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).
Here is Kamal Lohia's excellent explanation:
Let the number be a²-1 as it's 1 short of a perfect square.
Now, a²-1 = (a-1)(a+1) is product of exactly two prime numbers. So, both (a-1) and (a+1) should be prime numbers.
Also a is multiple of 10 as the final number ends in '99'.
Checking
40²-1 = 39•41 but 39 isn't prime,
50²-1 = 49•51 where both aren't prime,
60²-1 = 59•61 SATISFIES
70²-1 = 69•71 but 69 isn't prime,
80²-1 = 79•81 but 81 isn't prime,
90²-1 = 89•91 but 91 isn't prime and
100²-1 = 99•101 but 99 isn't prime.
Only one number satisfies all the conditions: 60²-1 = 59•61 = 3599.
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 |