澳洲AMC最后一道求救，真做不出

在澳大利亚中学教育



类似题型还出现过不少次，这类题目怎么解法？
答案：765

评论

汗，看到楼主好几个类似的帖子
一直没有回复，因为真不想做圣母。不过看你这么挣扎，还是忍不住了。
这些题目详解在AMT官网的书籍里面都有的，孩子真喜欢数学，想要参与竞赛的话，我强烈建议你买几本书吧。也支持一下AMT是非盈利机构，靠卖书回点本。绝对物超所值，我们没有去补习社的，都是靠自己做题看详解（有时候好几个解法，都是智慧结晶）一步步走下来的，实在做不来的题目有时候自己看明白了比别人喂收获更大。

当然如果楼主只是想放到论坛上赚点分，大家讨论讨论，可以忽略我的发言~

评论

這東西沒有配方，熟悉基本結果，然後順著每條思路想一想，看看那條有用。
要自己做才能懂，光是聼或看解答很容易一聼就懂，一看就會，一做就錯。

這次給些提示吧。
a) 假設3個數位是 A,B,C
b) 跟 222 與其倍數有關
c) 跟 222 的倍數和 3231 差有關

想“系統”地學要花時間，把這裏高中的排列組合學透。
然後把基本數論内容學透例如 https://refkol.ro/matek/mathbook ... y-Number-Theory.pdf 1-9 章



评论
接着楼上白虎和路人的话说两句，纯个人观点。

AMC最后5题难度（及以上）题目的解答，是需要前置一些超纲知识与解题技巧的（这里不讨论天生圣体）。

这里的超纲知识与解题技巧，不是指具体的某道题目怎么做，而是一些经常会用到的东西，它们未必能直接帮你找到答案，但能让你在寻求答案时事半功倍。比如：
- 3，5，7，9，11这些数的倍数有哪些特点。
- 除以3，5，7都余/差1的数有什么特点。
- 点线依次连接，什么时候点比线多1，什么时候点线一样多。
等等

以这题为例,能不能解出这道题并不重要，重要的是：

1、学会如何用数学表示这六个数
100a+10b+c
100a+10c+b
100b+10a+c
100b+10c+a
100c+10a+b
100c+10b+a
（把这6个数加起来就能明白路人说的222是什么意思了）

2、知道WLOG，算5个数的和的时候可以去掉上面6个数的任意一个（比如去掉100a+10b+c），那么
--- 6个数的和 = 222a + 222b + 222c = 222(a+b+c)
--- 5个数的和 = 122a + 212b + 221c

如果娃卡在这里了，你可以和娃讨论一下，是从6个数的和继续还是从5个数的和继续，以及为什么这么选择。

如果娃连如何表示这6个数以及它们的和都不会，那意味着这个题目对娃来说太难了。一般来说，（不放弃的话）个人建议三种选择：
选择一：先找更简单/容易一些的题练习，或者像楼上建议的找适合的书籍自学。
选择二：父母有能力的话，每道题都尽可能将题目以及相关知识全部讲透。根据个人经验，初期一道题目半小时到一小时是家常便饭。
选择三：报个奥数/AMC/奖学金之类会教一些基本套路的补习班。（不要说套路化方法不好，那是熟悉套路以后才有资格说的话）

再次，以上仅为个人观点。

评论


再说几句：

------------------------------------------------

上贴说到，六个数字的和一定是222的倍数
222 × 15 = 3330
3330 - 3231＜100
六个数字的和一定比3330大
（让孩子想一想，为什么从 × 15开始）

222×16 = 3552
3552 - 3231 = 321
经检验，321不是答案
检验方法一：321+312+213+231+123+132<3552
检验方法二：321*6<3552
检验方法三：3+2+1<16
（让孩子想一想三个方法各是什么原理，哪个方法最优）

222×17 = 3774
3774 - 3231 = 543
经检验，543不是答案

222×18 = 3996
3996 - 3231 = 765
经检验，765是答案

（下面的两步计算非必要，但可作为验算的一部分）
222×19 = 4218
4218 - 3231 = 987
经检验，987不是答案

222×20 = 4440
4440 - 3231 > 987
六个数的和不可能大于等于4440

以上的部分，让孩子独立探索（尽可能少地给予提示）是相当必要的，这有助于培养良好的数感。

------------------------------------------------

PS：现在回头看下路人给出的提示：

a) 假設3個數位是 A,B,C（脑内部分）
六个数的表示略

b) 跟 222 與其倍數有關（脑内部分，A，B，C在三个位数各出现两次）
六个数的和=222A+222B+222C=222(A+B+C)

c) 跟 222 的倍數和 3231 差有關（脑内部分）
第六个数等于222*15=3330-3231=99+222=321（16×）+222=543（17×）+222=765（18√）

路人水平解这道题，草稿纸上至多只需要红色部分这么多的步骤。

评论

1.
设3个非0数为a, b, c， 其排列共6种，组成的三位数的和可表示为：
abc+acb+bac+bca+cba+cab=222(a+b+c),

为何是 222(a+b+c)？

步骤如下：
6个数字总和的百位是（2a+2b+2c)x100， 十位是（2a+2b+2c)x10， 个位就是（2a+2b+2c）
加起来就是（2a+2b+2c)x100 +（2a+2b+2c)x10 +（2a+2b+2c）=2（a+b+c)(100+10+1)=222 x（a+b+c）



2.
因其中的五个三位数的和为3194，这六个三位数的和的范围是：（第6个数最少是123，最大为987）
3231+123＜  222 x（a+b+c）＜  3231+987

该数的范围是（3354，4218）之间并且是222的倍数，且3354÷222＜a+b+c＜4218÷222
即15.1＜a+b+c＜19,

a+b+c只可能是16，17，18，
16x222=3552
17x222=3774
18x222=3996

3.
在这个区间内是222的倍数的只有3552，3774，3996
由于这是6个数的和，因此必须减去5个数和3231，得到第6个数字
就是3552-3231=321， 3774-3231=543，3996-3231=765

4.
由于a+b+c只可能是16，17，18（见上）
321 的3个数字总和3+2+1=6（排除掉）
543 的3个数字总和5+4+3=12 （排除掉）

5.
唯一符合答案的就只有765 的3个数字总和7+6+5=18 （排除掉）
所以第6个数只能是765

评论
你自己看吧，看懂了再去教小孩子

评论
a+b+c=0 mod 3,guess &check,7,6,5

评论
忍不住AI了一下。不同的AI平台居然不同答案。有知道哪个对的吗？

评论
GPT o1:

**Solution Explanation**

Suppose the three distinct nonzero digits are \(a\), \(b\), and \(c\).  The six different 3-digit numbers formed from these digits are:

\[
\begin{aligned}
&abc = 100a + 10b + c,\\
&acb = 100a + 10c + b,\\
&bac = 100b + 10a + c,\\
&bca = 100b + 10c + a,\\
&cab = 100c + 10a + b,\\
&cba = 100c + 10b + a.
\end{aligned}
\]

If we sum *all* six of these numbers, note how many times each digit appears in the hundreds, tens, and units places:

- Each of \(a, b, c\) appears exactly twice in the hundreds place (contributing \(2 \times 100 = 200\) each).
- Each of \(a, b, c\) appears exactly twice in the tens place (contributing \(2 \times 10 = 20\) each).
- Each of \(a, b, c\) appears exactly twice in the ones place (contributing \(2 \times 1 = 2\) each).

Hence, the total contribution of each digit \(a, b, c\) is
\[
200 + 20 + 2 = 222,
\]
and the sum of *all six* permutations is
\[
\text{Sum of all 6} \;=\; 222(a + b + c).
\]

We are told that the *sum of five* of these numbers is 3231.  Let \(M\) be the *missing* (i.e., sixth) number.  Then

\[
M \;=\; \Bigl[\text{Sum of all 6}\Bigr] \;-\; 3231 \;=\; 222(a + b + c) \;-\; 3231.
\]

Since \(M\) itself must be one of the 3-digit permutations \(\{abc, acb, \dots\}\), it must be between 100 and 999, and its digits must add up to \(a + b + c\).  

Through checking feasible values of \(a+b+c\) (and ensuring the difference is indeed a 3-digit permutation of those digits), one finds:

- When \(a + b + c = 18\),
  \[
  222 \times 18 = 3996,
  \quad
  M = 3996 - 3231 = 765.
  \]
  The digits of 765 are 7, 6, and 5, which indeed sum to 18.  

Therefore, the three digits are \(5, 6, 7\), and the missing number is

\[
\boxed{765}.
\]

To confirm, the six permutations of \(5,6,7\) are:
\[
567,\; 576,\; 657,\; 675,\; 756,\; 765,
\]
whose total sum is \(3996\).  Omitting \(765\) leaves \(3996 - 765 = 3231\), exactly as stated in the problem. Hence \(\,765\) is the required sixth number.

评论
three different non-zero-digits are used to form six different 3-digit numbers. the sum of five of them is 3231. what is the sixth number?
Understanding the Problem
We are given:

Three different non-zero digits: Let's denote them as
A
A,
B
B, and
C
C.

Six different 3-digit numbers: These are formed by arranging
A
A,
B
B, and
C
C in all possible orders. Since there are 3 digits, the number of permutations is
3
!
=
6
3!=6.

Sum of five of these numbers is 3231: We need to find the sixth number.

Our goal is to determine the value of the sixth 3-digit number.

Step 1: Listing All Possible 3-Digit Numbers
First, let's list all six possible 3-digit numbers formed by
A
A,
B
B, and
C
C:

A
B
C
ABC (which represents
100
A
+
10
B
+
C
100A+10B+C)

A
C
B
ACB (
100
A
+
10
C
+
B
100A+10C+B)

B
A
C
BAC (
100
B
+
10
A
+
C
100B+10A+C)

B
C
A
BCA (
100
B
+
10
C
+
A
100B+10C+A)

C
A
B
CAB (
100
C
+
10
A
+
B
100C+10A+B)

C
B
A
CBA (
100
C
+
10
B
+
A
100C+10B+A)

Step 2: Calculating the Sum of All Six Numbers
Let's add all six numbers together:

A
B
C
+
A
C
B
+
B
A
C
+
B
C
A
+
C
A
B
+
C
B
A
=
(
100
A
+
10
B
+
C
)
+
(
100
A
+
10
C
+
B
)
+
(
100
B
+
10
A
+
C
)
+
(
100
B
+
10
C
+
A
)
+
(
100
C
+
10
A
+
B
)
+
(
100
C
+
10
B
+
A
)
=
(
100
A
+
100
A
+
10
A
+
A
+
10
A
+
A
)
+
(
10
B
+
B
+
100
B
+
100
B
+
B
+
10
B
)
+
(
C
+
10
C
+
C
+
10
C
+
100
C
+
100
C
)
=
(
222
A
)
+
(
222
B
)
+
(
222
C
)
=
222
(
A
+
B
+
C
)
ABC+ACB+BAC+BCA+CAB+CBA
​
  
=(100A+10B+C)+(100A+10C+B)
+(100B+10A+C)+(100B+10C+A)
+(100C+10A+B)+(100C+10B+A)
=(100A+100A+10A+A+10A+A)
+(10B+B+100B+100B+B+10B)
+(C+10C+C+10C+100C+100C)
=(222A)+(222B)+(222C)
=222(A+B+C)
​

So, the sum of all six numbers is
222
(
A
+
B
+
C
)
222(A+B+C).

Step 3: Relating the Given Sum to Find the Sixth Number
We are told that the sum of five of these numbers is 3231. Let's denote the sixth number as
N
N. Therefore:

Sum of five numbers
=
3231
Sum of all six numbers
=
3231
+
N
Sum of five numbers=3231
Sum of all six numbers=3231+N
From Step 2, we know:

3231
+
N
=
222
(
A
+
B
+
C
)
3231+N=222(A+B+C)
Our goal is to find
N
N. To do this, we need to determine
A
+
B
+
C
A+B+C.

Step 4: Solving for
A
+
B
+
C
A+B+C
Let's rearrange the equation:

N
=
222
(
A
+
B
+
C
)
−
3231
N=222(A+B+C)−3231
Since
A
A,
B
B, and
C
C are digits (1 through 9),
A
+
B
+
C
A+B+C must be an integer between 3 and 27.

Let's find possible values of
A
+
B
+
C
A+B+C such that
N
N is a positive 3-digit number.

Let's denote
S
=
A
+
B
+
C
S=A+B+C. Then:

N
=
222
S
−
3231
N=222S−3231
We need
N
N to be a 3-digit number, so:

100
≤
N
≤
999
100
≤
222
S
−
3231
≤
999
100≤N≤999
100≤222S−3231≤999
Let's solve the inequalities:

222
S
−
3231
≥
100
222S−3231≥100

222
S
≥
3331
S
≥
3331
222
S
≥
15.0
222S≥3331
S≥
222
3331
​

S≥15.0
222
S
−
3231
≤
999
222S−3231≤999

222
S
≤
4230
S
≤
4230
222
S
≤
19.05
222S≤4230
S≤
222
4230
​

S≤19.05
Since
S
S must be an integer,
S
S can be 15, 16, 17, 18, or 19.

Step 5: Testing Possible Values of
S
S
Let's test each possible value of
S
S to find a valid
N
N.

Case 1:
S
=
15
S=15
N
=
222
×
15
−
3231
=
3330
−
3231
=
99
N=222×15−3231=3330−3231=99
But
N
=
99
N=99 is not a 3-digit number. So,
S
≠
15
S

=15.

Case 2:
S
=
16
S=16
N
=
222
×
16
−
3231
=
3552
−
3231
=
321
N=222×16−3231=3552−3231=321
N
=
321
N=321 is a valid 3-digit number.

Let's check if
S
=
16
S=16 is possible with three distinct digits from 1 to 9.

Possible combinations for
A
+
B
+
C
=
16
A+B+C=16:

7, 8, 1 → Sum = 16

6, 9, 1 → Sum = 16

5, 9, 2 → Sum = 16

4, 9, 3 → Sum = 16

5, 8, 3 → Sum = 16

6, 7, 3 → Sum = 16

4, 8, 4 → Not valid (digits must be distinct)

etc.

There are multiple valid combinations, so
S
=
16
S=16 is feasible.

Case 3:
S
=
17
S=17
N
=
222
×
17
−
3231
=
3774
−
3231
=
543
N=222×17−3231=3774−3231=543
N
=
543
N=543 is a valid 3-digit number.

Check for
S
=
17
S=17:

Possible combinations:

8, 7, 2 → Sum = 17

9, 6, 2 → Sum = 17

7, 6, 4 → Sum = 17

etc.

Again, multiple valid combinations exist.

Case 4:
S
=
18
S=18
N
=
222
×
18
−
3231
=
3996
−
3231
=
765
N=222×18−3231=3996−3231=765
N
=
765
N=765 is a valid 3-digit number.

Check for
S
=
18
S=18:

Possible combinations:

9, 8, 1 → Sum = 18

9, 7, 2 → Sum = 18

8, 7, 3 → Sum = 18

etc.

Valid combinations exist.

Case 5:
S
=
19
S=19
N
=
222
×
19
−
3231
=
4218
−
3231
=
987
N=222×19−3231=4218−3231=987
N
=
987
N=987 is a valid 3-digit number.

Check for
S
=
19
S=19:

Possible combinations:

9, 8, 2 → Sum = 19

9, 7, 3 → Sum = 19

8, 7, 4 → Sum = 19

etc.

Valid combinations exist.

Step 6: Determining the Correct
S
S and
N
N
We have multiple possible values for
S
S and corresponding
N
N. To find the correct one, we need additional constraints or information. However, since the problem states that the sum of five numbers is 3231, and we've derived that
N
=
222
S
−
3231
N=222S−3231, and
N
N must be one of the six numbers formed by
A
A,
B
B, and
C
C, we can deduce that
N
N must be consistent with the digits
A
A,
B
B, and
C
C.

Given that all six numbers are formed by the same three digits, and their sum is
222
S
222S, and five of them sum to 3231, the sixth number
N
N must be such that when added to 3231, it equals
222
S
222S.

Since
N
N is one of the six numbers, and all six numbers are permutations of
A
A,
B
B, and
C
C,
N
N must be a permutation of these digits.

Given that, let's consider the smallest possible
N
N that fits our earlier cases.

From our cases:

S
=
16
S=16,
N
=
321
N=321

S
=
17
S=17,
N
=
543
N=543

S
=
18
S=18,
N
=
765
N=765

S
=
19
S=19,
N
=
987
N=987

We need to check which of these
N
N values can be formed by the digits
A
A,
B
B, and
C
C that sum to
S
S.

Let's take
S
=
16
S=16,
N
=
321
N=321:

Digits: 3, 2, 1

Sum:
3
+
2
+
1
=
6
≠
16
3+2+1=6

=16

This doesn't match
S
=
16
S=16.

Next,
S
=
17
S=17,
N
=
543
N=543:

Digits: 5, 4, 3

Sum:
5
+
4
+
3
=
12
≠
17
5+4+3=12

=17

Doesn't match.

Next,
S
=
18
S=18,
N
=
765
N=765:

Digits: 7, 6, 5

Sum:
7
+
6
+
5
=
18
7+6+5=18

This matches
S
=
18
S=18.

Finally,
S
=
19
S=19,
N
=
987
N=987:

Digits: 9, 8, 7

Sum:
9
+
8
+
7
=
24
≠
19
9+8+7=24

=19

Doesn't match.

Therefore, the only valid case is
S
=
18
S=18,
N
=
765
N=765.

Step 7: Verifying the Solution
Let's verify that with
A
=
7
A=7,
B
=
6
B=6,
C
=
5
C=5, and
S
=
18
S=18, the sum of all six numbers is
222
×
18
=
3996
222×18=3996, and the sum of five numbers is 3231, so the sixth number should be
3996
−
3231
=
765
3996−3231=765.

Listing all six numbers:

765
765

756
756

675
675

657
657

567
567

576
576

Sum of all six:

765
+
756
+
675
+
657
+
567
+
576
=
3996
765+756+675+657+567+576=3996
Sum of five numbers (excluding 765):

756
+
675
+
657
+
567
+
576
=
3231
756+675+657+567+576=3231
This matches the given condition.

Final Answer
The sixth number is 765.



评论
心算结果是567
不拿纸笔不行啊

评论
sum of total six numbers:
   100*(2a+2b+2c)+10(2a+2b+2c)+(2a+2b+2c)=222*(a+b+c)

Maxium total of six numbers :  3231+987=4218  
           -> max sum of 3 digits : 4218/222  =19
Minimum total of six numbers:  3231+123=3354
           -> min sum of 3 digits :  3354/222  ~=15.11->16


222*16=3552 -3231  = 321     6*222 X
222*17=3774  -3231 = 543     12*222 x
222*18=3996  -3231 = 765
222*19=4218  -3231 = 987     24*222 x
澳洲中文论坛热点