Question Details

On January 1st, 2023, a person saved ` 1. On January 2nd, 2023, he saved ` 2 more than that on the previous day. On January 3rd, 2023, he saved ` 2 more than that on the previous day and so on. At the end of which date was his total savings a perfect square as well a perfect cube?

Options

A

7th January, 2023

B

8th January, 2023

C

9th January, 2023

D

Not possible

Show Answer

Correct Answer :

Option B

8th January, 2023

Solution :

The correct option is 8th January, 2023.

Let us analyze the savings day by day to understand the pattern. The problem states that:
- On January 1st (Day 1), the person saved 1 rupee.
- On January 2nd (Day 2), he saved 2 more than on the previous day. So, he saved 1+2=3 rupees.
- On January 3rd (Day 3), he saved 2 more than on the previous day. So, he saved 3+2=5 rupees.
- This pattern continues, meaning the savings on each day form an arithmetic progression of odd numbers: 1,3,5,7,...

The sum of the first n odd natural numbers is well-known to be a perfect square, specifically n2. Let's verify this step-by-step:
- At the end of Day 1: Total savings = 1=12
- At the end of Day 2: Total savings = 1+3=4=22
- At the end of Day 3: Total savings = 1+3+5=9=32
- Generally, at the end of Day n (where n represents the date in January), the total savings will be n2 rupees.

We are looking for a date n such that the total savings, n2, is both a perfect square and a perfect cube.
For a number to be both a perfect square and a perfect cube, it must be of the form k6 for some positive integer k.
Thus, we set:

n2=k6

Taking the square root on both sides:

n=k3

This means that the day number n itself must be a perfect cube. Let's look at the smallest perfect cubes:
- For k=1, n=13=1 (January 1st). The savings would be 12=1, which is both a square and a cube. However, this is not in the options.
- For k=2, n=23=8 (January 8th). The savings at the end of the 8th day would be:

82=64

Let's check if 64 is a perfect square and a perfect cube:
- It is a perfect square: 82=64
- It is a perfect cube: 43=64
Since 64 satisfies both conditions, the date is indeed the 8th of January, 2023.

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