document.write( "Question 337068: If w,x,y and z are non negative integers, each less than 3, and w(33) + x(32) + y(3) + z = 34, then w+z=\r
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\n" ); document.write( "\n" ); document.write( "The answer must be 2, but i do not understand the reason why it should be 2. Can you please clarify.\r
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Algebra.Com's Answer #241575 by Jk22(389)\"\" \"About 
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If w,x,y and z are non negative integers, each less than 3, and w(33) + x(32) + y(3) + z = 34, then w+z= ?\r
\n" ); document.write( "\n" ); document.write( "we can rewrite as : 32(w+x)+3y+(w+z)=34, 0<=w,x,y,z<3\r
\n" ); document.write( "\n" ); document.write( "we need to have w+x=1. Indeed w+x>1 impossible since all are positive, and 32(w+x)>34\r
\n" ); document.write( "\n" ); document.write( "w+x=0 neither, since all are smaller than 3, but 34/5>3\r
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\n" ); document.write( "\n" ); document.write( "Hence we get : 32+3y+(w+z)=34\r
\n" ); document.write( "\n" ); document.write( "the only possibility is y=0, else we get 32+3y>34 which is impossible, since w+z>0\r
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\n" ); document.write( "\n" ); document.write( "we finally get : 32+w+z=34 => w+z=2 \n" ); document.write( "
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