Anonymous
Anonymous asked in Science & MathematicsMathematics · 1 month ago

The sum of three integers is 208. The two smaller are consecutive integers and the two larger are consecutive odd integers. Find the smallest number.

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• Sounds like LIBERAL JARGON #GodBless

• Let two smaller consecutive intgers numbers=x,x+1

Let larger odd consecutive intergers numbers=x+1,x+3

According question x+x+1+x+3=208

3x=208-4

3x=204

x=204÷3

x=68

x+1=69

x+3=71

So these three numbers=68,69,71

• Hi

Let, the two larger consecutive odd integers be 2n + 1, 2n + 3

The smallest number is then 2n

6n + 4 = 208

The smallest number is 204/3 = 68

68 + 69 + 71 = 208

• a+b+c = 208

a is even, b = a+1 is odd and a+b is odd

c = b+2

c = a+3

3a+4 = 208

a = 204/3 = 68

b = 69

c = 71

• Let the two larger consecutive odd integers be 2n + 1, 2n + 3

The smallest number is then 2n

6n + 4 = 208

The smallest number is 204/3 = 68

68 + 69 + 71 = 208

• Let the Integers by

n , n+1 & n+3

'n' & 'n+1' are consecutive

'n+1' & 'n + 3' are consecutive odd .

Hence their sum is

n + ( n+ 1 + (n+ 3) = 208

3n + 4 = 208

3n = 204

n = 68  The smallest number.

The other being '69' & '71'

• Whats up anonymous you big red ****, lets get mathy, as I always say at the start of my math lessons. As you know, of course, an integer is the sum of the fraction between two sides of a rhomboid. So, applying this basic principle to this fraction, we can see that x = rhomboid, y = 208, ☺ = 34, and the amount of consectuvie integers = 4. So, it's simple really, x Xx X x = y xX Xxx + (☺☺ looktheyre like a little couple hehe)* 56

what you really wanna be looking out for here is that answer doesn't protrude more than 5/8ths of an inch off the page, otherwise that would be disgusting. With the equation laid out above as I did for you, you should get it in no time. Id tell you the answer, but I don't wanna do all your homework for you! hehe

Source(s): math teacher for 29 years, father for 18, gamer for 2, meth & math addicted leftist
• I would ignore that the second two integers are odd, except that it means they differ by two.  I can check for oddness, later.

If the smallest integer is a, then the next integer is consecutive, so it's a+1.  And the next integer after that is a+3.

a + a+1 + a+3 = 208

3a + 4 = 208

3a = 204

a = 68

quick check, yes, the second and third integers are odd

---

Alternate thinking, not faster in this case

The numbers are all almost the same, so divide 208 by 3 as a guess for the middle one.  210 is the closest thing I can easily divide by 3, which would give 70 as the middle one.  But 70 is even, 210 is too much, so I'll take it down a notch and guess 69 for the middle one.  (If 210 was too little, I'd try 71 as the middle one.).  Check the guess.

• The sum of three integers is 208.

The two smaller are consecutive integers and

the two larger are consecutive odd integers.

Find the smallest number.

x + (x + 1) + (x + 3) = 208

The smallest number is 68

• The sum of three integers is 208:

x + y + z = 208

The two smaller are consecutive integers:

y = x + 1

And the two larger are consecutive odd integers:

z = y + 2  (if y is odd, then the next odd is (y + 2))

We now have a system of three equaitons and three unknowns.  The way I set up the equations:

x < y < z

The smallest number is then "x".

Let's substitute the last equation in terms of y for z in the first equation, then simplify:

x + y + z = 208

x + y + y + 2 = 208

x + 2y + 2 = 208

x + 2y = 206

Now we can substitute the expression from the second equation into the above and solve for x:

x + 2(x + 1) = 206

x + 2x + 2 = 206

3x + 2 = 206

3x = 204

x = 68

The smallest number is 68.

If you want to solve for y and z, substistute into the other equations and solve and you should get:

y = 69 and z = 71