If a b c d are consecutive integers then a + b2+ c3 is divisible by d?

Question

Yes, i know i asked this already, but i need a more clear understanding of how you get the answer?


-questions i have-

Is there an actual answer? Like since its consecutive numbers ex. 1,2,3,4...or is it more of a why answer?

Help! Thankes!

Answer 1

Since a, b, c, d are consecutive integers,

a = a
b = a+1
c = a+2
d = a+3

a+b^2+c^3 =
(a)+(a+1)^2+(a+2)^3 =
(a)+(a^2+2a+1)+(a^3+6a^2+12a+8) =
a^3+7a^2+15a+9

Is this divisible by (a+3)? yes.

(a^3+7a^2+15a+9) / (a+3) = (a^2+4a+3)

Q E D !

Answer 2

a + b²+ c³ is divisible by d

Since they are consecutive then
b = a + 1
c = a + 2
d = a + 3

(a + (a + 1)² + (a + 2)³)/(a + 3)

(a + a² + 2a + 1 + a³ + 6a² + 12a + 8)/(a + 3)

(a³ + 7a² + 15a + 9)/(a + 3)

= a² + 4a + 3 with 0 remainder

So this is a proof that
If a, b, c, & d are consecutive integers then a + b²+ c³ is divisible by d

Answer 3

When you say b2, do you mean 2b, as in 2 times b?
Look at n, n+1, n+2, n+3.
Let's see if n+3 divides n + 2n + 2 + 3n + 6 = N.
N = 6n + 8.
(6n + 8)/(n + 3) = ???
Do some long division, and see what you get. If you can find some integer for n, you'll have your answer.

Answer 4

if there isn't any obtrusive answer, such an equation is clearly impossible. a good thought is first to seem what occurs with low value of a. you need to use a calculator. you will see there isn't any answer. yet another thought is to apply congruence. employing a calculator, you will see. It would not artwork too. in actuality, this 2d thought is only too complicate. the belief is to evaluate the selection that looks. you have : b=a+one million, c=a+2, d=a+3. b^5=(a+one million)^5 is going speedier than d^3=(a+3)^3 to +oo in +oo and to -oo in -oo. if a is gigantic, b is only too super to make available the equality. if a is incredibly adverse, b is only too adverse to make available the equality. So the belief is only too locate 2 numbers n>0 and m<0 such that : if a>n then a^3+b^5+c^3>d^3 if a

Answer 5

Yes it's true, and Ian worked the proof out for you in answer to your previous question. Read thru his answer one line at a time and think about each piece. It's a "why" answer that works for all consecutive numbers, not just a set you happen to choose.

Answer 6

To make things more confusing, how about a
proof without algebra? The smaller consecutive
numbers a,b,c leave remainders of -3,-2,-1
when divided by d. Plug these into a+b^2+c^3
and you get (-3)+(-2)^2+(-1)^3 = 0 which is a
remainder of zero, and you're done.

Answer 7

dno