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Question

By taking three different values of n verify the truth of the following statements:
(i) If n is even , then n3 is also even.
(ii) If n is odd, then n3 is also odd.
(iii) If n leaves remainder 1 when divided by 3, then n3 also leaves 1 as remainder when divided by 3.
(iv) If a natural number n is of the form 3p + 2 then n3 also a number of the same type.

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Solution

(i)
Let the three even natural numbers be 2, 4 and 8.
Cubes of these numbers are:
23=8, 43=64, 83=512
By divisibility test, it is evident that 8, 64 and 512 are divisible by 2.
Thus, they are even.
This verifies the statement.

(ii)
Let the three odd natural numbers be 3, 9 and 27.
Cubes of these numbers are:
33=27, 93=729, 273=19683
By divisibility test, it is evident that 27, 729 and 19683 are divisible by 3.
Thus, they are odd.
This verifies the statement.

(iii)
Three natural numbers of the form (3n + 1) can be written by choosing n=1,2,3... etc.
Let three such numbers be 4,7 and 10.
Cubes of the three chosen numbers are:
43=64, 73=343 and 103=1000

Cubes of 4,7 and 10 can expressed as:
64=3×21+1, which is of the form (3n + 1) for n = 21
343=3×114+1, which is of the form (3n + 1) for n = 114
1000=3×333+1, which is of the form (3n + 1) for n = 333

Cubes of 4,7, and 10 can be expressed as the natural numbers of the form (3n + 1) for some natural number n. Hence, the statement is verified.

(iv)
Three natural numbers of the form (3p + 2) can be written by choosing p=1,2,3... etc.
Let three such numbers be 5,8 and 11.
Cubes of the three chosen numbers are:
53=125, 83=512 and 113=1331

Cubes of 5,8, and 11 can be expressed as:
125=3×41+2, which is of the form (3p + 2) for p = 41
512=3×170+2, which is of the form (3p + 2) for p = 170
1331=3×443+2, which is of the form (3p + 2) for p = 443

Cubes of 5,8, and 11 could be expressed as the natural numbers of the form (3p + 2) for some natural number p. Hence, the statement is verified.

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