Inequalities #2 Find a four-digit number such that it equals to the forth power of the sum of all of its integers
2401.
Method I used: look at fourth powers with four digits, there are only a few.
Yup... there are only a few... but how to do it (not by trial and error) ?
\[ 1000w+100x+10y+z=n^4\\ w+x+y+z=n\\ n\in\mathbb{N} \] Maybe. Dunno.
1000a+100b+10c+d = (a+b+c+d)^4 5< a+b+c+d < 10
1634 8208 9474
the forth power of the sum of all of its integers:( i am sorry! i find the the sum of the forth power of all of its integers
Never mind. :) So... it seems easier now.. but still by trial and error :(
Hmm... I wish I could take back my words... Actually, solving it by trial and error is not the problem. But I wish to find a better method only. Thanks all :)
9999^1/4<10 1000^1/4>5 So, a+b+c+d=(6,7,8,9) 6^4=1296 7^4=2401 8^4=4096 9^4=6561 a=2 b=4 c=0 d=1
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