1) show that the square of any positive integer cannot be the form 6m+2 or 6m+5 for ant integer m.
2) show that the cube of a positive integer is in the form 6q+r where q is an integer and r = 0,1,2,3,4,5.
3) show that one and only one out of n , n+4 , n+8 , n+12 and n+16 is divisible by 5 , where 'n' is any positive integer.
4) show that the square of an odd positive integer can be of the form 6q+1 or 6q+3 for some integer q.
please help!!
Answers
x = 6q +2
Squaring both sides
Xsquare = 36(q)square + 4 + 24q
x square = 36(q) square - 6+ 2 + 24q
xsquare = 6 [ 6(q) square - 1 + 4q] + 2
put [ 6(q) square - 1 + 4q] = m
Therefore x = 6m + 2
2 .
let a be any +ve integer
b=6
by euclid 's division lemma,
a=bq+r, 0=< r < b
a= 6q+r,0=< r
when
r=0,a = 6q= (6q)3 ---> a3 = 216q3 =6(36q3 )=6q(where m is = 6q3 )
by similar manner u can prove for
r=1,2,3,4,5
and u will get the proof..(if you need it as 1,2,3,4,5ask me)...
3.If n = 5q
n is divisible by 5
Now, n = 5q
⇒ n + 4 = 5q + 4
The number (n + 4) will leave remainder 4 when divided by 5.
Again, n = 5q
⇒ n + 8 = 5q + 8 = 5(q + 1) + 3
The number (n + 8) will leave remainder 3 when divided by 5.
Again, n = 5q
⇒ n + 12 = 5q + 12 = 5(q + 2) + 2
The number (n + 12) will leave remainder 2 when divided by 5.
Again, n = 5q
⇒ n + 16 = 5q + 16 = 5(q + 3) + 1
The number (n + 16) will leave remainder 1 when divided by 5.
Case II:
When n = 5q + 1
The number n will leave remainder 1 when divided by 5.
Now, n = 5q + 1
⇒ n + 2 = 5q + 3
The number (n + 2) will leave remainder 3 when divided by 5.
Again, n = 5q + 1
⇒ n + 4 = 5q + 5 = 5(q + 1)
The number (n + 4) will be divisible by 5.
Again, n = 5q + 1
⇒ n + 8 = 5q + 9 = 5(q + 1) + 4
The number (n + 8) will leave remainder 4 when divided by 5
Again, n = 5q + 1
⇒ n + 12 = 5q + 13 = 5(q + 2) + 3
The number (n + 12) will leave remainder 3 when divided by 5.
Again, n = 5q + 1
⇒ n + 16 = 5q + 17 = 5(q + 3) + 2
The number (n + 16) will leave remainder 2 when divided by 5.
Similarly, we can check the result for 5q + 2, 5q + 3 and 5q + 4.
In each case only one out of n, n + 2, n + 4, n + 8, n + 16 will be divisible by 5.
4.Let a be any positive integer and b = 6.
Then, by Euclids algorithm,
a = 6q + rfor some integer q ≥ 0, and r = 0, 1, 2, 3, 4, 5 because 0 ≤ r
Therefore, a = 6q or 6q + 1 or 6q + 2 or 6q + 3 or 6q + 4 or 6q + 5
Also, 6q + 1 = 2 × 3q + 1 = 2k 1 + 1, where k 1 is a positive integer
6q + 3 = (6q + 2) + 1 = 2 (3q + 1) + 1 = 2k 2 + 1, where k 2 is an integer
6q + 5 = (6q + 4) + 1 = 2 (3q + 2) + 1 = 2k 3 + 1, where k 3 is an integer
Clearly, 6q + 1, 6q + 3, 6q + 5 are of the form 2k + 1, where k is an integer.
Therefore, 6q + 1, 6q + 3, 6q + 5 are not exactly divisible by 2. Hence, these expressions of numbers are odd numbers.
And therefore, any odd integer can be expressed in the form 6q + 1, or 6q + 3, or 6q + 5....
Hope this helps you...
Please mark it as brainliest answer...☺☺☺☺
Answer:
1.Let the square of any posititve integer be in the form 6q + 2, 6q +5
x = 6q +2
Squaring both sides
Xsquare = 36(q)square + 4 + 24q
x square = 36(q) square - 6+ 2 + 24q
xsquare = 6 [ 6(q) square - 1 + 4q] + 2
put [ 6(q) square - 1 + 4q] = m
Therefore x = 6m + 2
2 .
let a be any +ve integer
b=6
by euclid 's division lemma,
a=bq+r, 0=< r < b
a= 6q+r,0=< r
when
r=0,a = 6q= (6q)3 ---> a3 = 216q3 =6(36q3 )=6q(where m is = 6q3 )
by similar manner u can prove for
r=1,2,3,4,5
and u will get the proof..(if you need it as 1,2,3,4,5ask me)...
3.If n = 5q
n is divisible by 5
Now, n = 5q
⇒ n + 4 = 5q + 4
The number (n + 4) will leave remainder 4 when divided by 5.
Again, n = 5q
⇒ n + 8 = 5q + 8 = 5(q + 1) + 3
The number (n + 8) will leave remainder 3 when divided by 5.
Again, n = 5q
⇒ n + 12 = 5q + 12 = 5(q + 2) + 2
The number (n + 12) will leave remainder 2 when divided by 5.
Again, n = 5q
⇒ n + 16 = 5q + 16 = 5(q + 3) + 1
The number (n + 16) will leave remainder 1 when divided by 5.
Case II:
When n = 5q + 1
The number n will leave remainder 1 when divided by 5.
Now, n = 5q + 1
⇒ n + 2 = 5q + 3
The number (n + 2) will leave remainder 3 when divided by 5.
Again, n = 5q + 1
⇒ n + 4 = 5q + 5 = 5(q + 1)
The number (n + 4) will be divisible by 5.
Again, n = 5q + 1
⇒ n + 8 = 5q + 9 = 5(q + 1) + 4
The number (n + 8) will leave remainder 4 when divided by 5
Again, n = 5q + 1
⇒ n + 12 = 5q + 13 = 5(q + 2) + 3
The number (n + 12) will leave remainder 3 when divided by 5.
Again, n = 5q + 1
⇒ n + 16 = 5q + 17 = 5(q + 3) + 2
The number (n + 16) will leave remainder 2 when divided by 5.
Similarly, we can check the result for 5q + 2, 5q + 3 and 5q + 4.
In each case only one out of n, n + 2, n + 4, n + 8, n + 16 will be divisible by 5.
4.Let a be any positive integer and b = 6.
Then, by Euclids algorithm,
a = 6q + rfor some integer q ≥ 0, and r = 0, 1, 2, 3, 4, 5 because 0 ≤ r
Therefore, a = 6q or 6q + 1 or 6q + 2 or 6q + 3 or 6q + 4 or 6q + 5
Also, 6q + 1 = 2 × 3q + 1 = 2k 1 + 1, where k 1 is a positive integer
6q + 3 = (6q + 2) + 1 = 2 (3q + 1) + 1 = 2k 2 + 1, where k 2 is an integer
6q + 5 = (6q + 4) + 1 = 2 (3q + 2) + 1 = 2k 3 + 1, where k 3 is an integer
Clearly, 6q + 1, 6q + 3, 6q + 5 are of the form 2k + 1, where k is an integer.
Therefore, 6q + 1, 6q + 3, 6q + 5 are not exactly divisible by 2. Hence, these expressions of numbers are odd numbers.
And therefore, any odd integer can be expressed in the form 6q + 1, or 6q + 3, or 6q + 5....
Hope this helps you...
Please mark it as brainliest answer...☺☺☺☺