000th '5'. Let's use a simpler way to count how many numbers past 15000 we need. We're looking for the 400th additional '5'. The '5' in the thousands place of 15000 counts as 1. The '5' in the thousands place of 15001 counts as 1. ... The '5' in the thousands place of 15334 counts as 1. (This is 334−0+1=335 occurrences from the thousands place). Let's quickly re-sum: Up to 15330, we had 4994 '5's. 15331: adds one '5' (thousands place) = 4995. 15332: adds one '5' (thousands place) = 4996. 15333: adds one '5' (thousands place) = 4997. 15334: adds one '5' (thousands place) = 4998. 15335: This number has two '5's. The '5' in the thousands place makes it the 4999th '5'. The '5' in the units place makes it the 5000th '5'. You would have written the digit ‘5’ for the 5000th time when you write the number 15335