N = (1!)*(2!)*(3!)*(4!)*.....(19!)*(20!).
Precisely 1 of the 20 factorials needs to be removed from N to make it a perfect square.
What factorial needs to be removed? Provide adequate reasoning for your answer.
To be a perfect square the overall product must have an even number of each prime factor in its prime factorization.
We can ignore the 1!, so
The product of the 20 factorial numbers will be the product of all the numbers below:
2
2 3
2 3 4
2 3 4 5
2 3 4 5 6
2 3 4 5 6 7
2 3 4 5 6 7 8
2 3 4 5 6 7 8 9
2 3 4 5 6 7 8 9 10
2 3 4 5 6 7 8 9 10 11
2 3 4 5 6 7 8 9 10 11 12
2 3 4 5 6 7 8 9 10 11 12 13
2 3 4 5 6 7 8 9 10 11 12 13 14
2 3 4 5 6 7 8 9 10 11 12 13 14 15
2 3 4 5 6 7 8 9 10 11 12 13 14 15 16
2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17
2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18
2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19
2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
In the product, the prime factorization will have a number of 2's equal to
19 for the 19 2's +
2*17 for the 17 4's +
15 for the 15 6's +
3*13 for the 13 8's +
11 for the 11 10's +
2*9 for the 9 12's +
7 for the 7 14's +
4*5 for the 5 16's +
3 for the 3 18's +
2 for the 20
That adds up to 168 as the number of 2's in the prime factorization of the product of these factorials.
Similarly for the number of 3's:
18 for the 18 3's
15 for the 15 6's
12*2 for the 12 9's
9 for the 9 12's
6 for the 6 15's
3*2 for the 3 18's
That's 78 prime factors that are 3.
For 5's:
16 for the 16 5's
11 for the 11 10's
6 for the 6 15's
1 for the 20
That's 34 5's
Then the 7's:
14 for the 14 7's
7 for the 7 14's
That's 21 7's.
For the 11's:
10 for the 10 11's, and that's it.
Similarly
8 13's
4 17's
2 19's
Only 7 appears an odd number of times in the prime factorization of the product of the first 20 factorials. We need to remove (divide by) a factorial that has an odd number of 7's in its prime factorization, but an even number of every other prime factor.
Here, the computer can help us out:
for i=2:20
disp([i factor(factorial(sym(i)))])
end
The first number in each line is n. The remaining numbers constitute the prime factorization of n!
2: 2
3: 2 3
4: 2 2 2 3
5: 2 2 2 3 5
6: 2 2 2 2 3 3 5
7: 2 2 2 2 3 3 5 7
8: 2 2 2 2 2 2 2 3 3 5 7
9: 2 2 2 2 2 2 2 3 3 3 3 5 7
10: 2 2 2 2 2 2 2 2 3 3 3 3 5 5 7
11: 2 2 2 2 2 2 2 2 3 3 3 3 5 5 7 11
12: 2 2 2 2 2 2 2 2 2 2 3 3 3 3 3 5 5 7 11
13: 2 2 2 2 2 2 2 2 2 2 3 3 3 3 3 5 5 7 11 13
14: 2 2 2 2 2 2 2 2 2 2 2 3 3 3 3 3 5 5 7 7 11 13
15: 2 2 2 2 2 2 2 2 2 2 2 3 3 3 3 3 3 5 5 5 7 7 11 13
16: 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 3 3 3 3 3 3 5 5 5 7 7 11 13
17: 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 3 3 3 3 3 3 5 5 5 7 7 11 13 17
18: 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 3 3 3 3 3 3 3 3 5 5 5 7 7 11 13 17
19: 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 3 3 3 3 3 3 3 3 5 5 5 7 7 11 13 17 19
20: 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 3 3 3 3 3 3 3 3 5 5 5 5 7 7 11 13 17 19
(punctuation has been modified in the above output to improve readability)
The first one that has an odd number of 7's is 7!, but that also has an odd number of 5's, ruling that out.
The same is true of 8! and 9!.
But 10! has an even number of each of the other prime factors; Eureka.
We easily see that the rest all have an odd number (i.e., 1) of 11's and some other prime factors.
Of course we could have used the computer from the beginning:
prod=sym(1);
for i=2:20
prod=prod*factorial(sym(i))
end
for i=2:20
sr=sqrt(prod/factorial(sym(i)));
disp([i sr])
end
gives us the square roots:
[2, 213342473868862078312112010189602111501420783492883096482611200000000000000000*14^(1/2)]
[3, 71114157956287359437370670063200703833806927830961032160870400000000000000000*42^(1/2)]
[4, 35557078978143679718685335031600351916903463915480516080435200000000000000000*42^(1/2)]
[5, 7111415795628735943737067006320070383380692783096103216087040000000000000000*210^(1/2)]
[6, 7111415795628735943737067006320070383380692783096103216087040000000000000000*35^(1/2)]
[7, 7111415795628735943737067006320070383380692783096103216087040000000000000000*5^(1/2)]
[8, 1777853948907183985934266751580017595845173195774025804021760000000000000000*10^(1/2)]
[9, 592617982969061328644755583860005865281724398591341934673920000000000000000*10^(1/2)]
[10, 592617982969061328644755583860005865281724398591341934673920000000000000000]
[11, 53874362088096484422250507623636896843793127144667448606720000000000000000*11^(1/2)]
[12, 8979060348016080737041751270606149473965521190777908101120000000000000000*33^(1/2)]
[13, 690696949847390825926288559277396113381963168521377546240000000000000000*429^(1/2)]
[14, 49335496417670773280449182805528293812997369180098396160000000000000000*6006^(1/2)]
[15, 9867099283534154656089836561105658762599473836019679232000000000000000*10010^(1/2)]
[16, 2466774820883538664022459140276414690649868459004919808000000000000000*10010^(1/2)]
[17, 145104401228443450824850537663318511214698144647348224000000000000000*170170^(1/2)]
[18, 48368133742814483608283512554439503738232714882449408000000000000000*85085^(1/2)]
[19, 2545691249621814926751763818654710723064879730655232000000000000000*1616615^(1/2)]
[20, 1272845624810907463375881909327355361532439865327616000000000000000*323323^(1/2)]
showing only the elimination of 10! resulting in an integer square root. But of course that method completely obscures the role of 7.
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Posted by Charlie
on 2021-12-01 11:50:44 |
solution with a little help from the computer
