document.write( "Question 479514: if a,b,c and d are in set of natural numbers(N) with a=bcd, b=cda and d=abc then,what is the value of (a+b+c+d)^2 ?
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Algebra.Com's Answer #328557 by ikleyn(52780) $\"\"$ $\"About$

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\n" ); document.write( "\n" ); document.write( "Since $\"a=+bcd\"$ and $\"b+=+cda\"$ , we can substitute b into the first equation and obtain $\"a+=+%28c%2Ad%2Aa%29%2Ac%2Ad\"$ , which can be simplified into $\"a+=+c%5E2%2Ad%5E2%2Aa\"$ . Since a,b,c,d are natural numbers (not equal to zero), we can divide both sides of the last equation by a and obtain $\"1+=+c%5E2%2Ad%5E2\"$ . Again, using the fact that a,b,c,d are natural numbers we conclude that both c and d are equal to 1. \r
\n" ); document.write( "\n" ); document.write( "Now we have to find values of a and b. We can use the last equation $\"d+=+abc\"$ and substitute c = 1 and d = 1 to obtain $\"1+=+ab\"$ . Again, a and b are natural numbers, and thus both equal to 1. So, the only solution to the system of equations for a,b,c,d among the set of natural numbers is all of the being equal to 1. \r
\n" ); document.write( "\n" ); document.write( "Thus $\"%28a+%2B+b+%2B+c+%2B+d%29%5E2\"$ is equal to 16. \n" ); document.write( "

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