dvinoth86 wrote:
Is x > 10^10 ?
(1) x > 2^34
(2) x = 2^35
My working is just a tad different.
Statement 2, enough
Statement 1,
x > 10^10
x > 2^10.5^10
So x > 2^34
x > 2^10.2^24
Now, my thinking was 5 is 2.5 times more than 2, so 5^10 will be closer to 2^20 but w/o working i cant say higher or lower , whereas 2^21 > 5^10 (definite as its 1 more than double), we have 2^24. So sufficient.
I know its not perfect but saved time for me, of course Bunuel's method is flawless !!.
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