NCERT Exemplar Solutions - Class 10 - Mathematics - Unit 1: Real NumbersExercise 1.4
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Question. 1
1. Show that the cube of a positive integer of the form \(6q + r\), where \(q\) is an integer and \(r = 0, 1, 2, 3, 4, 5\), is also of the form \(6m + r\).
Answer
Yes. The cube of such an integer is again of the form \(6m + r\).
Step by Step Solution
Step 1: Express the integer.
Any integer can be written as
\(n = 6q + r\),
where \(q\) is some integer and \(r\) is the remainder when dividing by 6.
So \(r\) can only be one of the values 0, 1, 2, 3, 4, or 5.
Step 2: Consider the cube.
We need to check the form of \(n^3\). For this, it is enough to look at the values of \(r^3\) modulo 6.
Step 3: Check each remainder one by one.
• If \(r = 0\):
\(0^3 = 0 \equiv 0 \pmod{6}\).
• If \(r = 1\):
\(1^3 = 1 \equiv 1 \pmod{6}\).
• If \(r = 2\):
\(2^3 = 8\).
Since \(8 = 6 + 2\), we have \(2^3 \equiv 2 \pmod{6}\).
• If \(r = 3\):
\(3^3 = 27\).
Since \(27 = 6 \times 4 + 3\), we have \(3^3 \equiv 3 \pmod{6}\).
• If \(r = 4\):
\(4^3 = 64\).
Since \(64 = 6 \times 10 + 4\), we have \(4^3 \equiv 4 \pmod{6}\).
• If \(r = 5\):
\(5^3 = 125\).
Since \(125 = 6 \times 20 + 5\), we have \(5^3 \equiv 5 \pmod{6}\).
Step 4: Observe the pattern.
In each case, we found that
\(r^3 \equiv r \pmod{6}\).
Step 5: Write the final form.
If \(n = 6q + r\), then
\(n^3 = (6q + r)^3\),
and when divided by 6, its remainder is again \(r\).
This means
\(n^3 = 6m + r\) for some integer \(m\).
Conclusion. The cube of any integer of the form \(6q + r\) is again of the form \(6m + r\).
Question. 2
2. Prove that one and only one out of n, n + 2 and n + 4 is divisible by 3, where n is any positive integer.
Answer
Exactly one out of n, n + 2 and n + 4 is divisible by 3.
Step by Step Solution
Step 1: Think about remainders when dividing by 3.
Every integer \(n\) must leave one of three possible remainders when divided by 3:
\(n \equiv 0 \pmod{3}\), or \(n \equiv 1 \pmod{3}\), or \(n \equiv 2 \pmod{3}\).
Step 2: Case when \(n \equiv 0 \pmod{3}\).
If \(n\) is divisible by 3, then:
\(n \equiv 0\)
\(n+2 \equiv 2\)
\(n+4 \equiv 1\)
So, only \(n\) is divisible by 3.
Step 3: Case when \(n \equiv 1 \pmod{3}\).
If the remainder is 1, then:
\(n \equiv 1\)
\(n+2 \equiv 3 \equiv 0 \pmod{3}\)
\(n+4 \equiv 5 \equiv 2\)
So, only \(n+2\) is divisible by 3.
Step 4: Case when \(n \equiv 2 \pmod{3}\).
If the remainder is 2, then:
\(n \equiv 2\)
\(n+2 \equiv 4 \equiv 1\)
\(n+4 \equiv 6 \equiv 0 \pmod{3}\)
So, only \(n+4\) is divisible by 3.
Step 5: Combine all cases.
No matter what remainder \(n\) gives when divided by 3, exactly one of the numbers \(n, n+2, n+4\) will be divisible by 3.
Therefore, proved.
Question. 3
3. Prove that one of any three consecutive positive integers must be divisible by 3.
Answer
Yes, in every set of three consecutive integers, one of them is divisible by 3.
Step by Step Solution
Step 1: Represent three consecutive integers.
Let the three consecutive positive integers be:
\(n,\; n+1,\; n+2\)
Step 2: Recall the property of division by 3.
When any integer is divided by 3, the possible remainders are:
\(0,\; 1,\; \text{or } 2\)
Step 3: Check each case for \(n\).
Case A: If \(n \equiv 0 \pmod{3}\), then \(n\) itself is divisible by 3.
Case B: If \(n \equiv 1 \pmod{3}\), then
\(n+2 \equiv 0 \pmod{3}\)
So, \(n+2\) is divisible by 3.
Case C: If \(n \equiv 2 \pmod{3}\), then
\(n+1 \equiv 0 \pmod{3}\)
So, \(n+1\) is divisible by 3.
Step 4: Conclude.
In every group of three consecutive integers, one of them must be divisible by 3. This proves the statement.
Question. 4
4. For any positive integer \(n\), prove that \(n^3 - n\) is divisible by 6.
Answer
Yes. For every positive integer \(n\), the number \(n^3 - n\) is divisible by \(6\).
Step by Step Solution
Idea. Show that \(n^3 - n\) is divisible by \(2\) and by \(3\). Since \(\mathrm{lcm}(2,3) = 6\), divisibility by both \(2\) and \(3\) implies divisibility by \(6\).
Step 1: Factorise.
Write
\(n^3 - n = n(n^2 - 1)\)
\(= n(n - 1)(n + 1)\).
This is the product of three consecutive integers: \(n - 1\), \(n\), \(n + 1\).
Step 2: Divisible by \(2\).
Among any three consecutive integers, at least one is even. Therefore, \(n(n - 1)(n + 1)\) is divisible by \(2\).
Step 3: Divisible by \(3\).
Among any three consecutive integers, exactly one is a multiple of \(3\). Hence, \(n(n - 1)(n + 1)\) is divisible by \(3\).
Step 4: Combine the results.
Since the same number is divisible by \(2\) and by \(3\), it is divisible by \(6\).
Therefore, for every positive integer \(n\),
\(6 \mid (n^3 - n)\).
Optional check (small values).
\(n = 1\): \(1^3 - 1 = 0\), divisible by \(6\).
\(n = 2\): \(8 - 2 = 6\), divisible by \(6\).
\(n = 3\): \(27 - 3 = 24\), divisible by \(6\).
These examples agree with the proof.
Question. 5
5. 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.
Answer
Exactly one and only one out of n, n + 4, n + 8, n + 12 and n + 16 is divisible by 5.
Step by Step Solution
Step 1: Write the numbers.
We are given five numbers: \(n, n+4, n+8, n+12, n+16\).
Step 2: Consider division by 5.
Any integer \(n\), when divided by 5, will leave one of the remainders 0, 1, 2, 3, or 4.
We write this as \(n \equiv r \pmod{5}\), where \(r\) is one of 0, 1, 2, 3, 4.
Step 3: Calculate the remainders of each term.
If \(n \equiv r \pmod{5}\), then:
- \(n \equiv r \pmod{5}\)
- \(n + 4 \equiv r + 4 \pmod{5}\)
- \(n + 8 \equiv r + 8 \equiv r + 3 \pmod{5}\)
- \(n + 12 \equiv r + 12 \equiv r + 2 \pmod{5}\)
- \(n + 16 \equiv r + 16 \equiv r + 1 \pmod{5}\)
Step 4: Observe the pattern.
The five numbers give remainders \(r, r+1, r+2, r+3, r+4\) (all taken modulo 5).
This set covers all possible remainders: 0, 1, 2, 3, 4.
Step 5: Conclusion.
Since exactly one of these remainders is 0, it follows that exactly one of the numbers is divisible by 5.