We say that an integer a is a type 0 integer if there exists an integer n such that a = 3n. An integer a is a type 1 integer if there exists an integer n such that a = 3n + 1. An integer a is a type 2 integer if there exists an integer n such that a = 3n + 2. Prove that if a is a type 1 integer, then a 2 is a type 1 integer\

Proof below

Step-by-step explanation:

Let's assume a is a type 1 integer. By definition, it means we can find an integer n such that

a=3n+1

We need to prove is a type 1 integer

Expanding

If is a type 1 integer, then we should be able to find an integer m such as

Equating

solving for m

Since we know n is an integer, then the expression of m gives an integer also. Having found the required integer m, the assumption is proven

hello from mrbilldoesmath!

answer:

(7 x + 6 y) (49 x^2 - 42 x y + 36 y^2)

discussion:

the trick is to recognize that 343 = 7^3 and 216 = 6^3 and, in general,

a^3 + b^3 =   (a + b) (a^2 - a b + b^2).

in our case

343x^3+216y^3   =   (7x)^3 + (6y)^3

so take a = 7x and b = 6y in the above equation to get the answer.

you,

mrb

13664

step-by-step explanation:

=cross multiply

=100 x = 13664