Mathematical Excalibur, Volume 9, Number 5

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Volume 9, Number 5 January 2005 – February 2005

例析數學競賽中的計數問題 (一)

費振鵬（江蘇省鹽城市城區永豐中學 224054 )

Olympiad Corner

The 7th_{China Hong Kong Math}

Olympiad took place on December 4, 2004. Here are the problems. Problem 1. For n ≥ 2, let a1, a2, …,

be positive and a 1 , n n a a ₊ 2 – a1 = 3 2 n 1 n 0. a −a =L=a + −a ≥ Prove that 2 2 2 3 1 1 1 n a + a +L + a2 1 2 1 2 1 1 . 2 n n n n a a a a n a a a a + + + − ≤ ⋅ 1

Determine when equality holds.

Problem 2. In a school there are b

teachers and c students. Suppose that (i) each teacher teaches exactly k students; and

(ii) for each pair of distinct students, exactly h teachers teach both of them.

Show that _. ) 1 ( ) 1 ( − − = k k c c h b

Problem 3. On the sides AB and AC of

triangle ABC, there are points P and Q respectively such that ∠APC =∠AQB = 45°. Let the perpendicular line to side

AB through P intersects line BQ at S.

Let the perpendicular line to side AC through Q intersects line CP at R. Let D be on side BC such that AD ⊥ BC .

(continued on page 4)

Editors: ஻ Ի ஶ(CHEUNG Pak-Hong), Munsang College, HK

ଽ υ ࣻ (KO Tsz-Mei)

గ ႀ ᄸ (LEUNG Tat-Wing)

؃ ୊ ፱ (LI Kin-Yin), Dept. of Math., HKUST

֔ ᜢ ݰ (NG Keng-Po Roger), ITC, HKPU

Artist: ྆ ؾ ़ (YEUNG Sau-Ying Camille), MFA, CU

Acknowledgment: Thanks to Elina Chiu, Math. Dept.,

HKUST for general assistance.

On-line:

http://www.math.ust.hk/mathematical_excalibur/ The editors welcome contributions from all teachers and students. With your submission, please include your name, address, school, email, telephone and fax numbers (if available). Electronic submissions, especially in MS Word, are encouraged. The deadline for receiving material for the next issue is March 31, 2005.

For individual subscription for the next five issues for the 03-04 academic year, send us five stamped self-addressed envelopes. Send all correspondence to:

Dr. Kin-Yin LI Department of Mathematics

The Hong Kong University of Science and Technology Clear Water Bay, Kowloon, Hong Kong

Fax: (852) 2358 1643 Email: [email protected] 組合數學中的計數問題，數學競賽題中的熟面孔，看似司空見慣，不足為奇．很多同學認為只要憑藉單純的課內知識就可左右逢源，迎刃而解．其實具體解題時，卻會使你挖空心思，也無所適從．對於這類問題往往首先要通過構造法描繪出對象的簡單數學模型，繼而借助在計數問題中常用的一些數學原理方可得出所求對象的總數或其範圍． 1 運用分類計數原理與分步計數原理分類計數原理與分步計數原理（即加法原理與乘法原理）是關於計數的兩個基本原理，是解決競賽中計數問題的基礎．下面提出的三個問題，注意結合排列與組合的相關知識，構造出相應的模型再去分析求解．例 1 已知兩個實數集合

{

1, , ,2 100

}

A= a a La 與B=

{

b b₁, , ,₂Lb₅₀

}

，若從A到B的映射f使得B中每個元素 都有原象，且 f (a1) ≤ f (a2) ≤ …≤ f(a100)，則這樣的映射共有（）個．（A）C50（B）C （C）C （D）C 100 48 99 49 100 49 99 解答設按從小到大排列為（因集合元素具有互異性，故其中不含相等情形）． 1, , ,2 50 b b _Lb 1 2 5 c <c <L<c0 1 將A 中元素 分成 50 組，每組依次與B中元素 對應．這裏，我們用，表示 1, , ,2 100 a a _L a 1, , ,2 50 c c _Lc 1 2 3 1 4 5 2 a a a c a a c L 1 2 3 ( ) ( ) ( ) f a =f a =f a =c ， f a( )₄ = f (a5) = c2，… 考慮f a( )₁ ≤ f a( )₂ ≤_L≤ f a( ₁₀₀)，容易得到f a( ₁₀₀)=c₅₀，這就是說只能寫在的右邊，故只需在之間的 99 個空位 “□ ＂中選擇 49 個位置並 50 c 100 a 1 2 3 98 99 100 50 a a□ □ □ □a _L a □a □a c 按從左到右的順序依次填上 9．從而構成滿足題設要求的映射共有個．因此選D． 1, , ,2 4 c c L c 49 99 C 例2 有人要上樓，此人每步能向上走 1 階或 2 階，如果一層樓有 18 階，他上一層樓有多少種不同的走法？解答1 此人上樓最多走 18 步，最少走9 步．這裏用分別表示此人上樓走18 步，17 步，16 步，…， 9 步時走法（對於任意前後兩者的步數，因後者少走 2 步 1 階，而多走 1 步2 階，計後者少走 1 步）的計數結果．考慮步子中的每步2 階情形，易得，，，…，． 18, 17, 16, , 9 a a a La 0 18 C18 a = 1 17 C17 a = 2 16 C16 a = 9 9 C a = 9 綜上，他上一層樓共有 0 1 2 9 18 17 16 9 C +C +C +L+C = +1 17 120+ +L+1 = 4181 種不同的走法．解答2 設表示上n階的走法的計數 結果． n F 顯然，，F （2 步 1 階； 1 步 2 階）．對於起步只有兩種不同走法：上1 階或上 2 階． 1 1 F = ₂=2 3, , ,4 F F L 因此對於F₃，第1 步上 1 階的情形，還剩階，計種不同的走法；對於第1 步上 2 階的情形，還剩 3 1 2− = F2 3 2 1− = 階，計F₁種不同的走法．總計 3 2 1 2 1 3 F =F +F = + = ．同理， F₄=F₃+F₂= + =3 2 5 ， 5 4 3 5 3 8 F =F +F = + = ， … ， 18 17 16 2584 1597 4181 F =F +F = + = ．例3 在世界盃足球賽前，F國教練為 了考察A A₁, , ,₂…A₇這七名隊員，準備讓他們在三場訓練比賽（每場90 分鐘）都上場．假設在比賽的任何時刻，這

些隊員中有且僅有一人在場

上

並且

每人上場的

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Mathematical Excalibur, Vol. 9, No. 5, Jan. 05- Feb. 05 Page 2 些隊員中有且僅有一人在場上，並且 1, , ,2 3 4 A A A A每人上場的總時間（以分鐘為單位）均被7 整除，A A A5, ,6 7每人上場的總時間（以分鐘為單位）均被 13 整除．如果每場換人次數不限，那麼按每名隊員上場的總時間計算，共有多少種不同的情況．解答設A i_i

(

=1,2, ,7L

)

上場的總時間分別為a i_i

(

=1,2, ,7L

)

分鐘．根據題意，可設

1 2

)

(

2 3

)

(

3 1

)

1,

105 =⎡⎢_⎣₁₀₅240⎤⎥_⎦=2. 從而此時有 3 5 ( ) [ ( ) ( ) (

card U −card A +card A +card A

15 21 35

2[card A( ) card A( ) card A( )]

7)] + + + 105 4card A( ) 136 − = 名同學面對教練．如果我們借助威恩圖進行分析，利用上面所得數據分別填入圖 1，注意按從內到外的順序填． 55 ③ 14 9 2 4 28 19 ⑦ ⑤ 109 圖1 如圖1，此時面對教練的同學一目了然，應有109+14+4+9=136 名． (continued on page 4)

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Problem Corner

We welcome readers to submit their solutions to the problems posed below for publication consideration. The solutions should be preceded by the solver’s name, home (or email) address and school affiliation. Please send submissions to Dr. Kin Y. Li,

Department of Mathematics, The Hong Kong University of Science & Technology, Clear Water Bay, Kowloon, Hong Kong. The deadline for

submitting solutions is March 31,

2005.

Problem 216. (Due to Alfred Eckstein, Arad, Romania) Solve the equation

6 2

4x −6x +2 2 0= .

Problem 217. Prove that there exist

infinitely many positive integers which cannot be represented in the form

3 5 7 9 11

1 2 3 4 5,

x +x +x +x +x

where x1, x2, x3, x4, x5 are positive

integers. (Source: 2002 Belarussian

Mathematical Olympiad, Final Round) Problem 218. Let O and P be distinct

points on a plane. Let ABCD be a parallelogram on the same plane such that its diagonals intersect at O. Suppose P is not on the reflection of line AB with respect to line CD. Let M and N be the midpoints of segments AP and BP respectively. Let Q be the intersection of lines MC and ND. Prove that P, Q, O are collinear and the point Q does not depend on the choice of parallelogram ABCD. (Source: 2004

National Math Olympiad in Slovenia, First Round)

Problem 219. (Due to Dorin Mărghidanu, Coleg. Nat. “A.I. Cuza”, Corabia, Romania) The sequences a0,a1,a2,… and b0,b1,b2,… are defined

as follows: a0,b0 > 0 and 1 1 , 2 n n n a a b + = + 1 1 2 n n n b b a + = + for n = 1,2,3,…. Prove that 2004 2004 max{a ,b }> 2005.

Problem 220. (Due to Cheng HAO, The Second High School Attached to Beijing Normal University) For i =

1,2,…, n, and k ≥ 4, let Ai = (ai1, ai2, … ,

aik) with aij = 0 or 1 and every Ai has at least 3 of the k coordinates equal 1.

Define the distance between Ai and Aj to be 1 | | k im jm m a a = −

∑

.

If the distance between any Ai and Aj (i ≠ j) is greater than 2, then prove that

n ≤ 2k–3 _{– 1.} *****************

Solutions

****************

Problem 211. For every a, b, c, d in [1,2], prove that 4 . a b c d a c b c d a b d + ₊ + _≤ + + + +

(Source: 32nd_{Ukranian Math Olympiad)}

Solution. CHEUNG Yun Kuen

(HKUST, Math Major, Year 1), Achilleas

P. PORFYRIADIS (American College of Thessaloniki “Anatolia”, Thessaloniki, Greece) and HUDREA Mihail (High School “Tiberiu Popoviciu” Cluj-Napoca

omania). R

Since 0 < b + d ≤ 4, it suffices to show

.

a b c d _a _c

b c d a

+ ₊ + _{≤ +}

+ +

Without loss of generality, we may assume 1 ≤ a ≤ c, say c = a + e with e ≥ 0. Then 1 1 a b c d e b c d a d a + ₊ + _{≤ +}⎛ ₊ ⎞ ⎜ ⎟ + + ⎝ + ⎠ ≤ 2 + e ≤ 2a+ e = a + c. In passing, we observe that equality holds if and only if e = 0, a = c = 1, b = d = 2.

Other commended solvers: CHENG Hei

(Tsuen Wan Government Secondary School, Form 5), LAW Yau Pui (Carmel Divine Grace Foundation Secondary School, Form 6) and YIM Wing Yin (South Tuen Mun Government Secondary School, Form 5).

Problem 212. Find the largest positive integer N such that if S is any set of 21 points on a circle C, then there exist N arcs of C whose endpoints lie in S and each of the arcs has measure not exceeding 120°.

Solution.

We will N = 100. To see that N ≤ 100, consider a diameter AB of C. Place 11 points close to A and 10 points close to B. The number of desired arcs is then

1 1 1 0 1 0 0 . 2 2 ⎛ ⎞₊⎛ ⎞ ₌ ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠

To see that N ≥ 100, we need to observe that for every set T of k = 21 points on

C, there exists a point X in T such that

there are at least [(k – 1)/ 2] arcs XY (with Y in T, Y ≠ X) each having measure not exceeding 120°. This is because we can divide the circle C into three arcs C1, C2, C3 of 120° (only

overlapping at endpoints) such that the common endpoint of C1 and C2 is a

point X of T. If X does not have the required property, then there are 1 + [(k – 1)/2] points of T lies on C3 and

any of them can serve as X.

Next we remove X and apply the same argument to k = 20, then remove that point, and repeat with k = 19, 18, … , 3. We get a total of 10 + 9 + 9 + 8 + 8 + … + 1 + 1 = 100 arcs.

Problem 213. Prove that the set of all positive integers can be partitioned into 100 nonempty subsets such that if three positive integers a, b, c satisfy a + 99b = c, then at least two of them belong to the same subset.

Solution. Achilleas P. PORFYRIADIS

(American College of Thessaloniki Anatolia”, Thessaloniki, Greece).

“

Let f(n) be the largest nonnegative integer k for which n is divisible by 2k_. Then given three positive integers a, b, c satisfying a + 99 b = c at least two of f(a),

f(b), f(c) are equal. To prove this, if f(a)

= f(b), then we are done. If f(a) < f(b), then f(c) = f(a). If f(a) > f(b), then f(c) =

f(b).

Therefore, the following partition suffices:

Si = {n | f (n) ≡ i (mod 100)} for 1≤ i ≤100.

Problem 214. Let the inscribed circle of triangle ABC be tangent to sides AB,

BC at E and F respectively. Let the

angle bisector of ∠ CAB intersect segment EF at K. Prove that ∠CKA is a right angle.

Solution. CHENG Hei (Tsuen Wan

Government Secondary School, Form 5), HUDREA Mihail (High School “Tiberiu Popoviciu” Cluj-Napoca Romania), Achilleas P. PORFYRIADIS (American College of Thessaloniki “Anatolia”, Thessaloniki, Greece), YIM

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Mathematical Excalibur, Vol. 9, No. 5, Jan. 05- Feb. 05 Page 4 Government Secondary School, Form

5) and YUNG Ka Chun (Carmel Divine Grace Foundation Secondary School, Form 6).

Most of the solvers pointed out that the problem is still true if the angle bisector of ∠CAB intersect line EF at K outside

the segment EF. So we have two

figures. I A B _C E F K F E B A I K C

Let I be the center of the inscribed circle. Then A, I, K are collinear. Now ∠CIK = ½(∠BAC+∠ACB). Next, BE = BF implies that ∠BFE = 90° – ½ ∠CBA = ½ (∠BAC+∠ACB) = ∠CIK. (In the second figure, we have ∠CFK = ∠BFE = ∠CIK.) Hence C, I, K, F are concyclic. Therefore, ∠CKI = ∠CFI = 90°.

Other commended solvers: CHEUNG Yun Kuen (HKUST, Math Major, Year 1).

Problem 215. Given a 8×8 board. Determine all squares such that if each one is removed, then the remaining 63 squares can be covered by 21 3×1 rectangles.

Solution. CHEUNG Yun Kuen

(HKUST, Math Major, Year 1).

Let us number the squares of the board from 1 to 64, with 1 to 8 on the first row, 9 to 16 on the second row and so on. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64

Using this numbering, a 3×1 rectangle will cover three numbers with a sum divisible by 3. Since 64 ≡ 1 (mod 3), only squares with numbers congruent to 1 (mod 3) need to be considered for our problem. If there is a desired square for the problem, then considering the left-right symmetry of the board and the up-down symmetry of the board, the images of a desired square under these symmetries are also desired squares. Hence they must also have numbers congruent to 1 (mod 3) in them. However, the only such square and its image squares having this property are the squares with numbers 19, 22, 43 and 46. Finally square 19 has the required property (and hence also squares 22, 43, 46 by symmetry) by putting 3×1 rectangles as shown in the following figure (those squares having the same letter are covered by the same 3×1 rectangle). A A A B B B F G C C C D D D F G H I E E E F G H I J J J K K K H I L L L M M M N O P Q R S T U N O P Q R S T U N O P Q R S T U

Other commended solvers: HUDREA Mihail (High School “Tiberiu Popoviciu” Cluj-Napoca Romania), NG Siu Hong (Carmel Divine Grace Foundation Secondary School, Form 6) and Achilleas

P. PORFYRIADIS (American College of Thessaloniki “Anatolia”, Thessaloniki, Greece).

Olympiad Corner

(continued from page 1)

Problem 3. (cont.) Prove that the lines PS, AD, QR meet at a common point

and lines SR and BC are parallel.

Problem 4.

Let S = {1, 2, …, 100}. Determine the number of functions

:

f S → S satisfying the following

conditions. (i) f(1)=1;

(ii) f is bijective (i.e. for every y in S, the equation f(x) = y has exactly one solution);

(iii) f(n) = f(g(n)) f(h(n)) for every n in

S.

Here g(n) and h(n) denote the uniquely determined positive integers such that

g(n) ≤ h(n), g(n) h(n) = n and h(n) − g(n)

is as small as possible. (For instance,

g(80) = 8, h(80) = 10 and g(81) = h(81)

= 9.)

例析數學競賽中的計數問題 (一)

(continued from page 2)

⑵用上面類似的方法可算得自左至右第1 號至第 105 號同學中面對教練的有60 名．考慮所報的數不是 3，5，7 的倍數的同學沒有轉動，他們面對教練；所報的數是3，5，7 中的兩個數的倍數的同學經過兩次轉動，他們仍面對教練；其餘同學轉動了一次或三次，都背對教練．作如下分析：106，107，108（3 的倍數），109，110（5 的倍數），111（3 的倍數），112（7 的倍數），113，114 （3 的倍數），115（5 的倍數），116， 117（3 的倍數），118，119（7 的倍數），120（3、5 的倍數），……，可知面對教練的第66 位同學所報的數是118. (to be continued)

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1: Let d = a, -a,_, > 0. Note that u: > u: - d* = (a, + d)(u, -d) = at_,ak+, for k > 1. So

1 1

7<-=

‘k %lak+l q+l

Then

= L.

sw,+, +

44an+I - 4a2an+, -44a,

2d _al=za,u,+, 1 ada,+,-d+ada, -4) =-. 2d _=,4a,a,, G J_. (n - WGV-J, + 4d 2d QP*%~*+, n - 1 =-. w, + a,a,, 2 _44a,a,,

Hence equality holds if and only if d = 0.

(Alternative Solution by Wong Chiu Wai) We proceed by induction. Again let d be the common ditference. For n k 2, we have

which is clearly valid, and the equality holds if and only if d = 0. Consider now

+-.&+_

_u2:n+,)-5k&+

_d

₁

1 I L II II+, a2un+lan+2

=tj(-&+&-(%+,

-4X1+

1 II %%%+I a,a.L.+, ))

=- _;t, _udu _+L+ 111 ₎

2 a+1 a+* a,=,+, a2a,+,a,+,

‘( a1 =- +L) 2 a2u,+lu,+2 a,=,+, =‘(L+ 4 2 %+I%+* an%+1 =-

G+,

2a,+, a.+2 n n+l

+‘(‘+&z_)

1

_{+L(d_d)}

=C, %+,

a,+,a, a,+,a,+, 1 d* 2 1 =-+ U :+I 2-, 2=,+, %%+1%+2 a:+,

with the. equality holds if and only if d = 0. The statement is then valid by the method of induction.

2. We first show that each student is taught by the same number h(c-1)

k-l

of teachers. Indeed, we fix a student A, and suppose that he is taught by r teachers. We count the number of pairs (x, Y) of students and teachers s&h that teacher y teaches both students A and x (where x is ditferent from A), By first choosing x and then y, there are (c-1)h pairs. By first choosing y and then x, there are r(k-1) pairs. Hence (c-l)h=r(k-l),i.e.

h(c -1) r=-

k-l

as we claimed. Now, if we consider the total number of students taught by the teachers (counting multiplicities), we have

kk _ c k(c-1) _{- .-*} k-l Therefore b c(c-1) -=-* h k(k-1)

(Alternative Solution by Chung On Yip) Construct a graph with two set of vertices, the first set B consisting of b elements, correspond to the teachers, and the second set P

c(c - 1) consisting of -

2 elements, correspond to all possible distinct pairs of students. Link an element in B to an element in P if the teacher happens to teach that particular pair of

k(k - 1) students. By (i) each element in B is connected to -

2 element in P, hence there are altogether b k(k - ‘) - edges. By (ii) each element in P is connected to h elements in B,

2 c(c - 1) hence altogether h- have b k(k - 1) = h c(c2- ‘)

edges are drawn. The two quantities must be equal, hence must

2 -> 2 and the result follows.

(Alternative Solution by Poon Ming Fung) Construct a graph witb two set of vertices, the first set B consisting of b elements, correspond to the teachers, and the second set S, consisting of c elements, correspond to the students. Link au element in B to an element in S if the teacher happens to teach that particular student. By (i), altogether bk edges may be drawn. Now there are y distinct pairs of students, by (ii), we may drawn 2h- c(c - 1)

2 edges. However consider a particular edge, corresponds to teacher X teaching student Y, say. Then X also teaches k - 1 other students, Y,, Y, ,-.s, Yk_, , say, besides Y. Hence when

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we consider the pairs (Y,Y,),(Y,Y,),...,(Y,Y,_,), relating to X, we draw the edge connecting X and Y.k - 1 times. Therefore fmlly we have bk = 2h - c(c-I) - 1 , and the

2 k-l

result follows.

3. Let hues QR and AD intersect at E. We will show that lines PE and PS coincide. Note mQ = 135’ - LBAC = LACP. So B, P, Q, Care concyclic. Then LAPQ = LACB. Next

LAEQ =90’- LEAC = LACB .

Hence A, P, E, Q are concyclic. Then

w1E=180’ -LAQE=90’ Hence

lines PE and PS are both

perpendicular to AB. Therefore they coincide and lines PS, AD, QR are

concurrent. Efio

From LBQE = 90’ - 45’ = LCPE, _{i ,,’},i ,’ ,’

we get P, Q, R, S are concyclic. : ,I

Hence LQPR= LQSR. Since we Ei,’

showed that B, P, Q, C are

concyclic, so LQPR = LQBC. Therefore LQBC = LQSR, which implies lines SR and BC are parallel.

4. The answer is 2! x 2 I x 4! x lo! = 348364800. The proof goes as follows,

(a) We can prove by induction on the number of prime divisors (counting multiplicities) thatfis completely multiplicative, i.e.

f@~P~~~~~P~)=rf~*)l~[f(P2)1~~~~[f(P,)l’~

(b) Since 100 2 f (64) = f (2’) = v(2)]“, f(2) 5 2. But f (1) = 1, so this forces f (2) = 2 .

By the same argument, we can show that f (3) = 3. Consequently, we see that f(n) = n for n = 1,2,3,4,6, 8,9.

(c) Since f (n’) =Lf(n)12, perfect squares must be mapped to perfect squares. As 1,4,9,

16,64,81 are already used, f (25) may be 25,49 or 100. But

100 2 f(75) = f(3)f(25) = 3f(25), so f(25) = 25, i.e. f(5) = 5. Similarly, f (7) = 7 and f(lO)=lO.So f(n)=n for l<nzGlO.

(d) We claim that f(p) is prime if and only if p is prime. Indeed, since f is bijective and f(l)=l, f(q)>1 for q>l. If p is not prime, then g(p),h(p)>l, so f(p) = f (g(p)) f (h(p)) is not prime. If f(p) is not prime for some primep, then there

exists at least one composite number r for which f(r) is prime, contradicting what we just proved This establishes the claim.

(e) For 115 n s 100, we must have g(n) ,< 10 and h(n) is either smaller than 10 or is some prime between 11 and 100. Therefore, it remains to determine the value of f(p) for primes between 11 and 100.

(f) Since 1002 f(99)= f(g)f(ll)=gf(ll), we must have f(li)=ll. Likewise, f(13) = 13.

(g) Since 100 2 f(85) = f (5)f (17) = 5f(17), f (17) can only be 17 or 19. The same is true for f(19). There are 2! ways to fix the values of f (17) and f (19) ;

(h) Since~lOO 2 f (92) = f(4)f (23) = 4f (23)) we must have f (23) = 23.

(i) By essentially the same argument as in (6), there are 2! ways to tix the values of f(29) and f(31).

6) Applying the same argument again, there are 4! ways to permute the function values of

{37,41,43,47} among themselves and lo! ways to permute the function values of (53, 59,61,67,71,73,79,83,87,89} among themselves.

(k) Inview of (7), (9) and (lo), there are 2!x2!x4!xlO! possrbilities forf.

(I) Finally, we must chock that all 2!x 2!x 4! x lo! such possibilities indeed have the desired properties. That f is well-defined is clear: our constructions in steps (f) to (j) guarantees that f(n) will not exceed 100. Properties (i) and (iii) clearly hold for every suchf; so it remains to check the bijectivity. Since S is finite, we need only show that every such f is surjective, and this follows readily from the Runiamental Theorem of Arithmetic and de fact thatf maps the subset containing all primes in S to itself.

Mathematical Excalibur, Volume 9, Number 5

例析數學競賽中的計數問題 (一)

Olympiad Corner

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