341Muh thinking suh:
Let the two powers be 2a and 2b with a>b. Then 9|2a-2b = 2b(2a-b-1).
This means 9|(2a-b-1)
Remember that 2a and 2b got to have the same number of digits (as the last digit of 2a cannot be zero)
Since log 2 = 0.3010, a-b is at most 3 and so 9 cannot divide 2a-b-1
Hence this fantasy though enticing cannot be made concrete
341
Hari Shankar
·2010-08-07 07:04:17
Awaiting some comment on this.
62awesome thought process.. to an awesome question [1]
11Phenominal soln, sir, and surely that's not a surprise from u.
Actually one of my college seniors asked me this qsn, and I gave a diffrent soln, I'm not absolutely sure whether it's correct or not......
Since 2a and 2b (lets call them like that), have the same digits in different orders, we have 2a≡2b(mod 9)......
Now let a=6k+c, and b=6l+d.... for the congruence condition to be satisfied, it's obvious c=d is the only soln.....with 0≤c,d≤5.
Further we use the fact that 2a & 2b have the same number of digits, which gives (a-b)≤4......which in turn gives 6(k-l)≤4, which contradicts the fact that such powers exist.....
341
Hari Shankar
·2010-08-07 10:34:48
Its the essentially the same solution. I rewrote the congruence relation as 2a-b≡1 and first itself noted the condition on a-b (not very accurately though!).
11
Devil
·2010-08-07 23:18:18
Oh sir,It's gr8.....! [1]
21soln:
let A be a power of two
numbers that are power of two and permutation of A can be 2A , 4A , 8A
but A , 3A and 7A are not divisible by 9
so No such number exists ;)
