Maths 問題 (急)(10點)

P.16
27) Let the four-digit odd number be abcd.
We have
i) abcd > 9000 => a = 9
ii) 9bcd is divisible by 13 => 9 - bcd = 13A, where A is an integer.
iii) Case 3.1: If d ≧ 5, when rounded off to 3 sig. fig., it becomes 9b(c+1)0
Divisible by 4 => (c+1)0 = 4B, where B is an integer.
Case 3.2: If d ≦ 4, when rounded off to 3 sig. fig., it becomes 9bc0
Divisible by 4 => c0 = 4B', where B' is an integer.
iv) Case 4.1: If c ≧ 5, when rounded off to 2 sig. fig., it becomes 9(b+1)00
Divisible by 11 => [9 + 0] - [(b+1) + 0] = 11C => 8 - b = 11C, where C is an integer.
Case 4.2: If c ≦ 4, when rounded off to 2 sig. fig., it becomes 9b00
Divisible by 11 => [9 + 0] - [b + 0] = 11C' => 9 - b = 11C', where C' is an integer.
v) d is odd.

From (iv), case 4.1: If c ≧ 5, b = 8; case 4.2: If c ≦ 4, b = 9.
From (iii),
case 3.1: if (c+1)0 is a multiple of 4, then c+1 must be a multiple of 2
i.e. If d ≧ 5, c must be odd.
case 3.2: similarly, If d ≦ 4, c must be even.
From (ii),
of course, d = 1,3,5,7 or 9
Now, we have the number = 98cd or 99cd
Case A: 98cd
From (ii), 8cd = 9 - 13A, since 899 ≧ 8cd ≧ 800, we have the possibilities:
8cd = 802 (when A = -61), 815, 828, 841, 854, 867, 880 or 893 (when A = -68)
If c ≧ 5, b = 8, only the numbers 802, 815, 828 and 841 are possible.
But if c ≦ 4, b = 9, therefore, no numbers in this case are possible.
Case B: 99cd
From (ii) 9cd = 9 - 13A, since 999 ≧ 9cd ≧ 900, we have the possibilities:
9cd = 906 (when A = -69), 919, 932, 945, 958, 971, 984 or 997 (when A = -76)
If c ≧ 5, b = 8; if c ≦ 4, b = 9, so only the number 919 and 958 are possible.
But how we check...9919 is the answer
9919 = 13A
9920 = 4B
9900 = 11C
*[We get two numbers in the last step because when 9958 is rounded off to 2 sig. fig., it becomes a 5-digit number, which we omitted in case 4.1.]
*[I wonder if (iii) is necessary as we didn't use it, maybe it can help us eliminate the impossible answers faster.]

For the 2nd question, do I have to login to view it?

2007-09-01 15:08:28 補充：
Btw hkedcity 數據庫搬緊...睇唔到@@