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| 1. |
Find three consecutive positive integer which sum is 16 times it\'s product |
| Answer» Let\xa0three consecutive positive integers be\xa0x, x + 1 and x\xa0+ 2As per given condition product of\xa0three consecutive positive integers is equal to sixteen times their sum.{tex}\\therefore{/tex} (x)(x +1)(x\xa0+ 2) = 16(x\xa0+ x + 1 + x + 2){tex}\\Rightarrow{/tex}(x2 + x)(x\xa0+ 2) = 16(3x\xa0+ 3){tex}\\Rightarrow{/tex}x3 + 2x2 + x2 + 2x = 48x\xa0+ 48{tex}\\Rightarrow{/tex}\xa0x3 + 3x2 - 46x\xa0- 48 = 0When x = -1, we haveLHS = (-1)3 + 3(- l)2 - 46 (-1) -48= -1 + 3 + 46-48 = 0{tex}\\therefore{/tex}(x + 1) is a factorFor other factors\xa0{tex}\\therefore{/tex}x3 + 3x2 - 46x\xa0- 48 = 0{tex}\\Rightarrow{/tex}(x + 1)(x2 + 2x\xa0- 48) = 0{tex}\\Rightarrow{/tex}\xa0(x + 1) (x + 8)(x - 6) = 0{tex}\\Rightarrow{/tex}\xa0x\xa0= -1, x\xa0= -8, x\xa0= 6.Rejecting -ve values,therefore, x = 6{tex}\\therefore{/tex}Positive integers are 6, 7 and 8. | |