3√x−2y+3√x−2y=103(√x−2y)2−10√x−2y+3=0,x≠2y√x−2y=13 or √x−2y=3x−2y=19 or x−2y=9(a−2)y+b=19 or (a−2)y+b=9Assuming the above equations are true for all values of y,(a,b)=(2,19) or (2,9)
Sum=∞∑n=17n−110n=17∞∑n=1(710)nsum of GS=17(7101−710)=13
From x + y + z = 8 and x - y + z = 4,
I skipped a step. I will show it here.
Infinitely many. Only those with a prime number of occurence of '1's are primes.
I have not been online for quite a while, and I have seldom logged on this site, so I don't know what's happening here.
There are not much difference since I last logged on.
Oh, yes. Considering only the cases that satisfy the condition is quite troublesome, so I want to make sure that each kid gets one candy at the very beginning, so there will be less constraints to consider.
2)
Consider the pairwise sum of terms.