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More about 1729 (Posted on 2012-01-04) Difficulty: 3 of 5
Much was said about 1729 a.k.a. Ramanujan number.or a taxicab number.
1729 is also a Carmichael number and the first absolute Euler pseudoprime, a sphenic number, a Zeisel number etc etc

  Please note the following (and solve):
  1. 1729 is one of four positive integers (with the others being A, B and the trivial case 1) which, when its digits are added together, produces a sum which, when multiplied by its reversal, yields the original number:
1 + 7 + 2 + 9 = 19 ; 19*91 = 1729
Find A and B.

2. 1729= xyz, where x,y and z are integer members of an arithmetic progression.
Find these members.

3. 1729 can be expressed (in more than one way) as a sum of distinct Fibonacci numbers
List all the expressions.
 

See The Solution Submitted by Ady TZIDON    
Rating: 5.0000 (3 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
Solutions | Comment 1 of 8
1.)  8+1 = 9; 9*9 = 81
1+4+5+8 = 18; 18*81 = 1458
[Brute force, there aren't that many to check.]

2.) 1729 = 7*13*19
Just factor it.  You can't miss it.

3.)  List the Fibonacci numbers less than 1729:
1,1,2,3,5,8,13,21,34,55,89,144,233,377,610,987,1597
Pick the largest possible values to get the target sum.  Then each of these can be broken into two smaller numbers, you just have to watch for repeats:
1597+89+34+8+1 = 1729
1597+89+34+5+3+1 = 1729
1597+89+21+13+8+1 = 1729
1597+89+21+13+5+3+1 = 1729
1597+55+34+21+13+8+1 = 1729
1597+55+34+21+13+5+3+1 = 1729
Replace 1597 with 987+610 in each of the six above.
Replace 1597 with 987+377+233 in each of the six above.
Replace 1597 with 987+377+144+89 in the last two of the six above.
For a total of 20 ways.

This is an interesting exercise and there is a good Perplexus problem in there somewhere.

  Posted by Jer on 2012-01-04 11:27:31
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