標題:
發問:
1.Given any integers x, show that the values of ax^2+bx+c is always a perfect square number.Prove：a) 2a, 2b, and c are integers. b) a, b, c are integers and c is a perfect square.2.There are exactly k perfect squares which are divisors of 1996^1996. Find k.3.A 4-digit number which is a perfect square is... 顯示更多 1. Given any integers x, show that the values of ax^2+bx+c is always a perfect square number. Prove：a) 2a, 2b, and c are integers. b) a, b, c are integers and c is a perfect square. 2. There are exactly k perfect squares which are divisors of 1996^1996. Find k. 3. A 4-digit number which is a perfect square is created by writing Anne's age in years followed by Ben's age in years. Similarly, in 31 years, their ages in the same order will again form a 4-digit perfect square. Determine the present age of Anne and Ben. 4. The amount of budget just happens to be the smallest number of cents (other than 1 cent) that is a perfect square, perfect cube, and a perfect fifth power. How much is the amount of budget? 最佳回答會給5粒星，請大家盡力回答，要計算方法，多謝！ 更新: 1. b) prove c is a perdfect square. 更新 2: 3) 爲什麽 x-31=y?????
最佳解答:
1). a) 當 x = 0 , 原式 = c , 因此 c 為平方數即整數。. 當 x = 1 , 原式 = a + b + c = m^2 , m為整數 ......(1) 當 x = - 1 , 原式 = a - b + c = n^2 , n為整數 ......(2) (1) + (2) : 2a = m^2 + n^2 - 2c 為整數 (1) - (2) : 2b = m^2 - n^2 為整數 b) 當 x = 4 , 原式 = 16a + 4b + c = k^2 , k 為整數, 16a + 4b = k^2 - c , 由於已證 c 為平方數 , 設 c = r^2 , r 為整數 , 16a + 4b = (k - r)(k + r) 8(2a) + 2(2b) = (k - r)(k + r) 因 (k - r) 與 (k + r) 奇偶性同 , 又左方為 2 的倍數 , 故 (k - r) 與 (k + r) 同為偶數 , 從而 (k - r)(k + r) 為 4 的倍數。 即 8(2a) + 2(2b) = 4(2(2a) + b) 為 4 的倍數 , 所以 2(2a) + b 必為整數 , 已證 (2a) 為整數 , 故 b 必為整數。 至此已證原式 ax^2 + bx + c 中 b , c 均為整數 , 由(1) , a + b + c = m^2 a = m^2 - b - c 亦為整數。 2) 1996^1996 = 4^1996 x 499^1996 4是平方數可取 0 至 1996 個 , 共 1997種 499非平方數可取 0 , 2 , 4 , .... 1996 個 , 共(1996/2)+1 = 999種 (取0種代表4^0 或 499^0 = 1) 平方數因子共 999 x 1997 = 1995003 種 3) 設兩人相隔 31 年 的年齡合併平方數分別為 x^2 及 y^2 , 由題設知 x + 31 = y 亦為兩位數 , 故 x^2 - y^2 = 3131 (x - y)(x + y) = 31 x 101 x - y = 31 x + y = 101 兩式相加/減 : x = 66 , x^2 = 4356 y = 35 , y^2 = 1225 31年前後兩人年齡分別 12 , 25 及 43 , 56 。 4) 因 2 , 3 , 5 互質 , 除 1 以外 最小合乎數值 = ((2^2)^3)^5 = 2^30 = 1073741824 cents. 2010-05-01 12:35:07 補充： 請見 1a) 第1行已證了。 2010-05-04 01:04:10 補充： 3)爲什麽 x-31=y????? 是弄錯了 , 「由題設知 x + 31 = y 亦為兩位數 ,」 改為 「由題設知 x^2 的首/末兩位數 - 31 = y^2 的首/末兩位數,」才對。 Sorry!!
其他解答:
2) 1996 = 499 x 2 x 2 1996^1996 = 499^1996 x 2^1996 x 2^1996 = 499^1996 x 2^3992 Powers of 499 can be all even from 0 to 1996 (total = 999) and powers of 2 can be all even from 0 to 3992 (total = 1997). So no. of perfect squares that can be formed = 999 x 1997 2010-05-01 01:02:58 補充： 3) Let x be the 4-digit values formed by their present age, then 31 years later the 4-digit values will be x + 3131. So by the fact that successive square values are formed by adding successive odd values: x + 3131 = x + a + (a + 2) + ... + [a + 2(n - 1)] 2010-05-01 01:03:04 補充： na + n(n - 1) = 3131 n(a + n - 1) = 31 x 101 n = 31, a + n - 1 = 101, giving a = 71 So with a = 71, we have: x = 1 + 3 + 5 + ... + 69 = 1225 x + 3131 = 4356 which are both perfect squares. Anne's age = 12, Ben's age = 25 2010-05-01 01:04:45 補充： 4) Since the value is a perfect square, cube and fifth power and is an integer, its smallest possible value is 2^30 since 30 is the LCM of 2, 3 and 5. Budget = 2^30 = 1073741824 cents
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perfect square 奧數4條，15分！數學高手們進發問:
1.Given any integers x, show that the values of ax^2+bx+c is always a perfect square number.Prove：a) 2a, 2b, and c are integers. b) a, b, c are integers and c is a perfect square.2.There are exactly k perfect squares which are divisors of 1996^1996. Find k.3.A 4-digit number which is a perfect square is... 顯示更多 1. Given any integers x, show that the values of ax^2+bx+c is always a perfect square number. Prove：a) 2a, 2b, and c are integers. b) a, b, c are integers and c is a perfect square. 2. There are exactly k perfect squares which are divisors of 1996^1996. Find k. 3. A 4-digit number which is a perfect square is created by writing Anne's age in years followed by Ben's age in years. Similarly, in 31 years, their ages in the same order will again form a 4-digit perfect square. Determine the present age of Anne and Ben. 4. The amount of budget just happens to be the smallest number of cents (other than 1 cent) that is a perfect square, perfect cube, and a perfect fifth power. How much is the amount of budget? 最佳回答會給5粒星，請大家盡力回答，要計算方法，多謝！ 更新: 1. b) prove c is a perdfect square. 更新 2: 3) 爲什麽 x-31=y?????
最佳解答:
1). a) 當 x = 0 , 原式 = c , 因此 c 為平方數即整數。. 當 x = 1 , 原式 = a + b + c = m^2 , m為整數 ......(1) 當 x = - 1 , 原式 = a - b + c = n^2 , n為整數 ......(2) (1) + (2) : 2a = m^2 + n^2 - 2c 為整數 (1) - (2) : 2b = m^2 - n^2 為整數 b) 當 x = 4 , 原式 = 16a + 4b + c = k^2 , k 為整數, 16a + 4b = k^2 - c , 由於已證 c 為平方數 , 設 c = r^2 , r 為整數 , 16a + 4b = (k - r)(k + r) 8(2a) + 2(2b) = (k - r)(k + r) 因 (k - r) 與 (k + r) 奇偶性同 , 又左方為 2 的倍數 , 故 (k - r) 與 (k + r) 同為偶數 , 從而 (k - r)(k + r) 為 4 的倍數。 即 8(2a) + 2(2b) = 4(2(2a) + b) 為 4 的倍數 , 所以 2(2a) + b 必為整數 , 已證 (2a) 為整數 , 故 b 必為整數。 至此已證原式 ax^2 + bx + c 中 b , c 均為整數 , 由(1) , a + b + c = m^2 a = m^2 - b - c 亦為整數。 2) 1996^1996 = 4^1996 x 499^1996 4是平方數可取 0 至 1996 個 , 共 1997種 499非平方數可取 0 , 2 , 4 , .... 1996 個 , 共(1996/2)+1 = 999種 (取0種代表4^0 或 499^0 = 1) 平方數因子共 999 x 1997 = 1995003 種 3) 設兩人相隔 31 年 的年齡合併平方數分別為 x^2 及 y^2 , 由題設知 x + 31 = y 亦為兩位數 , 故 x^2 - y^2 = 3131 (x - y)(x + y) = 31 x 101 x - y = 31 x + y = 101 兩式相加/減 : x = 66 , x^2 = 4356 y = 35 , y^2 = 1225 31年前後兩人年齡分別 12 , 25 及 43 , 56 。 4) 因 2 , 3 , 5 互質 , 除 1 以外 最小合乎數值 = ((2^2)^3)^5 = 2^30 = 1073741824 cents. 2010-05-01 12:35:07 補充： 請見 1a) 第1行已證了。 2010-05-04 01:04:10 補充： 3)爲什麽 x-31=y????? 是弄錯了 , 「由題設知 x + 31 = y 亦為兩位數 ,」 改為 「由題設知 x^2 的首/末兩位數 - 31 = y^2 的首/末兩位數,」才對。 Sorry!!
其他解答:
2) 1996 = 499 x 2 x 2 1996^1996 = 499^1996 x 2^1996 x 2^1996 = 499^1996 x 2^3992 Powers of 499 can be all even from 0 to 1996 (total = 999) and powers of 2 can be all even from 0 to 3992 (total = 1997). So no. of perfect squares that can be formed = 999 x 1997 2010-05-01 01:02:58 補充： 3) Let x be the 4-digit values formed by their present age, then 31 years later the 4-digit values will be x + 3131. So by the fact that successive square values are formed by adding successive odd values: x + 3131 = x + a + (a + 2) + ... + [a + 2(n - 1)] 2010-05-01 01:03:04 補充： na + n(n - 1) = 3131 n(a + n - 1) = 31 x 101 n = 31, a + n - 1 = 101, giving a = 71 So with a = 71, we have: x = 1 + 3 + 5 + ... + 69 = 1225 x + 3131 = 4356 which are both perfect squares. Anne's age = 12, Ben's age = 25 2010-05-01 01:04:45 補充： 4) Since the value is a perfect square, cube and fifth power and is an integer, its smallest possible value is 2^30 since 30 is the LCM of 2, 3 and 5. Budget = 2^30 = 1073741824 cents
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