show that the cube of any positive integer is of the form 4m, 4m+1, or 4m+3 for some integer m.
Answers
Answer:
Let a be any +ve integer
a=bq+
b=4
Step-by-step explanation:
Solution : put r=0,1,2,3
Case 1 : put r=0
a=4q+o.
a=4q
Cubing both side
a3=(4q) 3
a3=64q3
a3=4(16q3)
Put 16q3=m
a3=4m
Case 2 : put r=1
a=4q+r
a=4q+1
Cubing both side
a3=(4q+1)3
a3=64q3+1+8q
a3=(64q3+8q)+1
a3=4(16q3+2q)+1
Put 16q3+2q=m
a3=4m+1
Case 3 :
Put r=2
a=4q+2
Cubing both side
a3=(4q+2)3
a3=64q3+8+16q
a3=64q3+16q+8
a3=4(16q3+4q+2)+2
Put 16q3+4q+2=m
a3=4m+2
Case 4 :
Put r=3
a=4q+3
Cubing both side
a3=(4q+3)3
a3=64q3+27+24q
a3=64q3+24q+27
a3= 4(16q3+6q+3)+3
Put 16q3+6q+3)+3=m
a3=4m+3
Step-by-step explanation:
Step-by-step explanation:
Let 'a' be any positive integer and b = 4.
Using Euclid Division Lemma,
a = bq + r [ 0 ≤ r < b ]
⇒ a = 3q + r [ 0 ≤ r < 4 ]
Now, possible value of r :
r = 0, r = 1, r = 2, r = 3
CASE 1:
If we take, r = 0
⇒ a = 4q + 0
⇒ a = 4q
On cubing both sides,
⇒ a³ = (4q)³
⇒ a³ = 4 (16q³)
⇒ a³ = 9m [16q³ = m as integer]
CASE 2:
If we take, r = 1
⇒ a = 4q + 1
On cubing both sides ;
⇒ a³ = (4q + 1)³
⇒ a³ = 64q³ + 1³ + 3 * 4q * 1 ( 4q + 1 )
⇒ a³ = 64q³ + 1 + 48q² + 12q
⇒ a³ = 4 ( 16q³ + 12q² + 3q ) + 1
⇒ a³ = 4m + 1 [ Take m as some integer ]
CASE 3:
If we take r = 2,
⇒ a = 4q + 2
On cubing both sides ;
⇒ a³ = (4q + 2)³
⇒ a³ = 64q³ + 2³ + 3 * 4q * 2 ( 4q + 2 )
⇒ a³ = 64q³ + 8 + 96q² + 48q
⇒ a³ = 4 ( 16q³ + 2 + 24q² + 12q )
⇒ a³ = 4m [Take m as some integer]
CASE 4 :
If we take, r = 3
⇒ a = 4q + 3
On cubing both the sides;
⇒ a³ = (4q + 3)³
⇒ a³ = 64q³ + 27 + 3 * 4q * 3 (4q + 3)
⇒ a³ = 64q³ + 24 + 3 + 144q² + 108q
⇒ a³ = 4 (16q³ + 36q² + 27q + 6) + 3
⇒ a³ = 4m + 3 [Take m as some integer]
Hence, the cube of any positive integer is in the form of 4m, 4m+1 or 4m+3.
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Identity used ;
∵ ( a + b )³ = a³ + b³ + 3ab ( a + b ) .
Hence, it is solved.