How do you find two consecutive odd integers whose sum is 196?

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Answer 1 · 2016-04-25T12:40:06

$97 and 99$

There is at least two ways of doing this: This is my approach

$color(blue)("Step 1")$ Define a number so that it is always even

$" "$ Let $n$ be any number then $2n$ is always even.

$color(blue)("Step 2")$ Modify the defined number so that the result is always odd

$" "$ If $2n$ is always even then $2n+1$ is always odd
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Let the first odd number be $2n+1$
Then the second odd number is $(2n+1)+2=2n+3$

Thus our given condition is such that:

$(2n+1)+(2n+3)=196$

$=>4n+4=196$

Subtract 4 from both sides

$=>4n=192$

Divide both sides by 4

$=>n=192/4=48$

But the first number is $2n+1 = 2(48)+1 = 97$

So the second number is $97+2=99$
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Check: $97+99= 196$

Answer 2 · 2016-04-25T12:47:34

Another way

Let the first number be $n$
Let the second number be $n+2$

Then $n+n+2=196$

$2n+2=196$

Subtract 2 from both sides

$2n=194$

Divide both sides by 2

$n=194/2= 97$

The first number is 97 so the second is 97+2=99

$color(red)("Notice that the " 2n + 2" is of the same format as that in my")$ $color(red)("other solution of "4n+4)$