In fact you could just write out the pattern, but I'll try to write down a "clean" solution: (please note: this question is meant for younger pupils ) Consider the reverse progress: from 25 to 11; move A is dividing by 2, move B is adding 3. And we need to prove this is impossible in 6 moves. (7 is possible: BAABABB) Looking at the problem mod 3 shows us that the number of A moves will be odd - so 1,3 or 5. 1 is impossible, since that cannot make the number drop to or below 11. 5 is impossible, since then we only have 1 B step and we must start with B for sure so 28 is not a multiple of 32. so there must be 3 A's and 3 B's. The pattern begins with a B (28). if a number is even we mustn't do B cuz that will take all twi B's to get it back even, and we ain't getting a multiple of 11.4^k ==> if the i'th term is even, we must do A (and of course if it's odd, B is the only option) So if there were a 6-move-pattern ending in 11 it would be: 25-28-14-7-10-5-8 and 8 is not 11, contradiction.

I had the same solution , have you already solved question 4 of the juniors?

dEUS wrote: I had the same solution , have you already solved question 4 of the juniors? Post that one and you'll find out

I allready did :-p

Peter wrote: In fact you could just write out the pattern, but I'll try to write down a "clean" solution: (please note: this question is meant for younger pupils ) Consider the reverse progress: from 25 to 11; move A is dividing by 2, move B is adding 3. And we need to prove this is impossible in 6 moves. (7 is possible: BAABABB) Looking at the problem mod 3 shows us that the number of A moves will be odd - so 1,3 or 5. 1 is impossible, since that cannot make the number drop to or below 11. 5 is impossible, since then we only have 1 B step and we must start with B for sure so 28 is not a multiple of 32. so there must be 3 A's and 3 B's. The pattern begins with a B (28). if a number is even we mustn't do B cuz that will take all twi B's to get it back even, and we ain't getting a multiple of 11.4^k ==> if the i'th term is even, we must do A (and of course if it's odd, B is the only option) So if there were a 6-move-pattern ending in 11 it would be: 25-28-14-7-10-5-8 and 8 is not 11, contradiction. can you explain how does mod 3 show that the number of A moves will be odd please?

my solution (also mathmax12's solution): $25+3=28, 28/2=14. 14/2=7. 7+3=10. 10/2=5. 5+3=8. 8+3=11.$