Find three consecutive integers whose product is 161 larger than the cube of the smallest integer

Question

 Find three consecutive integers whose product is 161 larger than the cube of the smallest integer 

Answer 1

Here's what we know:

abc = a³ + 161

b = a + 1

c = a + 2

 

a(a + 1)(a + 2) = a³ + 161

 

From here, we distribute and then FOIL:

(a² + a)(a + 2) = a³ + 161

a³ + 2a² + a² + 2a = a³ + 161

 

Next we combine like terms:

a³ + 3a² + 2a = a³ + 161

3a² + 2a = 161

3a² + 2a - 161 = 0

 

We now have a quadratic equation. There're a few different ways you can solve for this. In this case, I'll be using the quadratic formula:

x = (-b ± √(b² - 4ac)) / 2a

x = (-2 ± √(2² - 4 * 3 * 161)) / 2*3

 

simplifying:

x = (-2 ± √(4 + 1932)) / 6

x = (-2 ± √1936) / 6

 

Now we break up the square root values into noticeable squares:

x = (-2 ± √(4 * 484)) / 6

x = (-2 ± √(4 * 4 * 121)) / 6

x = (-2 ± √(4 * 4 * 11 * 11)) / 6

 

We have two 4s and two 11s, so that means we have two perfect squares:

x = (-2 ± 4 * 11) / 6

x = (-2 ± 44) / 6

x = -46/6, 42/6

 

We were told that we have integers, and -46 doesn't divide evenly by 6, so it has to be the other possibility:

x = 42/6

x = 7

 

Now to check:

7 * 8 * 9 = 7³ + 161

504 = 343 + 161

504 = 504

 

So our three consecutive integers are 7, 8, and 9.

Answer 2

Let call the smallest integer a.

 

So we have:  a*(a+1)*(a+2) - a^3 = 161 

=> a^3 + 3*a^2 + 2*a - a^3 = 161

=> 3*a^2 + 2*a = 161

=> a(3*a+2) = 161

 

161 = 23*7, so try 3*a + 2 = 23, we have a = 7, which works

 

so the three integer are 7,8,9

 

Plug it in if you want to check the answer