how woild do you do 7 + 48 ???
40 + 7 = 47, 47 +8 = 55
40 + 4 + 4 + 4 + 3
(24*2)+(5+2)
100-48-7 = 45, hence 100-45=48+7=55
7/11 + 48/11 = 5, so 5*11 = 55
3+4 = 7, 40 + 8= 48, 7+48 = 55
7^2=49, 48-40=8, -> 49+8=55
100+7 = 107 and 107 + 48 = 155 so 155 -100 is 55
another way
e^7 × e^48 ≈ 769478526514201713818274.55901293939920707675726508409, ln(769478526514201713818274.55901293939920707675726508409) ≈ 55
7 + 48 mod 5 = 0
7 + 48 mod 7 = 6
There is one number mod 35 with this property, and that is 20. So our solution is one of 20, 55, 90,...
But the only reasonable value is 55 by order of magnitude
Let f:ℝ->ℝ be given by f(x)=7+x and let ε>0 be given. Let δ=ε and consider x∈ℝ such that 0<|x-48|<δ. Then |7+x-55|=|x-48|<δ=ε. Hence f(x)->55 as x->48. Moreover f is continuous (constant functions and x are constinuous functions and addition of continuous functions results in a continuous function) we have that f(48) is the limit of f(x) as x->48, thus f(48)=7+48=55
You're all making this too complicated, the homotopy group of a bouquet of 7 circles wedged with a bouquet of 48 circles is F_55
til 55 is prime
