PROBLEMS Vol. 102(5), May 2002

• 4731: Proposed by: Stanley Rabinowitz, Westford, MA

Let ABCD be a quadrilateral, none of whose angles is a right angle. Prove

that:

 á         4732: Proposed by: Monte J. Zerger, Alamosa, CO

Let,

represent the n^th triangular number. Consider the set of alternate triangular numbers {T_2k-1} = {1,6,15,·}. Find the sum of their reciprocals.

• 4733: Proposed by: Jose Luis Diâaz-Barrero, Barcelona, Spain

            Let n be a positive integer. Prove that: 

 where F_n is the n^th Fibonacci number, i.e., F₀ = 0, F₁ = 1 and F_n+2 = F_n+1 + F_n ,for n> 2. 

• 4734: Proposed by: Monte J. Zerger, Alamosa, CO

What is the locus of centers of all circles externally tangent to each of two fixed non-congruent, non-intersecting circles? 

• 4735: Proposed by: Richard L. Francis, Cape Girardeau, MO

Consider the unending decimal x beginning with .1 so that each succeeding digit is the digital root of the sum of the preceding digits. Such numbers are called modular decimals and, in this case, .1 is the reference number. Is ~r an irrational number? 

• 4736: Let g(x¹²) = x⁵ + x⁴ + x³ + x² + x + 1. What is the remainder when the polynomial g(x¹²) is divided by the polynomial g(x)?

Solutions

• 4700: Let MN be the diameter of a semicircle with center C. Let P be a point on CN and let A and B be points on the circumference of the semicircle such

hat ÜCAP = ÜCBP = 10¡. If arc MA  = 40¡, what is the measure of arc BN ?

         Solution by: Irwin Feinstein, Wilmette, IL.  

Since ÜACM = 40¡, ÜCPA = 30¡. Designate ÜAPB by q. Now using the law of sines twice we have:

         Since AC = BC, we have sin 30¡ =  sin(30⁰ + q). So q = 120¡Þ ÜBCP = 20¡. Therefore, the measure of arc BN  = 20¡.

Also solved by: Michael Brozinsky, Central isiip, NY; Jia Shao, Queens NY, student of Michael Brozinsky;     Monte J. Zerger, AlamÐosa, CO.

• 4701: Let f be a function with the following properties: i) f(n) is defined for every positive integer n; ii) f(n) is an integer; iii) f(2) = 2; iv) f(mn) = f(m)f(n) for all m and n; and v) f(m) > f(n) whenever m > n. Prove that f(n) = n for n = 1,2,3,·.

         Solution by: Harry Sedinger, St. Bonaventure, NY. 

We prove that f(n) = n for every positive integer n by strong mathematical induction. First, note that 2f(1) = f(2)f(1) = f(2 x 1) = f(2) = 2, so f(1) = 1 and the statement is true for n = 1. For a fixed n> 1, assume that f(k) = k  for every positive integer k <n. If n is even, then n = 2k where k <n so that f(k) = k. Thus f(n) = f * 2k) = f(2)f *+(k) = 2k = n. If n is odd, then n +1 = 2k where k <n so that f(k) = k and f(n+1) = f(2k) = f(2)f(k) = 2k=n+1. Thus f(n) <f(n+1) = n+1.

Clearly f(n) >f(n- 1) = n-1 since (n -1 <n), and it follows that f(n) = n. By strong induction, f(n) = n for every positive integer n. 

Also solved by: Melfried Olson, Macomb, IL; N.J. Kuenzi, Oshkosh, WI; Michael Brozinsky, Central Islip, NY; Paul M. Harms, North Newton, KS; Armend Shabani, Prishtina, Kosova; Monte J. Zerger, Alamosa, CO; Brian Beasley, Clinton, SC. 

• 4702: Given polynomial P(x). If P(x) is divided by x² + x +1, a remainder of

3 is obtained. If P(x) is divided by (x+5) a remainder of 7 is obtained. What is the remainder if P(x) is divided by

(x² + x +1)(x + 5)?

         Solution by: Monte J. Zerger; Alamosa, CO.

The minimal degree possible for P(x) is 3. Then we would have P(x)= (x² + x + 1)(x + b) + 3 and P(x) = (x + 5)(x² + dx + e) + 7, so that for real numbers b, d and e, we must have:

  (x² + x + 1)(x + b) + 3 = (x + 5)(x² + dx + e) + 7 

x³ + (b+1)x² + (b+1) x + (b+3) =x³+(d+5)x²+(e+5d)x+5e+7        

Equating corresponding coefficients and solving the resulting system of equaÐtions yields

b= 209/21 , d = 25/21 and e = 5/21..Therefore P(x)=x³+(130/21)x²+(130/21)x+(172/21) and dividing this by (x² + x +1) (x ±5) produces a remainder of (4/21)x² + (4/21)x + (67/21). 

Also solved by: N.J. Kuenzi, Oshkosh, WI; Stanley Rabinowitz, Westford, MA; Paul M. Harms, North Newton, KS; Harry Sedinger, St. Bonaventure, NY. 

• 4703: Proposed by: Richard L. Francis, Cape Girardeau, MO

Let S be the set of even perfect numbers. Can a polynomial equation whose coefficients are positives or negatives of the numbers in S also have a solution in S? 

Solution by: Paul M. Harms, North Newton, KS.

The answer is no. Consider the polynomial P(x) = a_nxⁿ + a_n-1x^n-1+· +a₁x +  a₀ where ¸a_i¸ë S, i = 0, 1,2,.. n. Since P(x) has integer coefficients, the possible rational zeros of P(x) are integer factors of a_o, divided by integer factors of a_n. Division of odd prime numbers, 2^p ÷ 1 and 2^q ö 1 cannot be an integer unless p = q. The only possible rational zeros of P(x), whose absolute value is also in S, are ± (a₀/1)= ± a₀. Suppose P(a₀) = 0, then a_n(a₀)ⁿ +a_n-1(a₀)^n-1 +·+a₁ a₀+a₀ = a₀[a_n,(a₀)^n-1 + a _n-1 (a₀)^n-2+...+ a₁ +1] = 0.

Since a₀ Õ 0, [a_n(a₀)^n-1 + a _n-1(a₀)^n-2 +... + a₁ + 1] = 0. The left side of the last equation is odd since all a_iâs are even and, thus, the last equation cannot be true. It follow that P(a₀) Õ0. In a similar manner, P(-a₀) Õ0can be shown.

 Also solved by the proposer.            

• 4704*: Let three triangles be given with sides of lengths X, Y and Z; U, V and W; and X+U,Y+V and Z+W. Let angles A,B and Cbe measured in radians and be such that LA is opposite Z; LB is opposite W and LC is opposite Z + W. Prove or disprove that:

(X+Y-Z) ÜA+(U+V-W) ÜB £  ((x+u)+(Y+v)-(z±w)) Üc

No solution was received for this problem, so it remains open. 

• 4705: Proposed by: Richard L. Francis, Cape Girardeau, MO Professor Y had a 3-page digital print-out of the Mersenne prime 2^2976221 -1. These pages of roughly the same number of digits became shuffled. One page began with 6 and ended with 3, another began with 2 and ended with 9, and still another began with 4 and ended with 1. Put these pages in order without generating the entire number.

Solution by N.J. Kuenzi, Oshkosh, WI.

The units digits of successive powers of 2 follow the cyclic pattern 2,4,8,6, 2,4, 8,6,2,4,8,6, etc. (One can use Mathematical Induction to prove this.) So for integers n Ò 0, the units digit of 2⁴ⁿ⁺¹ is 2; the units digit of 2⁴ⁿ⁺² is 4; the units digit of 2⁴ⁿ⁺³  is 8; and the units digit of 2^[4(n+1)] is 6. 

Since 2976221 = 4(744055) + 1, it follows that the units digit of 22976221 is 2 and the units digit of the Mersenne prime 2^2976221 -1 is one. So the page that began with 4 and ended with 1 is the last page of the printout. 

Using base ten logarithms yields log 2^2976221 = 2976221 log 2 È895931.7947, log 2^2976221 È895931 + 0.7947. Now 2^2976221 È (10⁸⁹⁵⁹³¹)(10^0.7947) È (6.233)(10⁸⁹⁵⁹³¹). Thus the lead digit of the Mersenne prime 2^2976221 - 1 is 6. So the page that began with 6 and ended with 3 must be the first page. This leaves the page that began with 2 and ended with 9 as the middle page. 

In surmnary, page 1 is the page that began with 6 and ended with 3; page 2 is the page that began with 2 and ended with 9, and page 3 is the page that began with 4 and ended with 1.

Also solved by: Stanley Rabinowitz, Westford MA; Charles AshÐbacher, Hiawatha, IA; and the proposer.