 All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars  perplexus dot info  A Really Weird Math Problem (Posted on 2018-03-12) On The Big Bang Theory's episode "The Athenaeum Allocation", Sheldon mentions to Amy that for their proposed wedding date, May 12,

"The month squared equals the square of the sum of the members of the set of the prime factors of the day. Isn't that romantic?"

Amy's reply contorts a Shakespeare quote: "shall I compare thee to a day that's also a really weird math problem?"

It's indeed weird as the squaring is introduced for no reason at all.

But also, how rare is the circumstance? What other dates of the year exhibit the property? Remember that because of the set-membership criterion, each prime factor is included in the sum only once: 12 has prime factors 2 and 3, and those add up to 5, for May.

 See The Solution Submitted by Charlie No Rating Comments: ( Back to comment list | You must be logged in to post comments.) There are N dates Comment 4 of 4 | Three are 22 dates of the 365/366 dates of a year.

1.FEB  2 :       2 =  2 :    2    =  2
2.FEB  4 :     2×2 =  4 :    2    =  2

3.FEB  8
:   2×2×2 =  8 :    2    =  2
4.FEB 16 :
2×2×2×2 = 16 :    2    =  2
5.MAR  3 :       3 =  3 :    3    =  3
6.MAR  9 :     3
×3 =  9 :    3    =  3
7.MAR 27 :
3×3×3 = 27 :    3    =  3
8.MAY  5 :       5 =  5 :    5    =  5
9.MAY  6 :     2
×3 =  6 :  (2+3)  =  5
10.MAY 12
:   2×2×3 = 12 :  (2+3)  =  5
11.MAY 18
:   2×3×3 = 18 :  (2+3)  =  5
12.MAY 24
: 2×2×2×3 = 24 :  (2+3)  =  5
13.MAY 25
:     5×5 = 25 :    5    =  5
14.JUL  7 :       7 =  7 :    7    =  7
15.JUL 10 :     2
×5 = 10 :  (2+5)  =  7
16.JUL 20 :
2×2×5 = 20 :  (2+5)  =  7
17.AUG 15 :
3×5 = 15 :  (3+5)  =  8
18.SEP 14 :
2×7 = 14 :  (2+7)  =  9
19.SEP 28 :
2×2×7 = 28 :  (2+7)  =  9
20.OCT 21 :     3×7 = 21 :  (3+7)  = 10
21
.OCT 30 :
2×3×5 = 30 : (2+3+5) = 10
22.NOV 11 :      11 = 11 :   11    = 11

Edited on March 12, 2018, 5:49 pm
 Posted by Dej Mar on 2018-03-12 17:38:04 Please log in:

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