The original title is:
Twelve balls have the same characteristics, but only one ball has an abnormal quality. It is required to weigh the ball three times with a balance without weight to find out the ball with abnormal quality.
Solution:
Let the mass of the standard ball be W, representing any normal ball, and number 12 balls as a 1, a2, ..., a 12 in turn, and group them as follows:
A 1, A2, A3 and A4 are A 1 group.
A5, A6, A7 and A8 are group A2.
A9, A 10, A 1 1 and A 12 were group A3.
= = (first time) 1 Select any two groups-compare A 1 and A2, if
1 A 1=A2, then group A3 is an abnormal ball group.
Reorganized into:
B 1:a9 a 10
B2:a 1 1 w
B3:a 12 w
= = = = (the second time) take B2 B3-B2 and B 1 any group 1 for comparison, if
1. 1 B 1=B2, B 1 B2 is the normal group, B3(a 12, w) is the abnormal group, and the abnormal ball is a 12.
1.2 B 1！ = B2 B3 (a12, w) as the normal group, with b 1
Expression exp0: A9+A 10
= = = = = = (the third time) compare a9 a 10, if
1.2.1a9 = a10, then a 1 1 is an abnormal ball.
1.2.2 a9！ = a 10, then a 1 1 is a normal ball, and according to EXP0, A9+a 10,
Therefore, the mass of abnormal ball is less than that of normal ball, and lighter a9 and a 10 are abnormal balls.
2 A 1！ = A2, then A3 (A9, A 10, A 1 1, A 12) is the normal group; Use a 1
The expression1:exp1:a1+a2+a3+a4 < a5+a6+a7+A8.
Reorganized into:
B 1:a 1，a2，a3
B2:a4、a5、a6
B3:a7，a8，w
= = = = (Second time) Compare B3 and B2.
2. 1 B3=B2
According to EXP 1, a4 = a5 = a6 = a7 = A8 = w.
a 1+a2+a3 & lt； The abnormal ball mass of 3w is smaller than the standard ball.
= = = = = (third time) Compare a 1 a2, if
2.1.1a1= a2, then a3 is an abnormal ball.
2. 1.2 a 1！ = a2, a3 is a normal ball, and a 1 and a2 are light balls.
2.2 B3 & gt； B2 B 1 was the normal group.
According to EXP 1, a1= a2 = a3 = w.
3w+a4 & lt； a5 + a6 + a7 + a8
(B3 & lt； B2)a4+a5+a6 & lt； a7 + a8 + w
Add 3W+2A4+A5+A6.
2a4 & lta7+ a8
= = = = = (the third time) take a7 a8 for comparison.
2.2. 1 a7 =a8 a4 is an abnormal ball with a mass less than the standard ball.
2.2.2 a7！ =a8 a4 is an ordinary ball, so 2w
2.3 B2 & lt； B3 B 1 was the normal group.
According to EXP 1, a1= a2 = a3 = w.
3w+a4 & lt； a5 + a6 + a7 + a8
(B2 & lt； B3)a4+a5+a6 & gt； a7 + a8 + w
Transformation: -A4-A5-A6 <-a7-a8-w
Add 3w-a5-a6.
a5+a6 & gt； 2w
As we all know, the mass of special-shaped ball is greater than that of standard ball.
= = = = = = (the third time) compare a5 a6.
2.3. 1 a5 a6 The heavier ball is an abnormal ball.
As can be seen from the above sections, a 1, a2, ..., a 12 is an abnormal ball, and it is112.