Question:
If \(n = 1 + x\), where \(x\) is the product of 4 consecutive positive integers, then which of the following is/are true?
1. \(n\) is odd
2. \(n\) is prime
3. \(n\) is a perfect square
If \(n = 1 + x\), where \(x\) is the product of 4 consecutive positive integers, then which of the following is/are true?
1. \(n\) is odd
2. \(n\) is prime
3. \(n\) is a perfect square
1. \(n\) is odd
2. \(n\) is prime
3. \(n\) is a perfect square
Show Hint
Write \(x\) as \((k^2+3k)(k^2+3k+2)\) and notice this has the form \(m(m+2)\), so \(x+1\) is a perfect square.
Updated On: Aug 18, 2026
- 1 and 3 only
- 1 and 2 only
- 1 only
- None of these
Show Solution
Verified By Collegedunia
The Correct Option is A
Solution and Explanation
Step 1: Represent \(x\) algebraically.
Let the four consecutive positive integers be \(k, k+1, k+2, k+3\). Then
\[ x = k(k+1)(k+2)(k+3) \]
Step 2: Pair the terms cleverly.
Group the first and last terms, and the middle two terms, instead of multiplying left to right:
\[ x = [k(k+3)] \times [(k+1)(k+2)] = (k^2+3k)(k^2+3k+2) \]
Both brackets share the same core expression \(k^2+3k\); let \(m = k^2+3k\). Then
\[ x = m(m+2) = m^2 + 2m \]
Step 3: Check statement 1, is \(n\) odd?
Among any 4 consecutive integers, at least two are even, so their product \(x\) is always even. Since \(n = x+1\), adding 1 to an even number always gives an odd number. So \(n\) is always odd, statement 1 is always true.
Step 4: Check statement 3, is \(n\) a perfect square?
\[ n = x + 1 = m^2 + 2m + 1 = (m+1)^2 \]
This is a perfect square for every positive integer \(k\), since \(m = k^2+3k\) is always a whole number. So statement 3 is always true. Testing it numerically confirms this: for \(k=1\), \(x = 1\cdot2\cdot3\cdot4 = 24\), \(n = 25 = 5^2\). For \(k=2\), \(x = 2\cdot3\cdot4\cdot5 = 120\), \(n = 121 = 11^2\). For \(k=3\), \(x = 3\cdot4\cdot5\cdot6 = 360\), \(n = 361 = 19^2\). Every case lands on a perfect square, matching the general formula \(n=(m+1)^2\).
Step 5: Check statement 2, is \(n\) prime?
Since \(n = (m+1)^2\) is a perfect square bigger than 1 (because \(k \ge 1\) makes \(m \ge 4\), so \(m+1 \ge 5\)), it always has at least three divisors: 1, \(m+1\), and \(n\) itself. A number with more than two divisors can never be prime, so \(n\) is never prime, statement 2 is always false.
Final Answer:
Statements 1 and 3 always hold, while statement 2 never does.
\[ \boxed{\text{1 and 3 only}} \]
Let the four consecutive positive integers be \(k, k+1, k+2, k+3\). Then
\[ x = k(k+1)(k+2)(k+3) \]
Step 2: Pair the terms cleverly.
Group the first and last terms, and the middle two terms, instead of multiplying left to right:
\[ x = [k(k+3)] \times [(k+1)(k+2)] = (k^2+3k)(k^2+3k+2) \]
Both brackets share the same core expression \(k^2+3k\); let \(m = k^2+3k\). Then
\[ x = m(m+2) = m^2 + 2m \]
Step 3: Check statement 1, is \(n\) odd?
Among any 4 consecutive integers, at least two are even, so their product \(x\) is always even. Since \(n = x+1\), adding 1 to an even number always gives an odd number. So \(n\) is always odd, statement 1 is always true.
Step 4: Check statement 3, is \(n\) a perfect square?
\[ n = x + 1 = m^2 + 2m + 1 = (m+1)^2 \]
This is a perfect square for every positive integer \(k\), since \(m = k^2+3k\) is always a whole number. So statement 3 is always true. Testing it numerically confirms this: for \(k=1\), \(x = 1\cdot2\cdot3\cdot4 = 24\), \(n = 25 = 5^2\). For \(k=2\), \(x = 2\cdot3\cdot4\cdot5 = 120\), \(n = 121 = 11^2\). For \(k=3\), \(x = 3\cdot4\cdot5\cdot6 = 360\), \(n = 361 = 19^2\). Every case lands on a perfect square, matching the general formula \(n=(m+1)^2\).
Step 5: Check statement 2, is \(n\) prime?
Since \(n = (m+1)^2\) is a perfect square bigger than 1 (because \(k \ge 1\) makes \(m \ge 4\), so \(m+1 \ge 5\)), it always has at least three divisors: 1, \(m+1\), and \(n\) itself. A number with more than two divisors can never be prime, so \(n\) is never prime, statement 2 is always false.
Final Answer:
Statements 1 and 3 always hold, while statement 2 never does.
\[ \boxed{\text{1 and 3 only}} \]
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