Question

The product of three consecutive integers is 30 times the smallest of the three

integers. Find the two possible sets of integers that will satisfy this condition.

Answer 1

Hi,

Explanation:

If the "is" means is equal to

Let's say a-1, a, a+1 the 3 consecutive integers.

`(a-1)*a*(a+1)=30*(a-1)\\\\(a-1)*(a*(a+1)-30)=0\\\\(a-1)(a^2+a-30)=0\\\\(a-1)(a^2+6a-5a-30)=0\\\\(a-1)(a(a+6)-5(a+6))=0\\\\(a-1)(a+6)(a-5)=0\\\\a=1\ or\ a=-6\ or\ a=5\\`

`\begin{array}ca&a-1&a+1&Prod&30*(a-1)\\---&----&----&----&--------\\1&0&2&0&30*0=0\\5&4&6&120&4*30=120\\-6&-7&-5&-210&30*(-6)=-180\end{array}\\`

The two sets are : (0,1,2) and (4,5,6).

Answer 2

Let's assume , three Consecutive integers be x , ( x +1 ) & ( x+2 ) , such that , x < ( x +1 ) < ( x +2 ) .

⠀⠀⠀⠀⠀⠀
`\underline {\boldsymbol{\star\: According \: to \: the \: Given \: Question \::}}\\`

⠀⠀⠀⠀⠀— The product of three consecutive integers is 30 times the smallest of the three integers.

`\sf \dashrightarrow \:30\: \bigg\{ \: Smallest_((\:integer\:))\:\bigg\}\:\:=\: \:\bigg\{ \: I^(st) \:Integer\:\bigg\}\: \:\bigg\{\: II^(nd)\:Integer\:\bigg\}\: \:\bigg\{ \: III^(rd) \:Integer\:\bigg\}\:\:\: \\\\\sf \dashrightarrow \:30\: \bigg\{ \:x\:\bigg\}\:\:=\: \:\bigg\{ \: x\:\bigg\}\: \:\bigg\{ \: x + 1\:\bigg\}\: \:\bigg\{ \: x +2 \:\bigg\}\:\:\: \\\\ \sf \dashrightarrow \:30\: \:\:=\: \:\bigg\{ \: x + 1\:\bigg\}\: \:\bigg\{ \: x +2 \:\bigg\}\:\:\: \\\\ \sf \dashrightarrow \:30\: \:\:=\: \:x^2 + 3x + 2 \:\:\: \\\\ \sf \dashrightarrow \:x^2 + 3x - 28 \:= \:0\: \\\\ \sf \dashrightarrow \:\:\bigg\{ \: x - 2 \:\bigg\}\:\:\bigg\{ \: x + 7 \:\bigg\}\: \:= \:0\: \\\\ \pmb {\underline {\boxed {\purple {\:\frak{ \:x\:\:=\:-7\: \&\:2\:}}}}}\:\bigstar \: \\\\\\`

Therefore ,

When , x = 2 ,

• First Integer : x , 2 ,
• Second Integer : ( x +1 ) , 3 &
• Third Integer : ( x +2 ) = 4 .

When , x = –7 ,

• First Integer : x , – 7 ,
• Second Integer : ( x +1 ) , – 6 &
• Third Integer : ( x +2 ) = – 5 .