June 2025
Problem of the Month

4 Digit Number Problem
Submitted by Brijesh Dave
from Mumbai, India

Determine the four-digit numbers that satisfy all three of the following properties.

(I) The tens and ones digit are both 9.

(II) The number is 1 less than a perfect square.

(III) The number is the product of exactly two prime numbers

Solution to the Problem:

Here is the solution:

Only one four digit number satisfies all three properties: 3,599.

Since the last two digits are 99 (Clue #I) and the number is one less than a perfect square (Clue #II), that means the perfect square must end in 00.
The only four-digit perfect squares ending in 00 are 1600, 2500, 3600, 4900, 6400, and 8100.

Clue #II also tells us that the number is in the form n² - 1.
We can factor this as the Difference of 2 Squares: n² - 1 = (n + 1)(n - 1). Hence:

40² -1 = (39)(41)
50² -1 = (49)(51)
60² -1 = (59)(61)
70² -1 = (69)(71)
80² -1 = (79)(81)
90² -1 = (89)(91)

Of these numbers, the only one that is a product of two primes is: 3599 = 60² -1 = (59)(61).

Correctly solved by:

1. Davit Banana	Istanbul, Turkey
2. Colin (Yowie) Bowey	Beechworth, Victoria, Australia
3. Ivy Joseph	Pune, Maharashtra, India
4. Dr. Hari Kishan	D.N. College, Meerut, Uttar Pradesh, India
5. Kamal Lohia	Hisar, Haryana, India
6. Kushagra Chugh	Hisar, Haryana, India
7. K. Sengupta	Calcutta, India
8. Kelly Stubblefield	Mobile, Alabama, USA

Send any comments or questions to: David Pleacher