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Bài giải: Dãy số nguyên dương

[ Wednesday, 23 March, 2011 ]

Một dãy số nguyên dương 1, 2, … , n được viết trên bảng đen (n > 2). Cứ mỗi phút hai số được xoá đi và thừa số nguyên tố nhỏ nhất của tổng của hai số vừa xoá được viết lên bảng. Cuối cùng chỉ còn lại số 97. Tìm giá trị nhỏ nhất của n mà ta có thể làm được điều này.

(Đặc biệt dành cho các học sinh Cơ Sở Thuận An bây giờ – Bài thi giỏi toán quốc tế 2010 – Độ khó: Trung bình)

Solution

Noting that any prime divisor not equal to 2 is an odd number, the sum of two such prime numbers is an even number, and the prime divisor that is also an even number must be number 2.

To find the least possible n we should keep a free adder, a number originally on the board in the series 1, 2, . . ., n that is free from any calculations until the end to add to the last prime divisor. Since adding two odd numbers or even numbers creates a new even number that has 2 as its least prime divisor, the problem requires the addition of a prime number to number 2 to become another prime number.

Furthermore, unlike the additions of the even numbers, the total odd numbers in the series 1, 2, . . ., n have to be in pairs plus one more. The pairs of the odd numbers help adding up to exact even numbers in order for us to get 2 as their prime divisors. The two consecutive prime numbers under 97 that satisfy these conditions are 41 and 43, 59 and 61. We pick the pair 41 and 43 to give us the least n and n = 57.

Erasing the numbers this way: first erase all the odd numbers in pairs from 1 to 39, their pairings are of no significance, to get number 2 at the end, leave number 41 alone. Next erase the rest of the odd numbers from 43 to 57 to get another number 2 on the board (we should have a total of three numbers 2 now.) Then erase all the even numbers except number 54 which is the free adder. The remaining numbers on the board are 2, 41, and 54. Now erase 41 and 2 to get 43. Then erase 43 and 54 to get 97.

Ban Bien Tap

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