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Oodles of Factors II (Posted on 2010-10-11) Difficulty: 3 of 5
A. What is the lowest base 12 positive integer that has exactly 10 (base 12) distinct positive factors?

B. Exactly 1,000 (base 12) distinct positive factors?

C. Exactly 1,000,000 (base 12) distinct positive factors?

For example, the distinct positive factors of 40 (base 12) are the base 12 numbers 1, 2, 3, 4, 6, 8, 10, 14, 20, and 40. Accordingly, 40 (base 12) has precisely A (base 12) distinct positive factors.

See The Solution Submitted by K Sengupta    
Rating: 5.0000 (1 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
re: Part C | Comment 6 of 8 |
(In reply to Part C by Daniel)

There must be some mistake in your report of the factorization.

2^8*3^5*5^2*7^2*11^2*13*17*19*23*29*31*37*41*53 = 64,366,893,671,563,561,478,400, not 130,085,492,110,229,957,747,846,400.

Also, of course the product of the incremented exponents, 9*6*3^3*2^9 does not equal 12^6.

Some calculation shows you probably left 43 and 47 out of the factorization of 130,085,492,110,229,957,747,846,400.


  Posted by Charlie on 2010-10-12 11:36:12
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