Sir/Mam in Real Numbers chapter in the question" Use Euclid’s divisi

1.	Sir/Mam in Real Numbers chapter in the question" Use Euclid’s division lemma to show that the cube of any positive integer is of the form9m, 9m + 1 or 9m +8" why we take 3 as the value.
Answer» ANSWER: ■●■In proving this QUESTION we take 3 as[ (b=3) in bq+r ] the value because: ◇taking b=9 will be too long to solve because we will have to solve all 9 cases where r=0,1,2,3,4,5,6,7,8. ◇ CUBE of expressions of the FORM 3q,3q+1,3q+2 are easily divided by 9 leaving simple remainders. For example: ♤(3q)^3=27q^3= 9(3q^3) = 9m ♤(3q+1)^3 = 27q^3 +1 +27q^2+ 9q =9(3q^3 +3q^2 +q) +1 =9m+1 ♤ (3q+2)^3=27q^3 +8 +54 q^2 +36q =9(3q^3 + 6q^2+4q) +8 = 9m+8. Ask for further EXPLANATION; if any.

Sir/Mam in Real Numbers chapter in the question" Use Euclid’s division lemma to show that the cube of any positive integer is of the form9m, 9m + 1 or 9m +8" why we take 3 as the value.

Answer»

ANSWER:

■●■In proving this QUESTION we take 3 as[ (b=3) in bq+r ] the value because:

◇taking b=9 will be too long to solve because we will have to solve all 9 cases where r=0,1,2,3,4,5,6,7,8.

◇ CUBE of expressions of the FORM

3q,3q+1,3q+2 are easily divided by 9 leaving simple remainders.

For example:

♤(3q)^3=27q^3= 9(3q^3) = 9m

♤(3q+1)^3 = 27q^3 +1 +27q^2+ 9q

=9(3q^3 +3q^2 +q) +1

=9m+1

♤ (3q+2)^3=27q^3 +8 +54 q^2 +36q

=9(3q^3 + 6q^2+4q) +8

= 9m+8.

Ask for further EXPLANATION; if any.

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Sir/Mam in Real Numbers chapter in the question" Use Euclid’s division lemma to show that the cube of any positive integer is of the form9m, 9m + 1 or 9m +8" why we take 3 as the value.​

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Sir/Mam in Real Numbers chapter in the question" Use Euclid’s division lemma to show that the cube of any positive integer is of the form9m, 9m + 1 or 9m +8" why we take 3 as the value.