1. 已知 $a$, $b$, $c$ 为整数, 且 $a+b = 2006$, $c - a = 2005$. 若 $a < b$, 则 $a + b + c$ 之最大值是多少?
解答:
$ecause a, binmathbf{Z}$, $a + b = 2006$, $ herefore a le 1002$, 因此由 $$egin{cases}a + b = 2006\ c - a = 2005end{cases}$$ $$Rightarrow a + b + c = 2006 + a + 2005 = a + 4011 le 1002 + 4011 = 5013.$$ 即 $a + b + c$ 之最大值为 $5013$.
2. 解方程: $$[2x] + [3x] = 8x - {7 over 2}.$$ 解答: $$egin{cases}2x - 1 < [2x] le 2x\ 3x - 1 < [3x] le 3x end{cases} Rightarrow 5x - 2 < [2x] + [3x] le 5x$$ $$Rightarrow 5x - 2 < 8x - {7over2} le 5x Rightarrow {1over2} < x le {7over6}$$ $$Rightarrow {1over2} < 8x - {7over2} le {35over6} = 5{5over6} Rightarrow 8x - {7over2} = 1, 2, 3, 4, 5,$$ $$Rightarrow x = {9over16}, {11over16}, {13over16}, {15over16}, {17over16}.$$ 经检验, $x_1 = displaystyle{13over16}$, $x_2 = displaystyle{17over16}$ 是原方程的解.
3. 已知 $x = displaystyle{bover a}$, ($a$, $b$ 互素), 且 $a le 8$, $sqrt2 - 1 < x < sqrt3 - 1$. 求所有满足条件的 $x$.
解答: $$sqrt2 - 1 < x < sqrt3 - 1 Rightarrow sqrt2 - 1 < {b over a} < sqrt3 - 1$$ $$Rightarrow (sqrt2 - 1)a < b < (sqrt3 - 1)a$$ 分别将 $a = 1, 2, cdots, 8$ 代入验证, 可得所有满足题意的 $x$ 值如下: $${1over2}, {2over3}, {3over5}, {3over7}, {4over7}, {5over7}, {5over8}.$$
4. $n$ ($n > 1$)个整数 $a_1$, $a_2$, $cdots$, $a_n$ 满足 $a_1 + a_2 + cdots + a_n = a_1 a_2 cdots a_n = 2007$, 求 $n$ 的最小值.
解答:
易知 $a_i$ 均为奇数, 且 $n$ 为奇数.
若 $n = 3$, 即 $a_1 + a_2 + a_3 = a_1a_2a_3 = 2007$, 不妨设 $a_1 ge a_2 ge a_3$, 则 $$a_1 ge {2007 over 3} = 669, a_2a_3 le 3.$$ 若 $a_1 = 669$, $a_2a_3 = 3$, 则 $$a_2 + a_3 le 4 Rightarrow a_1 + a_2 + a_3 le 673 < 2007.$$ 若 $a_1 > 669$, 则 $a_1 = 2007$, $a_2a_3 = 1$ 且 $a_2 + a_3 = 0$, 矛盾.
由此可知 $n ge 5$. 当 $a_1 = 2007$, $a_2 = a_3 = 1$, $a_4 = a_5 = -1$ 时满足题意.
因此 $n$ 的最小值为 $5$.
5. 求三位数与它的数字和之比的最小值.
解答:
设三位数为 $N = overline{xyz}$, 则 $$lambda = {100x + 10y + z over x+ y+z} = 1 + {99x + 9y over x + y + z}ge 1 + {99x + 9y over x + y + 9}$$ $$= 1 + {9(11x + y) over x + y + 9} = 1 + 9left(1 + {10x - 9 over x + y + 9} ight)$$ $$ge 1 + 9left(1 + {10x - 9 over x + 9 + 9} ight) = 1 + 9left(1 + {10x - 9 over x + 18} ight)$$ $$= 10 + 9cdot{10x - 9 over x + 18} = 10 + 9cdot left(10 - {189 over x + 18} ight)$$ $$= 100 - {1701 over x + 18} ge 100 - {1701 over 1+18} = {199 over 19}.$$ 因此当三位数为 $199$ 时可取到最小比值 $lambda = displaystyle10{9 over 19}$.
6. 若 $a > 0$, $b > 0$, 且 $h = minleft(a, displaystyle{b over a^2 + b^2} ight)$, 求 $h$ 之最大值.
解答:
分类讨论即可.
若 $h = minleft(a, displaystyle{b over a^2 + b^2} ight) = a$, 则 $$h^2 = a^2 le {ab over a^2 + b^2} le {ab over 2ab} = {1over2}$$ $$Rightarrow h le {sqrt2 over 2}.$$ 若 $h = minleft(a, displaystyle{b over a^2 + b^2} ight) = displaystyle{b over a^2 + b^2}$, 则 $$h^2 = left({b over a^2 + b^2} ight)^2 le {ab over a^2 + b^2} le {ab over 2ab} = {1over2}$$ $$Rightarrow h le {sqrt2 over 2}.$$ 综上, $h$ 之最大值为 $displaystyle{sqrt2 over 2}$.