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find three consecutive intergers whose sum is 480
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x+x+1+x+2=480 3x+3=480 3x=477 x=159 159,160,161r the no.s
Call one k. The other would be k+1, then the other would be k+2 Sum these: \[k+k+1+k+2=3k+3=480\] Subtract 3. 3k=477 k=159 That means you have 159, 160, 161
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