For last digit problems, it is expected to focus the computation on the last digit of the base. 5! and other bigger factorials have last digit zero. So the lkast digit of the base is determined by the first four factorials in the sum.
For anyone wondering, (1! + 2! + ... + 1982!)^1982 has 11,250,101 digits. The _first_ digit is a 7. The digit before the final digit is a 6. I'm sure a certain contingent of number lovers will get a kick out of that. (All assuming I haven't made any mistakes.)
@@PrimeNewtons The last two digits perhaps, at least, depending how you feel about the meme status of that particular two digit number. The first digit is a lot harder. Stirling's approximation would be a good tool, for example, but it only works for finding approximations to logarithms of factorials, not the logarithm of a sum of factorials. While the first digit of the sum almost always equals the first digit of the largest factorial on account of the size difference, that small difference has an effect when raising to a large power. That is, the power of the largest factorial will almost certainly not have the same first digit as the power of the sum. I'm not sure what tricks can be used to mitigate this. As the expert, you might know something I don't. As to how I did it: I had something calculate the sum _in full_ before taking the logarithm - it's "only" 5677 digits after all. Then it was a matter of multiplying that by 1982 and raising 10 to the fractional part. More of a computer exercise than one for blackboard or pen and paper.
My thought process before watching the video: Finding the last digit means finding the value (mod 10), so only the last digit matters. noticing that all n! with n>5 will give us a multiple of 10 (2*5), so what really matters is the first 4 terms in the bracket, ie. 1!+2!+3!+4! = 1+2+6+24 = 33 which is congruent to 3 (mod 10), so the whole thing is congruent to 3^1982 (mod 10). Notice the final digits of 3's powers are in 3, 9, 7, 1 period, which means 3^4n+k (mod 10) is congruent to 3^k (mod 10), and thus 3^1982 (mod 10) is congruent to 3^2 (mod 10), which means the final digit must be 9
We had the same thought process for the first half. I didn’t understand how your second half work out though, I just took 1982(mod 5), which gave 2. And 3^2 = *9*
@@_Unknown420_ Why mod 5? I think, it must be mod 4, because the length of the cycle of last digits in the powers of 3 is 4. The last digit of 3^1972 is 1 and not 9 because 1972 mod 4 is 0 (or 4), so 3^1972 is equivalent 3⁴4 mod 10.
You being a math person your writing is very good and I appreciate that. As me myself do not have that good handwriting and I have seen no math person having such good handwriting ❤
This is one of my favorite of your videos. Especially when you corrected the equality sign and then wrote it out correctly to include the phrase "last digit of ". Thank you for this video :)
Excellent video! Another (faster) solution is by directly using Euler's theorem: Note that 3^(φ(10)) ≡ 1 mod 10, where φ(n) is Euler's totient function. So we have 3^1982 = 3^(4*495 + 2) = ((3^4)^495) * 3^2. By Euler' theorem that is ((3^4)^495) * 3^2 ≡ 1 * 3^2 ≡ 9 mod 10. Since you made a video about tetration, I would like to suggest that in a future video you could calculate 3 arrow arrow 21 mod 100 using Euler's theorem and the Chinese remainder theorem.
Simply thought this... Factorials -> 5 and beyond must end with a 0 -> Only focus on the ones below 5 We have 1 2 6 4, and we're left with only 3 after summing all of them Exponents of 3 ends with 3 9 7 1 that keeps repeating 1982 -> 1900 can be divided by 4 (Any multiple of 100 can be divided by 4), 80 is divisible by 4, and we're left with 2 The 2nd in that loop of last digits is 9
The thing about factorials is that for every 5 you increase the number by, you get one more zero on the end. For example, 5! ends in one zero, 10! ends with two zeroes, 15! ends with three. Essentially you can divide the number by 5, drop the remainder, and that's how many zeros are on the end. So 1982! would end with a whopping 396 zeroes.
DAMN, you went really wild with this one. We were just playing around with some interesting calculus and algebraic equations a while back, now full on modular mathematics.
I at least found out the number would be odd then I watched a bit and realized about the property of factorials greater than 4! thanks to the awesome explanation
Since all factorials greater than 5! have 0 as last digit (because they include a factor of 5 as well as a factor of 2 and therefor are dividable by 10), the last digit of the sum of that factorials is the same as the last digit of 1!+2!+3!+4!=1+2+6+4 mod 10=3. So we have to look for the last digit of 3¹1982 (the digits before the last digit does not have any effect on the last digit in the power). So let's look at the last digits of the powers of 3: 3¹1 mod 10=3 3²2 mod 10=9 3³3 mod 10=7 3⁴4 mod 10=1 For all further powers, the cycle starts from the beginning. So 3²2 mod 10=3^6 mod 10=...=3^1982 mod 10. And we know, the 3^2 mod 10=9 mod 10=9. So the answer is 9.
Sir instead of writing 1982mod4 we simply could have divided 1982 by 4 and the remainder would be the answer in this case it is 2. Its also an interesting way to solve it.
first of all a question not about the video itself: how do you have the time and ideas to make one/two videos a day? ok math offer enough topics but even than wow! 0:34 of caurse not! that is posible but not in a human live time :D exsept you would know how to simplify the faculties 0:55 you have ti figer out what is the last digit of the faculties(sum) and than you just have to finde out the last digit of this digit before up to 1982 2:55 that was not my first idea but i even thought that that case must happend becouse at the point were you have a number dividable by 2 and a 5 the value of the sume ends with at least one zero so it woun´t change the first digit! 4:51 i have an idea we have always the same last digits so wie just need the MOD 8:15 you never learn out! nice video again! LG K.Furry
@@PrimeNewtons thanks that is great to hear! one question: is it posible to ask you for an explaination for sth. with math and you give an answer + explaination? (just say yes if that would not make any difficulties) and i hope again my english is readable :D LG K.Furry
I have studied the question in Z/10Z .. each rest class is the last digit of the number.. the first element of the sum are 1, 2, 6, 24, 120.. after 120 al the number end with zero..therefore we have 1+2+6+4=13 wich ends with 3.. so we have 3^1982= 9^1991.. 9 is equivalent to -1.. so we have that the number is equivalent to -1, because the exponent is odd.. and -1 means that the last digit is 9
1! + 2! + 3! + 4! = 33 5! = 120 (5! and beyond can be ignored since they all end in 0 (not changing the last digit) 1982 = 2 (mod 4) So, 33² = 1089 Therefore, last digit is a 9
also 3^1982=(3^2)^991=9^991 9 is congruent to -1 mod 10 so this is the same as (-1)^991 since 991 is odd this is congruent to -1 which in turn is congruent to 9
The final digit of the inside will be the final digit of the sum of the first 4 digits, because 5! and above will all end in 0. 1 + 2 + 6 + 24 = 33, so the last digit is three. The final digit of powers of three go in a cycle. 3 9 7 1 And repeat. Because this cycle is 4 numbers long, we find the remainder when the power (1982) is divided by 4, which gives a remainder of 2. The final digit will therefore be the final digit of 3^2, being 9.
Before watching ... We are only interested in the last digit so 1) we only want the last digit of the sum and 2) we only want the last digit of the power. Above 4! all factorials end with a zero so have no effect on the last digit. Factorial 1 to 4 end with 1 2 6 4 so the sum ends with 3. powers of 3 end with: 3, 9, 7, 1, (3, ...) so it cycles every four powers 1982 mod 4 = 2 so it ends with 9.
Every factorial bigger or equal than 5 has 0 as a last digit and thelast digit of the sum 1!+2!+3!+4! Is 3 and since its 3 you divide the exponent by 4 and if the rest is 0 then the last digit is 1 if the reminder is 1 the last digit is 3 if the reminder is 2 the last digit is 9 and if the reminder is 3 the last digit is 7 since the reminder of 1982 in the division by 4 is 2 then the last digit is 9 ( this is before seeing the answer in the video
did this shi in my head im goated 5! and on ends in 0 so we add 4!+3!+2!+1! = 33 so it ends in 3. then we notice 3^1 ends in 3, 3^2 ens in 9, 3^3 ends in 7, and 3^4 ends in 1 and the cycle repeats. when u divide 1982 by 4 the remainer is 2 so you countinue 2 from 1 which would be 3^1981 to 3, and 3^1982 to 9
Can you solve the following problem... I've been trying solving it for months... If A , B , C , D are positive integers How many possibilities could A + B + C + D = 20 As A is always greater than B
last digit of n! Is always zero for n >= 0 Now 1 ! + 2 ! + 3 ! + 4! = 33 Again 1982 = 495 * 4 + 2 Cyclic nature of last digit in the power of 3 i.e ( 3, 9, 7, 1) implies last digit of ( 1 ! + 2 ! + .. + 1982 ! ) ^ 1982 = 9
my attempt to solve 1! = 1 2! = 2 3! = 6 4! = 24 5! = 120 6! = 720 ... the last digit of 1! + 2! + ... + 1982! would be the last digit of 1 + 2 + 6 + 4 + 0 + 0 + ... + 0. which would be 3 so we need to find the last digit of 3^1982 last digit of 3¹: 3; 3²: 9, 3³: 7, 3⁴: 1, 3⁵: 3, ... the digits 3,9,7,1 repeat 1982 mod 4 = 2 so 3^1982 mod 10 = 3² mod 10 = 9 I'm guessing it's 9 🤞
Another possible way Is to (1!+2!+3!+....+1982!)^1982 mod 10 =(1!+2!+3!+4!)^1982 mod 10 33^1982 mod 10 =3^1982 mod 10 =9^991 mod 10 =-1 mod 10 = 9 mod 10
@@comdo777 The idea is to eliminate everything not related to the final digit. Within the bracket, all terms larger than 5! must be a multiple of 10, so all their multiples must also be and thus they'll not affect the result. Then we can observe that when raising 3 to higher powers, the final digit will repeat in a period of 4 (3,9,7,1), which means we can take away as many 4s from the exponent as we want, which leaves us 3^2
