Answer:
See below
Step-by-step explanation:
If a is divisible by 3 then a is divisible by 9
<u>FALSE </u>
Counter-example
6 is divisible by 3 but not by 9
The subtraction of 2 rational numbers is rational.
<u>TRUE </u>
<em>Proof </em>
If a, b are two rational numbers
for some integers p, q, r, s.
Then
since ps-qr and qs are integers, a-b is rational
A sufficient condition for an integer to be divisible by 8 is that it is divisible by 2
<u>FALSE </u>
<em>Counter-example </em>
4 is divisible by 2 but not by 8
A sufficient condition for an integer to be divisible by 6 is that it is divisible by 2
<u>FALSE </u>
<em>Counter-example </em>
4 is divisible by 2 but not by 6
If a is divisible by 9 then a is divisible by 3.
<u>TRUE </u>
<em>Proof</em>
If a is divisible by 9, then a = 9k for some integer k, but 9=3*3, so a = 3*(3k).
Since 3k is integer, a is divisible also by 3.
The product of 2 consecutive integers is even.
<u>TRUE </u>
<em>Proof </em>
Let p, q be two consecutive integers, then either p is even or odd.
Suppose first p is even. Then
p = 2n and q = 2n+1, so p*q=2n(2n+1)=2n*2n+2n=2(n*2n+n)
since (n*2n+n) is integer p*q is even.
Suppose now p is odd
p = 2n+1 q = 2n+2, then
p*q=(2n+1)(2n+2)=2n*2n+2*2n+2n+2=2(n*2n+2n+n+1)
since (n*2n+2n+n+1) is integer p*q is also even.
Answer the following questions:
30 division 3 =
10
30 mod 3 =
0 (the remainder when dividing 30 by 3)
-26 division 5 =
-5 (plus remainder -1)
-26 mod 5 =
-1 (the remainder when dividing -26 by 5)
28 division 4 =
7
28 mod 4 =
0
-29 division 10 =
-2 (plus remainder -9)
-29 mod 10 =
-9
24 division 9 =
2 (plus remainder 6)
24 mod 9 =
6
-28 division 6=
-4 (plus remainder -4)
-28 mod 6 =
-4
965255471 mod 101 =
87 (the remainder when dividing 965255471 by 101)
630153353 mod 101 =
11 (the remainder when dividing 630153353 by 101)