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Factorial Removal Remix (Posted on 2024-11-29) Difficulty: 3 of 5
N = (1!)*(2!)*(3!)*(4!)*.....*(21!)*(22!).

Precisely 2 of the 22 factorials needs to be removed from N to make it a perfect square.

What two factorials need to be removed? Provide adequate reasoning for your answer.

No Solution Yet Submitted by Brian Smith    
Rating: 4.0000 (1 votes)

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Solution computer solution Comment 3 of 3 |
clearvars
digits 1000
idx=combinator(21,2,'c');
n=vpa(2):22;
sf=prod(factorial(n))
n= factorial(n(idx));
for i=1:length(n)
  q=sf/prod(n(i,:));
  sr=round(sqrt( q));
  if sr^2==q
    disp(n(i,:))
  end
end

The number before division (including all the factorials):

73185187344722251685917129061596359994436879170423331597220791224873045503248922
81148195654813491347532442912806883951929242763804569032720897710945101414400000
0000000000000000000000000000000000000

Pairs of factorial numbers to divide by:

[2, 39916800]
[6, 479001600]
[24, 479001600]

For reference as to which factorials these are:

>> for i=1:13 disp([i,factorial(i)])
end
    1           1
    2           2
    3           6
    4          24
    5         120
    6         720
    7        5040
    8       40320
    9      362880
   10     3628800
   11    39916800
   12   479001600
   13  6227020800
   
indicates factorials (2,11), (3,12) and (4,12) work


  Posted by Charlie on 2024-11-29 12:53:32
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