貨運百科

此文章來自奇摩知識+如有不便請留言告知

標題:

if the last three digits of the square of a positive integer are the same and nonzero,we say that the positive integer is good. for example,?

發問:

....since 38^2=1444,so 38 is good. If we arrange all good numbers in ascending order,what is the second number? the answer is 462 plz explain

最佳解答:

The last 3 digits of the number must be 1000n + 111 , 1000n + 444 , 1000n + 555 , 1000n + 666 or 1000n + 999. But 1000n + 111 (mod 8) ≡ 7 , 1000n + 555 (mod 8) ≡ 3 , 1000n + 666 (mod 4) ≡ 2 , 1000n + 999 (mod 8) ≡ 7 are NOT perfect square since odd square (mod 8) ≡ 1 and even square (mod 4) ≡ 0. Let (10x + 2)2 = 100n + 44 100x2 + 40x + 4 = 100n + 44 5x2 + 2x = 5n + 2 5(x2 - n) = 2(1 - x) 1 - x (mod 5) ≡ 0 ? x = n = 1 or x = 6 , n = 38 , we have 122 = 144 and 622 = 3844. Let (10x + 8)2 = 100n + 44 100x2 + 160x + 64 = 100n + 44 5x2 + 8x + 1 = 5n 8x + 1 = 5(n - x2 ) 8x + 1 (mod 5) ≡ 0 ? x = 3 , n = 14 or x = 8 , n = 77 , we have 382 = 1444 and 882 = 7744. So the second number at least 3 digits. Let (100a + 12)2 = 1000x + 444 10000a2 + 2400a + 144 = 1000x + 444 100a2 + 24a - 3 = 10x 10(10a2 - x) = 3 - 24a , no solution for a. Let (100a + 38)2 = 1000x + 444 10000a2 + 7600a + 1444 = 1000x + 444 100a2 + 76a + 10 = 10x 76a = 10(x - 10a2 - 1) 19a = 5(x - 10a2 - 1) ? a = 5 , x = 289 , we have 5382 = 289444 Let (100a + 62)2 = 1000x + 444 10000a2 + 12400a + 3844 = 1000x + 444 100a2 + 124a + 34 = 10x 124a + 34 = 10(x - 10a2) a = 4 , x = 213 , we have 4622 = 213444. No need to consider a = 9 > 4. Let (100a + 88)2 = 1000x + 444 10000a2 + 17600a + 7744 = 1000x + 444 100a2 + 176a + 73 = 10x 176a + 73 = 10(x - 10a2) , no solution for a. All in all 462 is the second number.

其他解答:6A560DB68EC65D60