Question

Find three consecutive odd integers with the sum of 51.

Answer 1

Well.. hmmmm

bear in mind that, if you multiply any integer, odd or even, by 2, you get an even, no matter how you slice it and dice it, you always get an even number, 17 *2, 34*2, or whatever * 2, is always an EVEN integer.

you can always get an ODD integer, by simply adding or subtracting 1 from any even,  say  4 - 1, or 4 + 1, is 3 and 5 respectively, both odd ones.

and you can always get another consecutive odd integer, by simply jumping 2 units from any odd, so 3, 5, 7, 9, 11 and so on, notice, you simply hop two spots from one, on either direction back or forth, and you get another ODD integer.

alrite, with that in mind, let's pick a reference value, say "a" is our reference integer.

so, 2*a or 2a, we know is an EVEN integer, however, we also  know that 2a + 1 is an ODD integer, well, there you have it, our first odd integer is 2a + 1.

now, let's hop twice from there to get our second consecutive odd one, (2a + 1) + 2, or 2a + 3, now, let's hop again to get the next consecutive odd one, (2a + 3) + 2, or 2a + 5.

there you have it, our reference integer is "a", our odd integers are those.

alrite... now, we know their sum gives 51, alrite

$\bf (2a + 1) + (2a + 3) + ( 2a + 5) = 51 \\\\\\ 2a + 1 + 2a + 3 + 2a + 5 = 51 \\\\\\ 2a + 2a + 2a +1 + 3 ++ 5 = 51\implies 6a+9=51\implies 6a=51-9 \\\\\\ 6a=42\implies a=\cfrac{42}{6}\implies \boxed{a=7}$

now, recall, "a" is only our reference integer, so... if you want to get the consecutive odd integers, well, they're just 2a + 1, 2a + 3 and 2a + 5, and since you already know what "a" is, well, just plug it in to get the integers.