求救…召喚數學師

part c 到底點計？ 有冇人諗到

2024 仲唔識用chatgpt

試下1+2024

1 ≡ 1
12 = 1*10+2 ≡ 1*1+2 = 3
123 = 12*10+3 ≡ 3+3 = 6
1234 = 123*10+4 ≡ 6+4 = 10

12345678910111213.....20232024 (mod9)
=1+2+3+4+5+6+7+8+9+1+0+1+1+....+2+0+2+4 (mod9)
=1+2+3+4+5+6+7+8+9+10+11+12+.....+2024 (mod9)
=(1+2024)*2024/2 (mod9)
=2049300 (mod9)
=2+0+4+9+3+0+0 (mod9)
=18 (mod9)
=0

1 2 3 4 5 6 7 8 9
除以9嘅remainder係
1 2 3 4 5 6 7 8 0

呢個pattern 會一直重覆

唔係summation …係成個數 123456789101112…去到2024

我都係咁諗

睇唔出同part c solution嘅關系

mod 9入面可以拆開佢做每個數獨立加埋 (part b)

你part b咪prove左，每個digit 都可以加埋mode9

Part a 咪引導緊你囉，諗下人地點解係同一題啦
10÷9, 100÷9, 1000÷9... 嘅remainder係一樣
所以
12345...20232024 (mod 9)
=1+2+3+4+5...+2023+2024 (mod 9)

可能就係睇唔明先解唔到

人地part a同part b都係引導緊佢點做part c
條傻仔咁都諗唔明

Ok I c 3q

part a都已經係好大提示, part b直頭係話比你知道可以咁做

小學奧數

明了明了 使唔使咁串呢大哥？

wtf 我睇唔明 完全無見過

你連小學生都不如

無讀呀奧數呀Ching

無讀過奧數呀Ching

你都係

To solve part (c), we need to find the remainder of the large number  when divided by 9. Instead of adding all the digits directly (which is impractical), we can use properties from part (b) of the question: the remainder of a number modulo 9 is equal to the remainder of the sum of its digits modulo 9.

Step-by-Step Solution:

1. Break Down the Problem:
The number  is formed by concatenating all integers from 1 to 2024. We need to calculate the sum of all the digits that appear in these numbers and find that sum modulo 9.
2. Analyze Digits from 1 to 9:
For the numbers 1 to 9, each digit appears once. The sum of the digits is:

\( 45 \equiv 0 \pmod{9} \).
3. Analyze Digits from 10 to 99:
• Tens Place: The digits 1 to 9 appear 10 times each in the tens place across all numbers (e.g., 10-19 has ‘1’ as the tens digit 10 times). The sum contributed by the tens place is:

\( 450 \equiv 0 \pmod{9} \).
• Units Place: Each digit from 0 to 9 appears 9 times in the units place for each group (e.g., 10-19, 20-29, …, 90-99). The sum for one group of 10 numbers is:

\( 405 \equiv 0 \pmod{9} \).
Therefore, the sum of the digits from 10 to 99 is also \( \equiv 0 \pmod{9} \).
4. Analyze Digits from 100 to 999:
This can be broken down similarly into hundreds, tens, and units places:
• Hundreds Place: Each digit from 1 to 9 appears 100 times. The sum is:

\( 4500 \equiv 0 \pmod{9} \).
• Tens and Units Places: The sum for each is calculated as for the tens and units places in the range 10-99, and they contribute \( \equiv 0 \pmod{9} \).
5. Analyze Digits from 1000 to 2024:
• Thousands Place: The digit ‘1’ appears 1000 times (for numbers 1000-1999), and the digit ‘2’ appears 25 times (for numbers 2000-2024).

\( 1050 \equiv 6 \pmod{9} \) (since ).
• Hundreds, Tens, and Units Places: These can be analyzed similarly to previous cases, contributing \( \equiv 0 \pmod{9} \).

Final Calculation:

The main non-zero contribution comes from the thousands place. Therefore, the sum of all digits from 1 to 2024 modulo 9 is:
\[
6 \pmod{9}
\]

Conclusion:

The remainder when the number  is divided by 9 is 6.