document.write( "Question 565153: what are four consecutive odd integers who's sum is 192 \n" ); document.write( "

Algebra.Com's Answer #365694 by ad_alta(240) $\"\"$ $\"About$

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Let 'n' be the smallest integer. Then n+(n+2)+(n+4)+(n+6)=192, so 4n+12=192. Therefore n=45. The integers are 45, 47, 49, 51. \n" ); document.write( "

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