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Toán đố đầu tuần

[ Wednesday, 20 January, 2010 ]

by Vo Duc Dien

Dù ta đã già nhưng vẫn làm toán “con nít” để tự ta thấy mình trẻ lại. Hơn nữa đây là sân trường trung học nên những bài toán ở đây chỉ ngang trình độ đó. Bài đố vui nguyên văn là một bài thi học sinh giỏi toán Canada như sau:  

Problem 7 of the Canadian Mathematical Olympiad 1971

Let n be a five digit number (whose first digit is non-zero) and let m be the four digit number formed from n by deleting its middle digit. Determine all n such that n / m is an integer.

Như các bạn biết vào Internet tìm thì ra question nhưng không ra solution. Bài toán có thể giải dùng một lập trình nhanh gọn như sau:  

int main (int argc, const char * argv[])

{

          unsigned int m;

          unsigned int c=0;

          for (unsigned int i = 10000; i<= 99999; i++)

          {

                   unsigned int n = i;

                   unsigned int tenThousands = n/10000;

                   n -= tenThousands*10000;

                   unsigned int thousands = n/1000;

                   n -= thousands*1000;

                   unsigned int hundreds = n/100;

                   n -= hundreds*100;

                   unsigned int tens = n/10;

                   n -= tens*10;

                   unsigned int ones = n;

                   m = (tenThousands*1000) + (thousands*100) + (tens*10) + (ones);

                   if ((i % m) == 0)

                   {

                             c += 1;

                             printf(“%d %d\n”,i,m);

                   }

          }

          printf(“total: %d\n”,c);

          return 0;

}

Answer is 90 numbers n to satisfy the problem.

 Hay là giải theo sự yêu cầu của ban giám khảo như sau:

Let  n = abcde  where a, b, c, d and e are positive integers from 0 to 9 and a ≠ 0.

 We then have  m = abde   and 

n = 10000a + 1000b + 100c + 10d + e

m = 1000a + 100b + 10d + e

If n / m = k is an integer, we have

10000a + 1000b + 100c + 10d + e  =  1000ak + 100bk + 10dk + ek         (1)

 Now assume k > 10  or  k = 10 + p where p is a positive integer; (1) becomes

10000a + 1000b + 100c + 10d + e  =  10000a + 1000b + 100d + 10e  +

1000ap + 100bp + 10dp + ep                                                                                      (2)

Now let’s find the possible value for p. We have

p = ( 100c – 90d – 9e) / ( 1000a + 100b + 10d + e)

but since a ≠ 0 and b, c, d and e are all non-negative integers, the denominator is then

≥ 1000 and the numerator is less than 1000, so p < 1, and  k > 10  is not possible.

Similarly, if k < 10,  p =  (90d + 9e – 100c) /(1000a + 100b + 10d + e)

With the same argument  k < 10 is not a possibility. Therefore k = 10.

Substitute k = 10 into (1), we have      100c = 90d + 9e  which requires product 9e

to be a multiple of 10 which is not possible. This equation has the only solution

c = d = e = 0.  So n = ab000 where a and b are positive integers and a = 1–>9 and

b = 0–>9. So numbers n are

10000, 11000, 12000, 13000, 14000, 15000, 16000, 17000, 18000, 19000,

20000, 21000, 22000, 23000, 24000, 25000, 26000, 27000, 28000, 29000,

30000, 31000, 32000, 33000, 34000, 35000, 36000, 37000, 38000, 39000,

40000, 41000, 42000, 43000, 44000, 45000, 46000, 47000, 48000, 49000,

50000, 51000, 52000, 53000, 54000, 55000, 56000, 57000, 58000, 59000,

60000, 61000, 62000, 63000, 64000, 65000, 66000, 67000, 68000, 69000,

70000, 71000, 72000, 73000, 74000, 75000, 76000, 77000, 78000, 79000,

80000, 81000, 82000, 83000, 84000, 85000, 86000, 87000, 88000, 89000,

90000, 91000, 92000, 93000, 94000, 95000, 96000, 97000, 98000, 99000

a total of 90 numbers.

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