Ling Xiang Ning, Allan (Age: 13) from Raffles Institution in Singapore has sent in an almost complete solution:

"The starting numbers that allow you to reach 50 are divisible by 5."

The question showed an example of how it is possible to start at 15 and reach 50; he has used this to prove his finding:

"You can subtract 5(s) from all the multiples of 5 until they reach 15 (if they are greater than 15), or use doubling until you reach 20 (if they are smaller than 15) and then subtract 5 to reach 15."

The last part of the question asked for a proof that no other numbers would allow you to reach 50. He has proved that this is true when the two operations are squaring (instead of doubling) and subtracting 5:

"All numbers that are not multiples of 5 are 1(modulo 5), 2(mod 5), 3(mod 5), or 4(mod 5), while 50 is 0(mod 5). For the four types of numbers, no mattter how many times they have 5 subtracted from them, they will still be their respective ?(mod 5). For 1(mod 5) numbers, its square and square and square.. only brings about numbers of 1(mod 5). For 2(mod 5) numbers, its square and square and square.. brings about numbers of 2(mod 5), 4(mod 5), 3(mod 5), 1(mod 5), 2(mod 5), with no 0(mod 5). For 3(mod 5) numbers, its square and square and square.. brings about numbers of 3(mod 5), 4(mod 5), 2(mod 5), 1(mod 5), 3(mod 5), with no 0(mod 5). For 4(mod 5) numbers its square and square and square.. brings about numbers of 4(mod 5), 1(mod 5), 4(mod 5), 1(mod 5), 4(mod 5), with no 0(mod 5). Therefore, only multiples of 5 can reach a 0(mod 5) number. Therefore, no other numbers allows me to reach 50."

Can somebody adapt this argument to show that only multiples of 5 will reach 50 if we are only allowed to double and subtract 5?