Find three consecutive integers whose sum is 201?

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Answer 1 · 2018-02-27T23:35:13

#66, 67, 68#

Let's start with one integer, and call it #x.#

We now want two integers each consecutive to #x#, or each #1# above x. These will each be #1# above #x#, because integers do not contain decimals.

So, for our first integer consecutive to #x#, we have #x+1#, as this is #1# above #x#.

For our second integer consecutive to #x,# we have #x+2,# as #x+2# is consecutive to #x+1#.

Let's set the sum our integers equal to #201# and solve for #x:#

#(x)+(x+1)+(x+2)=201#

Combine like terms. This means combining all regular integers and combining all #x.#

#3x+3=201#

Solve for #x:#

#3xcancel(+3-3)=201-3#

#3x=198#

#cancel3x/cancel3=198/3#

#x=66#

So, our first integer, #x#, is #66#. The integers consecutive to this are just #x+1# and #x+2,# or #66+1=67# and #66+2=68#.