8-digit numbers whose squares end with the same 8 digits

 
 
 
87109376^2   =   7588043387109376

12890625^2   =   166168212890625

 

A general form:    numbers whose squares end with the same…

… 2 digits:
x \; = \; 25 \, (4 \, n+1)
x \; = \; 4 (25 \, n+19)

… 3 digits:
x \; = \; 125 \,(8 \, n+5)
x \; = \; 8 \,(125 \, n+47)

… 4 digits:
x \; = \; 625 \, (16 \, n+1)
x \; = \; 16 \, (625 \, n+586)

… 5 digits:
x \; = \; 3125 \, (32 n+29)
x \; = \; 32 \, (3125 n+293)

… 6 digits:
x \; = \; 15625 \, (64 \, n+57)
x \; = \; 64 \, (15625 \, n+1709)

… 7 digits:
x \; = \; 78125 \, (128 \, n+37)
x \; = \; 128 \, (78125 \, n+55542)

… 8 digits:
x \; = \; 390625 \, (256 \, n+33)
x \; = \; 256 \, (390625 \, n+340271)

… 9 digits:
x \; = \; 1953125 \, (512 \, n+109)
x \; = \; 512 \, (1953125 \, n+1537323)

… 10 digits:
x \; = \; 9765625 \, (1024 \, n+841)
x \; = \; 1024 \, (9765625 \, n+1745224)

… 11 digits:
x \; = \; 48828125 \, (2048 \, n+373)
x \; = \; 2048 \, (48828125 \, n+39935112)

… 12 digits:
x \; = \; 244140625 \, (4096 \, n+3761)
x \; = \; 4096 \, (244140625 \, n+19967556)

… 13 digits:
x \; = \; 1220703125 \, (8192 \, n+8125)
x \; = \; 8192 \, (1220703125 \, n+9983778)

… 14 digits:
x \; = \; 6103515625 \, (16384 \, n+9817)
x \; = \; 16384 \, (6103515625 \, n+2446398139)

… 15 digits:
x \; = \; 30517578125 \, (32768 \, n+8517)
x \; = \; 32768 \, (30517578125 \, n+22585503757)

… 16 digits:
x \; = \; 152587890625 \, (65536 \, n+41025)
x \; = \; 65536 \, (152587890625 \, n+57069119066)

… 17 digits:
x \; = \; 762939453125 \, (131072 \, n+73741)
x \; = \; 131072 \, (762939453125 \, n+333710340783)

… 18 digits:
x \; = \; 3814697265625 \, (262144 \, n+67177)
x \; = \; 262144 \, (3814697265625 \, n+2837143256329)

… 19 digits:
x \; = \; 19073486328125 \, (524288 \, n+118293)
x \; = \; 524288 \, (19073486328125 \, n+14770012057852)

… 20 digits:
x \; = \; 95367431640625 \, (1048576 \, n+967377)
x \; = \; 1048576 \, (95367431640625 \, n+7385006028926)

 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 

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About benvitalis

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

4 Responses to 8-digit numbers whose squares end with the same 8 digits

  1. paul says:

    Here is a list of numbers up to 20 digits

    25^2 = 625
    76^2 = 5776
    625^2 = 390625
    9376^2 = 87909376
    109376^2 = 11963109376
    890625^2 = 793212890625
    12890625^2 = 166168212890625
    87109376^2 = 7588043387109376
    1787109376^2 = 3193759921787109376
    8212890625^2 = 67451572418212890625
    81787109376^2 = 6689131260081787109376
    918212890625^2 = 843114912509918212890625
    40081787109376^2 = 1606549657881340081787109376
    59918212890625^2 = 3590192236006259918212890625
    3740081787109376^2 = 13988211774267263740081787109376
    6259918212890625^2 = 39186576032079756259918212890625
    256259918212890625^2 = 65669145682477392256259918212890625
    743740081787109376^2 = 553149309256696143743740081787109376
    7743740081787109376^2 = 59965510454276227407743740081787109376
    92256259918212890625^2 = 8511217494096854352392256259918212890625
    2607743740081787109376^2 = 6800327413935747244982607743740081787109376
    7392256259918212890625^2 = 54645452612300005057477392256259918212890625

    Paul.

  2. paul says:

    Here’s my MMa code for the list above, its based on this

    Mod[3n^2 – 2n^3,10^(2k)] where n=5 or 6 and k=1, the result is a new n and inc k to 2 etc

    length=20;n=5;k=1;list={};
    While[IntegerLength[n]<=length,AppendTo[list,{a=Mod[3n^2-2n^3,10^(2k)],a^2}];n=a;k++];n=6;k=1;While[IntegerLength[n]<=length,AppendTo[list,{a=Mod[3n^2-2n^3,10^(2k)],a^2}];n=a;k++];list=SortBy[list,#[[1]]&];Do[Print[list[[q,1]],"^2 = ",list[[q,2]]],{q,1,Length[list]}]
    

    Paul.

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