中三數超難一題,都係10分

i solved it!!!
hundreth(a): tenth(b): unit (c) = 6:4:3

i) 6+4+3=13
ii) a +2=2b ---> 6+2=2*4 YES
iii) [c,b,a]( as a digit) +297= [a,b,c] (as a digit)
346+297=643 YES!

Therefore, the original three digit number is 643!!!!!

From: mathsaaa

Let:
a be the hundredth; b be the tenth and c be the unit.

The sum of the digits of a three-digit no. is 13
a + b + c =13 .....(1)
he hundreds digit is smaller than twice the tens digit by 2
a+2 = 2b .....(2)
b = (a + 2) / 2 .....(2a)
If we exchange the positions of the hundreds digit and the units digit,the no. obtained is smaller than the original by297.
(100a + 10b + c) - (100c +10b + a) = 297 .....(3)
Simplify (3):
99a - 99c = 297
(a - c) = 3
=> c = a - 3 ...... (3a)

Substitute (2a) and (3a) into (1):
a + (a + 2) / 2 + (a - 3) = 13
=> a + a/2 + 1 + a - 3 =13
=> 2a + a + 2 + 2a - 6 = 26
=> 5a - 4 = 26
=> a = 6

Therefore, from (2a):
b = (a + 2) / 2
b = (6 + 2) / 2
b = 4

From (3a):
c = a - 3
c = 6 - 3
c = 3

Hence, a = 6, b = 4 and c = 3; ie, 643

當百位數與個位數對調, 則數值會減少297=300-3, 所以, 個位數大於百位數3
十位數字x2 減百位數字等於2, 所以百位數字一定是大於3的雙數
可以列出所有可能性
一)431, 4+3+1=8
二)643, 6+4+3=13
三)855, 8+5+5=18
只有643符合所有的要求, 包括數字和等於13。

2007-02-23 00:27:07 補充：
個位數是少於百位數, sorry, 打錯字

Let the no. be XYZ

X+Y+Z=13................................................(1)
X=2Y-2.....................................................(2)
(100X+10Y+Z)-(100Z+10Y+X)=297..............(3)

From (3),
99X-99Z=297
99(X-Z)=297
X-Z=3
Z=X-3.......................................................(4)

Sub. (2) into (4)
Z=(2Y-2)-3
Z=2Y-5.....................................................(5)

Sub. (2) & (5) into (1)
X+Y+Z=13
(2Y-2)+Y+(2Y-5)=13
5Y-7=13
5Y=20
Y=4

Sub Y=4 in (2) & (5)
X=2Y-2
X=2*4-2
X=6

Z=2Y-5
Z=2*4-5
Z=3

Therefore, the no. is 643.
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Actually, the aim of these kind of questions are to express all those annoying variables in terms of one single variable. Then it would be as easy as 1,2,3! Hope this helps you! :)

Let the hundreds digit, tens digit and unit digit of the number be a, b and c respectively.

Then, a + b + c = 13 ----------------------------------(1)
2b - a = 2 --------------------------------- (2)
[100a+10b+c]-[100c+10b+a] = 297 ------(3)

By simplifying (3), we have 99(a-c) = 297 or a - c = 3 ----------------(4)

(1)+(2), we have 3b + c = 15 ------------------(5)
(2)+(4), we have 2b - c = 5 -------------------(6)

(5)+(6), we have 5b = 20 or b = 4
Then, by (2), we have a = 2b - 2 = 6.
By (4), we have c = a - 3 = 3.

Therefore, the original three-digit number is 643.