Question

the product of four consecutive integers is 1920 find the number​

Answer 1

Answer:

Consecutive even numbers will be :

n, n+2, n+4, n+6

Brute force method :

we know that n * (n+2) * (n+6) = 1920

n^3 + 8n^2 + 12n = 1920

lets guess that n is approximately \sqrt^[3]{1920} = 12.43\sqrt^[3]{1920} = 12.43

We know 12 is going to be too big (since our estimate ignore 8*n^2 and 12n terms): in fact is n = 12:

n^3 + 8n^2 + 12n = 3024

So lets try lower

n = 8 :

n^3 + 8n^2 + 12n = 1120

Lets try n = 10

n^3 + 8n^2 + 12n = 1920 - Bingo

Alternatively solving using prime factors

1920 = 2^7 * 3 * 5

that means one of the even numbers is divisible by 3, and another by 5

and then you have 7 2s to split between 4 numbers - we probably have to be looking at a split where one of the values (likely the smallest has only one factor of 2

But we can’t start with 2*3 = 6, as that makes the 4th value is 12 (the sequence would be 6, 8, 10, 12) and we only have one power of 3.

If the sequences starts with 8, the 4th value is 14, but 1920 doesn’t have 7 as a prime factor

what if the sequence starts with 10:

10=2∗5,12=22∗3,14=2∗7,16=2410=2∗5,12=22∗3,14=2∗7,16=24

Note that here the product of the 1st 2nd and 4th values is 2∗5∗22∗3∗24=27∗5∗3.