International Mathematical Olympiad

July 17, 2009

The Standard Prelude makes tasks like this easy; we will use list-of (fold-of), sum, square, digits, ilog and permutations.

The first task is just a list comprehension that generates three-digit numbers and tests them for the required properties:

> (list-of n
    (n range 100 1000)
    (zero? (modulo n 11))
    (= (/ n 11)
       (sum (map square (digits n)))))
(550 803)

The second task calls for us to dissect a number and stitch it back together, looping until we find the answer:

> (let loop ((n 6))
    (let ((m (+ (* 6 (expt 10 (ilog 10 n))) (quotient n 10))))
      (if (= (* 4 n) m) n
        (loop (+ n 10)))))

The third task is a bit harder. In-position and two-consecutive count the number of contestants in the specified positions; the list comprehension loops over the permutations of (A B C D E), testing each one.

(define (in-position actual predicted)
  (let loop ((a actual) (p predicted) (n 0))
    (cond ((null? a) n)
          ((equal? (car a) (car p))
            (loop (cdr a) (cdr p) (+ n 1)))
          (else (loop (cdr a) (cdr p) n)))))

(define (two-consecutive actual predicted)
  (let loop ((a actual) (p predicted) (n 0))
    (cond ((or (null? a) (null? p)) n)
          ((null? (cdr p)) n)
          ((null? (cdr a))
            (loop actual (cdr p) n))
          ((and (equal? (car a) (car p))
                (equal? (cadr a) (cadr p)))
            (loop (cddr a) p (+ n 1)))
          (else (loop (cdr a) p n)))))

> (list-of p
    (p in (permutations '(a b c d e)))
    (= (in-position p '(a b c d e)) 0)
    (= (two-consecutive p '(a b c d e)) 0)
    (= (in-position p '(d a e c b)) 2)
    (= (two-consecutive p '(d a e c b)) 2))
((e d a c b))

You can run the suggested solution at


Pages: 1 2

8 Responses to “International Mathematical Olympiad”

  1. iyo said

    Hi, thanks for problems. I think they are really simple to solve (you can use brute force to solve them without any difficulties :-)

    The last one I even solved on paper. I share my solution:

    When you took the second prediction and mark 2 consecutive contestants creating 2 disjoint pairs, you get 3 choices:

    DA EC and B is somewhere else
    DA CB and E is somewhere else
    AE CB and D is somewhere else

    Your goal is to place the “unpaired” contestant in good place.

    1st choice:

    B DA EC – wrong, none of them finished as it was predicted
    DA B EC – B can’t be after A
    DA EC B – all of them finished at it was predicted

    2nd choice:
    E DA CB – the correct answer!!! only C,B are at the right places
    DA E CB – all of them finished at it was predicted
    DA CB E – C is on the 3rd place – it can’t be because of the first prediction

    3rd choice:
    D AE CB – all of them finished at it was predicted
    AE D CB – A is on the 1st place – it can’t be because of the first prediction
    AE CB D – A is on the 1st place – it can’t be because of the first prediction

  2. swaraj said

    ruby solution for first problem (

  3. Vivek N said

    How do you solve these problems with a mathematical approach, for example if N is very large (something like 10^9, where bruteforce won’t work).

  4. Skuba said

    Problem 1 with Python:

    def main():
        upper_bound = 999
        lower_bound = 100
        count = lower_bound
        #while loop runs through all three digit numbers 
        while count <= upper_bound:
            #Only numbers divisible by 11 are evaluated
            if count%11==0:
                #convert number to string in order to seperate digits
                digit1 = num[0]
                digit2 = num[1]
                digit3 = num[2]
                sum_of_squares = int(digit1)**2 + int(digit2)**2 + int(digit3)**2
                if count//11 == sum_of_squares:
                    print (num)
                    count = count+1
                    count = count+1
                count = count+1

    Results should be 550 and 803.

  5. skuba713 said

    Problem 2 with Python:

    def main():
        rearranged = 0
        modified = 0
        check = 0
        #loop cycles through numbers until solution is found
        while check == 0:
            #put 6 in front and exclude last digit
            rearranged = "6" + str(number)[0:len(str(number))-1]
            modified = number * 4
            #Check if solution is true
            if int(rearranged) == int(modified):
                check = 1
                #adding ten gives the next number ending in six
                number = number + 10
        print("The original number is:",number)

    Solution should be 153846

  6. jack frost said

    I think both of those python approaches could be minified and simplified.

    ######### Problem 1
    for a in range(1,10) :
    for b in range(0,10) :
    for c in range(0,10) :
    N = a*100 + b*10 + c
    if N % 11 is 0 and N/11 == a**2 + b**2 + c**2 :

    ######### Problem 2
    def n() :
    exp = 2
    while True :
    start = 10**exp
    for i in range(start+6,start*10,10) :
    x = (i – 6)/10 + 6 * start
    if 4 * i == x :
    return i
    exp += 1


    I’ll try problem 3 at another time

  7. jack frost said

    Something happened to the formatting……smh

  8. George said

    Answers in Ruby 2.0

    Determine all three-digit numbers N having the property that N is divisible by 11, and N/11 is equal to the sum of the squares of the digits of N.
    def problem_1
      properties = []
      properties << proc do |number|
        number%11 == 0
      properties << proc do |number|
        number/11 == ( {|digit| digit.to_i**2}.inject(:+))
      collection = []
      number = 100
      until number > 999
        collection << number unless{|property| property[number]}.include?(false)
    Find the smallest natural number n which has the following properties:
    (a) Its decimal representation has 6 as the last digit.
    (b) If the last digit 6 is erased and placed in front of the remaining digits, the resulting number is four times as large as the original number n.
    def problem_2
      properties = []
      properties << proc do |number|
        number.to_s.end_with? '6'
      properties << proc do |number|
        digits = number.to_s.chars
        last_digit = digits.pop
        digits.join.to_i == number*4
      index = 0
      index += 1 while{|property| property[index]}.include?(false)
    problem_1 #=> [550, 803]
    problem_2 #=> 153846

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