Num3ers that are divisible by 2-digit numbers contained in them

For example,

132         132 / 12 = 11
135         135 / 15 = 9
612         612 / 12 = 51

and many others. They are not very interesting since they are only divisible by one 2-digit number contained in them.

 
 
More interesting cases :

1296      1296/12 = 108      1296/16 = 81
1326      1326/13 = 102      1326/26 = 51

1248   is divisible by three 2-digit numbers contained in it:
1248/12 = 104          1248/24 = 52          1248/48 = 26

And  
1368 :
1368/18 = 76      1368/36 = 38      1368/38 = 36

1632 :
1632/16 = 102      1632/32 = 51      1632/12 = 136

1995 :
1995/19 = 105      1995/95 = 21      1995/15 = 133

1155 :
1155/11 = 105      1155/15 = 77      1155/55 = 21

3774 :
3774/37 = 102      3774/74 = 51      3774/34 = 111

 

And this very cool number:

2184 :
2184/21 = 104      2184/28 = 78      2184/24 = 91      2184/12 = 182     
2184/14 = 156      2184/42 = 52      2184/84 = 26

And,

31248 :
31248/31 = 1008      31248/12 = 2604      31248/14 = 2232      31248/18 = 1736
31248/24 = 1302      31248/28 = 1116       31248/48 = 651       31248/84 = 372
31248/42 = 744

 
Find more examples.
 
 
 
 

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About benvitalis

math grad - Interest: Number theory
This entry was posted in Math Beauty, Number Puzzles and tagged . Bookmark the permalink.

2 Responses to Num3ers that are divisible by 2-digit numbers contained in them

  1. David a/k/a @InfinitelyManic says:

    #!/usr/bin/python3
    # David a/k/a @InfinitelyManic
    # Num3ers that are divisible by 2-digit numbers contained in them

    # pick an arbitrary range
    for number in range(31000,35000):
    # get each digit within the number
    for a in str(number):
    # get each digit within the number
    for b in str(number):
    #concatenate the strings
    c = a+b
    # exclude 0; 1; where a == b; we only want integers
    if int(c) != 0 and int(c) != 1 and a != b and number % int(c) == 0:
    #print everything out
    print(number,’/’,c,’=’,number/int(c))

    ######### some results:
    31257 / 23 = 1359.0
    31260 / 30 = 1042.0
    31260 / 12 = 2605.0
    31260 / 10 = 3126.0
    31260 / 20 = 1563.0
    31260 / 60 = 521.0
    31260 / 03 = 10420.0
    31260 / 02 = 15630.0
    31260 / 06 = 5210.0
    31264 / 32 = 977.0
    31264 / 16 = 1954.0
    31265 / 13 = 2405.0
    31265 / 65 = 481.0
    31269 / 21 = 1489.0
    31270 / 10 = 3127.0
    31270 / 02 = 15635.0
    31272 / 12 = 2606.0
    31272 / 12 = 2606.0
    31275 / 15 = 2085.0
    31275 / 25 = 1251.0
    31275 / 75 = 417.0
    31278 / 13 = 2406.0
    31278 / 78 = 401.0
    31279 / 31 = 1009.0
    31280 / 10 = 3128.0
    31280 / 23 = 1360.0
    31280 / 20 = 1564.0
    31280 / 80 = 391.0
    31280 / 02 = 15640.0
    31280 / 08 = 3910.0
    31284 / 12 = 2607.0
    31284 / 18 = 1738.0

  2. Paul says:

    In a condensed form we find below 1,000,000 the number’s with the most possible divisors are

    413280 with 17 of its 25 being integer and
    524160.

    Now below 10,000,000 we have 1 winner
    5416320 with 22 of its possible 36 being integer.

    Paul

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