Analyzing the Daily Savings Sequence

The problem describes a person's daily savings starting from January 1st, 2023. Let's look at the pattern:

• On January 1st, 2023 (Day 1), savings = Rs 1.
• On January 2nd, 2023 (Day 2), savings = Rs 1 + Rs 2 = Rs 3.
• On January 3rd, 2023 (Day 3), savings = Rs 3 + Rs 2 = Rs 5.
• And so on, each day's savings is Rs 2 more than the previous day.

This sequence of daily savings (1, 3, 5, ...) is an arithmetic progression (AP). An arithmetic progression is a sequence of numbers where the difference between consecutive terms is constant. In this case, the first term ($a_1$) is 1, and the common difference ($d$) is 2.

The saving amount on the $n$-th day ($a_n$) can be found using the formula for the $n$-th term of an AP:

\(a_n = a_1 + (n-1)d\)

Substituting $a_1 = 1$ and $d = 2$:

\(a_n = 1 + (n-1)2\)

\(a_n = 1 + 2n - 2\)

\(a_n = 2n - 1\)

So, the saving on the $n$-th day is \((2n - 1)\) rupees.

Calculating Total Savings Over Time

The total savings at the end of a certain date is the sum of the daily savings up to that date. At the end of the $n$-th day (which corresponds to the $n$-th of January), the total savings is the sum of the first $n$ terms of the arithmetic progression. The sum of the first $n$ terms of an AP ($S_n$) can be calculated using the formula:

\(S_n = \frac{n}{2}(2a_1 + (n-1)d)\)

Using $a_1 = 1$ and $d = 2$ again:

\(S_n = \frac{n}{2}(2(1) + (n-1)2)\)

\(S_n = \frac{n}{2}(2 + 2n - 2)\)

\(S_n = \frac{n}{2}(2n)\)

\(S_n = n^2\)

So, the total savings at the end of the $n$-th day (January $n$-th) is \(n^2\).

Condition for Perfect Square and Perfect Cube Savings

The question asks at the end of which date the total savings was both a perfect square and a perfect cube. We found that the total savings after \(n\) days is \(S_n = n^2\).

We need \(S_n\) to be a perfect square and a perfect cube.

First, let's check if \(S_n = n^2\) is a perfect square. Yes, by definition, \(n^2\) is always a perfect square for any integer \(n\).

Second, we need \(S_n = n^2\) to be a perfect cube. A number that is both a perfect square and a perfect cube must be a perfect sixth power. This is because if a number can be written as \(k^2\) and \(m^3\) for some integers \(k\) and \(m\), then its prime factorization must have exponents that are multiples of 2 and also multiples of 3. The least common multiple of 2 and 3 is 6. Therefore, the exponents in its prime factorization must be multiples of 6.

We have \(S_n = n^2\). For \(n^2\) to be a perfect cube, the exponents in the prime factorization of \(n^2\) must be divisible by 3.

Let the prime factorization of \(n\) be \(p_1^{e_1} p_2^{e_2} \dots p_r^{e_r}\). Then the prime factorization of \(n^2\) is \(p_1^{2e_1} p_2^{2e_2} \dots p_r^{2e_r}\).

For \(n^2\) to be a perfect cube, each exponent \(2e_i\) must be divisible by 3. Since 2 and 3 have no common factors other than 1, \(e_i\) must be divisible by 3. This means that \(n\) itself must be a perfect cube.

Let \(n = c^3\) for some integer \(c \ge 1\) (since \(n\) represents the number of days).

If \(n = c^3\), then the total savings is \(S_n = n^2 = (c^3)^2 = c^6\). A number of the form \(c^6\) is indeed both a perfect square (\((c^3)^2\)) and a perfect cube (\((c^2)^3\)).

So, the condition simplifies to finding a date (number of days, \(n\)) such that \(n\) is a perfect cube.

Finding the Date (Number of Days)

The date is the $n$-th day of January 2023. We need to find a date \(n\) such that \(n\) is a perfect cube and \(S_n = n^2\) is both a perfect square and a perfect cube.

We need to find an integer \(n\) from the options that is a perfect cube.

Let's consider the options:

• Option 1: 7th January, 2023. This corresponds to \(n=7\) days. Is 7 a perfect cube? \(1^3 = 1\), \(2^3 = 8\). No, 7 is not a perfect cube. Total savings \(S_7 = 7^2 = 49\). 49 is a perfect square (\(7^2\)) but not a perfect cube.
• Option 2: 8th January, 2023. This corresponds to \(n=8\) days. Is 8 a perfect cube? Yes, \(8 = 2^3\). This fits the condition that \(n\) must be a perfect cube.
• Option 3: 9th January, 2023. This corresponds to \(n=9\) days. Is 9 a perfect cube? No, 9 is not a perfect cube. Total savings \(S_9 = 9^2 = 81\). 81 is a perfect square (\(9^2\)) but not a perfect cube.
• Option 4: Not possible. This would be true if none of the dates or any future date could satisfy the condition.

Since \(n=8\) is a perfect cube (\(2^3\)), the total savings on January 8th, 2023, should be both a perfect square and a perfect cube.

Verifying the Perfect Square and Cube Date

Let's verify for \(n=8\), corresponding to January 8th, 2023.

Total savings \(S_8 = 8^2 = 64\).

• Is 64 a perfect square? Yes, \(64 = 8^2\).
• Is 64 a perfect cube? Yes, \(64 = 4^3\).

Since the total savings on January 8th, 2023, is 64, which is both a perfect square and a perfect cube, this date satisfies the condition.

Therefore, the correct date is 8th January, 2023.

Revision Table: Perfect Powers

Understanding perfect squares and perfect cubes is key to solving this problem. Here’s a quick look at small perfect powers:

We needed the total savings \(S_n\) to be a number found in the last column of this table (or a similar number). Since \(S_n = n^2\), we needed \(n^2\) to be a number like \(1, 64, 729, ...\). From \(S_n = n^2\), if \(S_n=1\), then \(n^2=1 \implies n=1\). If \(S_n=64\), then \(n^2=64 \implies n=8\). If \(S_n=729\), then \(n^2=729 \implies n=27\). These required values for \(n\) are 1, 8, 27, ..., which are \(1^3, 2^3, 3^3, ...\). This confirms our finding that \(n\) must be a perfect cube.

Additional Information: Arithmetic Progressions

An arithmetic progression (AP) is defined by a first term (\(a_1\)) and a common difference (\(d\)).

• The \(n\)-th term is given by \(a_n = a_1 + (n-1)d\).
• The sum of the first \(n\) terms is given by \(S_n = \frac{n}{2}(a_1 + a_n)\) or \(S_n = \frac{n}{2}(2a_1 + (n-1)d)\).

In this problem, the daily savings form an AP: 1, 3, 5, 7, 9, 11, 13, 15, ...

• \(a_1 = 1\)
• \(d = 3 - 1 = 2\)
• The daily saving on day \(n\) is \(a_n = 1 + (n-1)2 = 2n - 1\).
• The total savings after \(n\) days is the sum \(S_n = 1 + 3 + 5 + \dots + (2n - 1)\). This is the sum of the first \(n\) odd numbers.

The sum of the first \(n\) odd numbers is always \(n^2\). For example:

• Sum of first 1 odd number: 1 = \(1^2\)
• Sum of first 2 odd numbers: 1 + 3 = 4 = \(2^2\)
• Sum of first 3 odd numbers: 1 + 3 + 5 = 9 = \(3^2\)
• Sum of first 4 odd numbers: 1 + 3 + 5 + 7 = 16 = \(4^2\)

This property directly leads to the total savings being \(S_n = n^2\), simplifying the problem significantly to finding \(n\) such that \(n\) is a perfect cube.