ok so if we say that a is a natural number (N)

this

(a+30)(a+50)(a+70)

what can it be divided by?

2, 3, 5, 7? which one(help plz, btw soz maybe its bad math lang)

 

 

 

You could "suck it and see". By which I mean let a=1, then 2, then 3 etc. and see what the numbers can be divided by.

ok so if we say that a is a natural number (N)

this

(a+30)(a+50)(a+70)

what can it be divided by?

2, 3, 5, 7? which one(help plz, btw soz maybe its bad math lang)

 

Off the top of my head there is no absolute rule for all choices of a. If a is divisible by 2 then the expression will also be; if a is not divisible by two then neither will the expression be.

 

if you multiply out you can begin to see why

 

[math] (a+30)(a+50)(a+70) = a^3 + 150a^2 + 2060a +105000 [/math]

 

 

Scratch the above it is not a good answer

 

This is a nice question

 

You need to consider the modulo values of

 

[math] (a+30)(a+50)(a+70) = a^3 + 150a^2 + 2060a +105000 [/math]

 

and importantly how modulo of a compares with modulo of a^3 and a^2 and the prime factors of the coefficients

 

105000 is always divisible by 2,3,5,7

150a^2 is always divisible by 2,3,5

2060a is always divisible by 103,5,2

If you put B=a+50

It can be simplified to

[math] (B-20) B (B+20)[/math]

But I don't see any light to answering the OP question.

 

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2,3,5,7 are prime numbers.

So what...

Edited November 22, 201114 yr by michel123456

[math] (a+30)(a+50)(a+70) = a^3 + 150a^2 + 2060a +105000 [/math]

 

 

Expanding the brackets is an overcomplication

 

 

for [imath](a+30)(a+50)(a+70)[/imath] to be divisible by x then any one or more of the brackets must be divisible by x ie (bracket)_mod_x = 0

 

basis 2

(a+30)_mod_2 = (a)_mod_2 as 30_mod_2 = 0 similarly with (a+50) and (a+70) - so basis modulo 2 the brackets go to (a)(a)(a). and thus expression will only divide by 2 if and only if a is divisible by two

 

basis 5

the exact same apply as 30, 50, 70 are all divisible by 5 - so basis modulo 5 the bracket are again (a)(a)(a). ie expression will divide evenly iff a does

 

basis 7

(a+30)_mod_7 = (a+2) (a+50)_mod_7 = (a+1) (a+70)_mod_7 = (a) - so expression will divide evenly by 7 iff a or (a+1) or (a+2) is divisible by 7.

 

basis 3

this is the good one though (a+30)_mod_3 = (a) (a+50)_mod_3 = (a+2) (a+70)_mod_3 = (a+1).

 

it seems that we have same situation as 7 but it is soon obvious that all numbers are either divisible by three or 1 less or 2 less - so whihc ever a is chosen one of the brackets will be divisible by 3 and thus the product will be divisible by 3

 

 

Edited November 22, 201114 yr by imatfaal

ok so if we say that a is a natural number (N)

this

(a+30)(a+50)(a+70)

what can it be divided by?

2, 3, 5, 7? which one(help plz, btw soz maybe its bad math lang)

 

 

 

 

It can be divided by a+30, a+50, a+70, any prime number that divides one of those and products of such primes.