# RMO PRACTICE

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 ?

• Lokesh Verma ·

good question...

a2+b2+c2 remains constant at each step..
Hence the set can never be changed to the one that you have written

• Ricky ·

Nishant 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 " ?

• Lokesh Verma ·

This 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"



More hint: 3 â‰ 5

• Ricky ·

Sir , I solved it . Its for others . Maybe you ' re referring to Invariance Principle : ) ?

• Lokesh Verma ·

yup i am :)

• Shaswata Roy ·

a6 - a5 +a4 - a3 +a2-a1 = 6-5+4-3+2-1 = 3 remains invariant and hence we cannot form that sequence because 3 â‰ 4.

• Lokesh Verma ·

na.. 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

• Shaswata Roy ·

how 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).

• 