求救…召喚數學師

On9 GPT垃圾嚟 咁都計錯

5. 後半開始錯

由1加到2024既總和
好似係小學生奧數題黎

「好簡單」— 高斯 (1785)

我唔識a，但睇到b，唔係一定計到c咩

我諗a唔要求好完美嘅proof

始終畀小學生做

c part一開始用b part拆開做1+2+3+...就好直觀
係之後要睇得出+9+1+0+1+1+1+2+....可以合返做...+9+10+11+12..
唔好on99走去諗點數幾多個1幾多個2

過到呢步之後都係簡單野

On9左

以為大學數

唔係答問題
純粹想證明點解mod9可以拆
a_{n}…a_{1} 可以寫成an *10^n +…+ a1*1
而10^n又可以寫成（9+1）^n
呢個binomial嘅每個coef 都有9，除咗尾數1，所以一除9的話
mod 9係乜就視乎a_n係乜
而mod一樣就可以相加（另外證，leave it as exercise）
Q.E.D.

然後只需重復運用，#6已答

part a 咪證左

你個proof複雜過part a

也可以，modulo of multiplication + (part a)

呢題part b 我gen到7， 好似正路
兄弟得閒既幫手睇下

# Question 2(b)
# Find the remainder of (653 + 853) mod 49
# Hint: Use binomial theorem
# 653 = (7 - 1)^53
# 853 = (7 + 1)^53

# Step 1: Expand using binomial theorem
# (7 - 1)^53 = Σ [C(53, k) * 7^k * (-1)^(53-k)]
# (7 + 1)^53 = Σ [C(53, k) * 7^k * (1)^(53-k)]
# Add the two expansions:
# (7 - 1)^53 + (7 + 1)^53 = Σ [C(53, k) * 7^k * ((-1)^(53-k) + 1^(53-k))]

# Step 2: Analyze the parity of 53 - k
# (-1)^(53-k) + 1^(53-k):
# - When 53 - k is odd, (-1) + 1 = 0 => Terms vanish.
# - When 53 - k is even, 1 + 1 = 2 => Terms contribute.
# Only terms where k is odd contribute.

# Step 3: Reduce 7^k mod 49
# For k = 1, 7^1 = 7 (mod 49).
# For k >= 2, 7^k ≡ 0 (mod 49).

# Step 4: Evaluate the k = 1 term
# Only the k = 1 term contributes:
# C(53, 1) * 7^1 * 2
# = 53 * 7 * 2 = 742.

# Step 5: Simplify modulo 49
# 742 mod 49 = 742 - 49 * floor(742 / 49)
# = 742 - 49 * 15
# = 742 - 735
# = 7.

# Final Answer:
# The remainder of (653 + 853) mod 49 is: 7

係7

Steps correct?

啱 但睇落好ai

Nvm 啱就得 3q

兄弟 check 埋呢題先好訓

To solve this problem, we need to determine the remainder of f_m when divided by 4 and use that to show it cannot be expressed as the sum of two positive square numbers.

### Step 1: Determine f_m modulo 4

The number f_m consists of m ones. For example, if m = 2, f_m is 11; if m = 3, f_m is 111, and so on. This number can be written as:

f_m = (10^m - 1) / 9

To find the remainder of f_m when divided by 4, we first need to calculate 10^m modulo 4. We know that:

- 10 is congruent to 2 modulo 4
- For any positive integer m, 10^m is congruent to 2^m, and since 2^m (for m ≥ 2) is divisible by 4, we get 10^m congruent to 0 modulo 4.

Therefore, 10^m - 1 is congruent to 3 modulo 4.

Next, we calculate f_m as:

f_m = (10^m - 1) / 9

Since 10^m - 1 is congruent to 3 modulo 4, we divide this by 9. Since 9 is congruent to 1 modulo 4, dividing by 9 does not change the remainder, and we still get 3.

So, f_m is congruent to 3 modulo 4.

### Step 2: Sum of Two Squares

To determine whether a number can be written as the sum of two squares, we use the following rule:

- The sum of two squares modulo 4 can only result in 0, 1, or 2. Specifically:
- The square of 0 is 0 modulo 4.
- The square of 1 is 1 modulo 4.
- The square of 2 is 0 modulo 4.
- The square of 3 is 1 modulo 4.

Thus, if a number can be written as the sum of two squares, its remainder when divided by 4 must be 0, 1, or 2, but **never 3**.

Since f_m is congruent to 3 modulo 4, it cannot be written as the sum of two squares. No combination of squares will give a result of 3 modulo 4.

### Conclusion

We have shown that f_m, the number formed by m consecutive 1's, is congruent to 3 modulo 4. Since numbers congruent to 3 modulo 4 cannot be written as the sum of two squares, it follows that f_m cannot be written as the sum of two positive square numbers.

睇你覺得使唔使prove
- The sum of two squares modulo 4 can only result in 0, 1, or 2. Specifically:

overall都啱

我中間加左
the square of all integers modulo 4 :
0^2 = 0, 0 mod 4 = 0
1^2 = 1, 1 mod 4 = 1
2^2 = 4, 4 mod 4 = 0
3^2 = 9, 9 mod 4 = 1
So, a square number can just be congruent to either 0 or 1 modulo 4. ,
Then consider, the sum of two square numbers modulo 4,
Can only be :
0 + 0 = 0 mod 4,
0 + 1 = 1 mod 4,
1 + 0 = 1 mod 4,
1 + 1 = 2 mod 4,

So only {0,1,2} are possible remainders when adding up two square numbers.

樓主 讀數唔可以齋gen answer 你要真係明先得 除非你唔係for學習 咁我冇嘢講
題#46可以咁做
4 | 6^2 | 6^n n>=2
1*（4+6）^n = binomial, coef係除咗尾數之外都係4倍數，最後mod4係乜視乎尾數6^n係乜
又因為6^2 及以上可以被4除 所以只係睇n做0同1嘅case，modulo addition下為3

然後睇sum of perfect number
As 雙數^2+奇數^2，雙數^2+雙數^2，奇數^2+奇數^2 mod4 下無一個係3 mod 4，由此得知

嚴格黎講要用mi