show that cube of any positive integer is of the form 6q +r for some integer q (r = 1,2,3,4,5)
please show for r=2,3,4
thankyou
please be bit fast !
Answers
6q + r is a positive integer, where q is an integer and r = 0, 1, 2, 3, 4, 5
Then, the positive integers are of the form 6q, 6q+1, 6q+2, 6q+3, 6q+4 and 6q+5.
Taking cube of each term, we have,
(6q)3 = 216 q3 = 6(36q)3 + 0
= 6m + 0, where m is an integer
(6q+1)3 = 216q3 + 108q2 + 18q + 1
= 6(36q3 + 18q2 + 3q) + 1
= 6m + 1, where m is an integer
(6q+2)3 = 216q3 + 216q2 + 72q + 8
= 6(36q3 + 36q2 + 12q + 1) +2
= 6m + 2, where m is an integer
(6q+3)3 = 216q3 + 324q2 + 162q + 27
= 6(36q3 + 54q2 + 27q + 4) + 3
= 6m + 3, where m is an integer
(6q+4)3 = 216q3 + 432q2 + 288q + 64
= 6(36q3 + 72q2 + 48q + 10) + 4
= 6m + 4, where m is an integer
(6q+5)3 = 216q3 + 540q2 + 450q + 125
= 6(36q3 + 90q2 + 75q + 20) + 5
= 6m + 5, where m is an integer
Hence, the cube of a positive integer of the form 6q + r, q is an integer and r = 0, 1, 2, 3, 4, 5 is also of the form 6m + r.
Hope it helps.
Answer:
Euclid's division algorithm,
a=6q+r
where a q and r are non-negative integers 0≤r<6 i.e., r=0,1,2,3,4,5.
Cubing (i) both sides , we get
(a)
3
=(6q+r)
3
⇒a
3
=(6q)
3
+(r)
3
+3(6q)
2
(r)+3(6q)(r)
2
⇒a
3
=6
3
q
3
+r
3
+3(6)
2
q
2
r+6×3qr
2
⇒a
3
=6[36q
3
+18q
2
r+3qr
2
]+r
3
....(ii)
When r=0
a
3
=6[36q
3
+18q
2
×0+3q0
2
]+0
3
[From (ii)]
⇒a
3
=6[36q
3
]
⇒a
3
=6m is perfect cube for some integer m such that m=36q
2
When r=1
a
3
=6[36q
3
+18q
2
×1+3q1
2
]+1
3
⇒a
3
=6[36q
3
+18q
2
×1+3q]+1
⇒a
3
=6m+1 is perfect cube for some integer m such that m=(36q
2
+18q
2
+3q)
When r=2
a
3
=6[36q
3
+18q
2
×2+3q×2
2
]+2
3
[From(ii)]
⇒a
3
=6[36q
3
+36q
2
+12q]+6+2
⇒a
3
=6[36q
3
+36q
2
+12q+1]+2
⇒a
3
=6m+2 is perfect cube for some integer m such that m=36q
3
+36q
2
+12q+1
When r=3
a
3
=6[36q
3
+18q
2
×3+3q×3
2
]+3
3
[From(ii)]
⇒a
3
=6[36q
3
+54q
2
+27q]+24+3
⇒a
3
=6[36q
3
+54q
2
+27q+4]+3
⇒a
3
=6m+3
So , (6m+3) is perfect cube for some integer m such that m=36q
2
+54q
2
+27q+4
When r=4
a
3
=6[36q
3
+18q
2
(4)+3q4
2
]+4
3
[From(ii)]
⇒a
3
=6[36q
3
+72q
2
+48q]+60+4
⇒a
3
=6[36q
3
+72q
2
+48q+10]+4
⇒a
3
=6m+4
So , (6m+4) is perfect cube for some integer m such that m=36q
3
+72q
2
+48q+10
When r=5
a
3
=6[36q
3
+18q
2
(5)+3q(5)
2
]+(5)
3
[From(ii)]
⇒a
3
=6[36q
3
+90q
2
+75q]+120+5
⇒a
3
=6[36q
3
+90q
2
75q+20]+5
⇒a
3
=6m+5
So, ,(6m+5) is perfect cube for some integer of m=36q
3
+90q
2
+75q+20
Hence , cubes of positive integers of the form (6q+r) is also of the form (6m+r) where m is a specified integer and r=0,1,2,3,4,5tep explanation: