All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars    
perplexus dot info

Home > Numbers
3 and 4-digit numbers (Posted on 2017-07-17) Difficulty: 2 of 5
Consider the number 4808
Notice: 4^2+8^2=80
i.e. the sum of the squares of this number's first and last digits equals the number obtained when the first and last digits are erased.

How many numbers with such feature exist below 10000?

Clearly, no leading zeroes.

See The Solution Submitted by Ady TZIDON    
No Rating

Comments: ( Back to comment list | You must be logged in to post comments.)
Just counting (spoiler) | Comment 1 of 6
Well, if we allowed 5 digit numbers, then there would be 90, one for each number from 10 to 99.
For instance, consider 65.  6^2 + 5^2 = 36+25 = 61, so the number is 6615.

However, for some numbers 10a+b, a^2 + b^2 > 99.
The numbers with this feature are

86 and 68
87 and 78
88
95 and 59
96 and 69
97 and 79
98 and 89
99

Altogether there are 14 of these.

So, the requested answer is 90 - 14 = 76.

  Posted by Steve Herman on 2017-07-17 13:08:18
Please log in:
Login:
Password:
Remember me:
Sign up! | Forgot password


Search:
Search body:
Forums (0)
Newest Problems
Random Problem
FAQ | About This Site
Site Statistics
New Comments (3)
Unsolved Problems
Top Rated Problems
This month's top
Most Commented On

Chatterbox:
Copyright © 2002 - 2017 by Animus Pactum Consulting. All rights reserved. Privacy Information