Show that only one out of a, a+2 and a+4 is divisible by 3
Answers
Answer:
Step-by-step explanation:
case 1 if a is divisible by 3
then a+2 and a+4 are not divisible by 3
case 2 if a+2 is divisible by 3
then [a+2] -2 and a+2+2 are not divisible by 3
that is a and a+4 are not divisible by 3
case 3 if a+ 4 is divisible by 3
then a+4 -2 and a+4 -4 are not divisible by 3
then [a+2] and a are not divisible by 3
case 4 if a is not divisible by 3
then a gives remainder 1 or 2 when it is divided by 3
then a + 2 or a+1 is divisible by 3 correspondingly
The given divisor is 3.
Among the integers, there are three kinds of integers according to 3 as divisor.
1. Integers which leave remainder 0 on division by 3.
E.g.: ..., -6, -3, 0, 3, 6,...
2. Integers which leave remainder 1 on division by 3.
E.g.: ..., -5, -2, 1, 4, 7,...
3. Integers which leave remainder 2 on division by 3.
E.g.: ..., -4, -1, 2, 5, 8,...
Each kind of integers can be taken as 3n, 3n + 1 and 3n + 2 respectively.
Means,
→ Integers which leave remainder 0 can be taken as 3n.
→ Integers which leave remainder 1 can be taken as 3n + 1.
→ Integers which leave remainder 2 can be taken as 3n + 2.
We have to prove that only one among the three integers a, a + 2 and a + 4 is exactly divisible by 3.
First we are going to take a as 3n, like the numbers in the first kind of integers.
Thus the other two integers will be 3n + 2 and 3n + 4.
→ a = 3n leaves remainder 0 on division by 3, so it is exactly divisible by 3.
→ a + 2 = 3n + 2 leaves remainder 2 on division by 3, so it is not exactly divisible by 3.
→ a + 4 = 3n + 4 = 3n + 3 + 1 = 3(n+1) + 1 leaves remainder 1 on division by 3, thus neither is a + 4.
Here we found that only 'a' among the three ones is exactly divisible by 3.
Now take a as 3n + 1, like the second kind of integers mentioned above.
→ a = 3n + 1 leaves remainder 1 on division by 3. Thus it's not.
→ a + 2 = 3n + 1 + 2 = 3n + 3 = 3(n + 1) is exactly divisible by 3 as the remainder is 0.
→ a + 4 = 3n + 1 + 4 = 3n + 1 + 2 + 2 = 3n + 3 + 2 = 3(n + 1) + 2 leaves remainder 2 on division by 3.
Here, only 'a + 2' is exactly divisible by 3 while neither are the others.
Now, taking a as 3n + 2,
→ a = 3n + 2 leaves remainder 2.
→ a + 2 = 3n + 2 + 2 = 3n + 2 + 1 + 1 = 3n + 3 + 1 = 3(n + 1) + 1 leaves remainder 1.
→ a + 4 = 3n + 2 + 4 = 3n + 6 = 3(n + 2) is exactly divisible by 3 as the remainder is 0.
Here, only 'a + 4' is exactly divisible by 3 while neither are the other two.
In the three cases each, we found that only one among the three integers a, a + 2 and a + 4 is exactly divisible by 3.