# how to find the natural number n for which (15n²+8n+6)/n is a natural number.

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(15n²+8n+6)/n = 15(n^2)/n+8n/n+6/n

= 15n+8+6/n

Since n is a natural number 15n will be a natural number.

Then definitely 15n+8 will be natural.

Therefore to (15n²+8n+6)/n be natural, 6/n should be natural.

In other words 6 can be divided by n.

Also n<6

Now 6 can be divided fully by 1,2,3 and 6.

**So the answer is n=1,n=2,n=3 and n=6**

Since the number `(15n^2 + 8n + 6)/n` is a natural number, hence, the numerator 15n^2 + 8n + 6 is divisible by n, hence the reminder of division `(15n^2 + 8n + 6)/n` is equal to a natural constant.

You should use the reminder theorem such that:

`15n^2 + 8n + 6 = n(an + b)` + c

You need to open the brackets to the right side:

`15n^2 + 8n + 6 = an^2 + bn` + c

Equating the coefficients of the like powers yields:

`a = 15 and b = 8 , c = 6=gt 15n^2 + 8n + 6 = n(15n + 8) + 6=gt (15n^2 + 8n + 6)/n= 15n + 8 + 6/n`

The problem provides the information that the number 15n + 8 + 6/n is natural and n is also natural, hence, you should give values to n such that:

`n = 1 =gt 15 + 8 = 23`

Substituting 1 for n yields:

`(15 + 8 + 6)/1 = 29 in N`

`n =2 =gt30 + 8 = 38`

`(60 +16 + 6)/2 = 82/2 = 41 in N`

`n =3 =gt (15*9 +24 + 6)/3 = (135 + 30)/3 = 165/3 = 55 in N`

n = 4 => (15*16 + 32 + 6)/4 = 69.5 !in N

`n = 5 =gt (15*25 +40 + 6)/5 = 84.2 !in N`

`n = 6 =gt (15*36 +48 + 6)/6 = 99 in N`

**Since n cannot be larger than 6 and it needs to be natural, hence `n in {1,2,3,6}.` **