1Sir , I solved it . Its for others . Maybe you ' re referring to Invariance Principle : ) ?
Start with the set S - { 3 , 4 , 12 } . In each step , you may chose two numbers " a " and " b " from S and replace " ( . 6 a - . 8 b) " by " ( . 8 a + . 6 b ) " . Can you form a set P - { 4 , 6 , 12 } by finitely many steps ?
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8 Answers
62good question...
a2+b2+c2 remains constant at each step..
Hence the set can never be changed to the one that you have written
1Nishant Sir is right as always - : )
Here ' s another -
A circle has six sectors in which the numbers " 1 , 2 , 3 , 4 , 5 , 6 " are written counter - clockwise .
In each step I increase two adjuscent numbers by 1 .
Can I form a sequence " 10 , 12 , 13 , 15 , 16 , 17 " ?
62This one i had solved in one of the classes if you remember... (Infact a very very similar one)
on the 2nd floor at DAV.. late night...
Hint: These questoins have something very common... "there is something that does not change.. try to find that"
[3]
More hint: 3 ≠5
1
11a6 - a5 +a4 - a3 +a2-a1 = 6-5+4-3+2-1 = 3 remains invariant and hence we cannot form that sequence because 3 ≠4.
62na.. see the sum of squares
Shaswata.. i think i did tell something very similar to u long time back...
See what is the relation between the sum of squares of the original numbers and the two new numbers formed
11how does the sum of squares remain invariant??
Sir I had answered the second question given by Ricky( the one dealing with sectors in which numbers from 1 to 6 are written).
Answer to the first question:
(0.6a - 0.8b)2 + (0.8a + 0.6b)2 + c2 = a2 +b2 +c2
since 32 + 42 + 122 ≠42 + 62 + 122
It can never be reached
I remember the class in which you had talked about the invariance principle 1 year ago(though I did not quite get it that time).