I did the first question in my head by observing that 2023 = 28 mod 35, so 2023^2023 = 28^2023 mod 35, then the powers of 28 mod 35 cycle over 28, 14, 7 and 21. Since 2023 = 3 mod 4, the third in the cycle 7 was the correct answer
Oh boy did he overcomplicate his answer. I used virtually the same method as you, noticed the cycling as well, but for some reason missed I basically had the answer by finding 2023 mod 4. I did some further work breaking it down more to get to the answer, which I now see was unnecessary.
I'm guessing this is group theory and group theory needs too much explanation for beginners? It's the way I did it and yes it offers a solution that can be done in our heads.
The second solution to the first problem is closely related to the Chinese remainder theorem. You can calculate the final remainder directly once you know the remainders mod 7 and mod 5. By CRT, we get (2*3*7 + 0) (mod 35) which gives 7. Where 3 is the multiplicative inverse of 7 (mod 5) i.e. 7*3=21= 1 (mod 5). The second term is 0 because the number is divisible by 7.
If you went back way less than that, you would absolutely revolutionize the field of mathematics with an average understanding of it in this day and age. You're forgetting how much has been proven and solved by people much smarter than us
Hmm...that makes me wonder if I go to way back in the past and revolutionize mathematics with what little knowledge I have as you said how much would today's mathematics get affected by that@@HelloIAmAnExist
I'll admit that second question would have taken some time and a lot of head scratching as I think my way through it. I wasn't up for that big of a challenge today so I just watched you haha. I was thinking about finding and analyzing a series representation would be the approach, and you did just that. Well done!
Using the Euler’s totient function and associated Euler’s theorem, you only need to consider the exponent mod 24 (the factors of 35 minus one, then multiplied). So 2023 to power 2023 mod 35 is the same as 28 to power 7 mod 35, which is 7. This is a useful trick in public key cryptography, especially RSA.
I did the first question by the mod7-mod5 method. 2023≡23-2 mod 7, since 7|1001, and 7|21, so I can simply ignore 7 and focus on 5. Then 3^3=27, so the answer is congruent to 2 mod 5 and 0 mod 7, which is obviously 7.
Basically the way I did it as well. Only thing I did different was how to figure out that 7|2023: Obviously 7|2100, so 2023≡-77 mod 7, and since 7|77, we're done.
Can we also do it like this:2023 is the multiple of 7. The remaining factor is 5. The units digit of 2023^2023 is 7. The factors of 5 have 0 /5 in their ones place. When divided by 5 it will remainder 2 but then it will not be able to divide by 7. But if we look at the next factor it is divisible by both 5 and 7 and leaves remainder of 7
A slightly different solution is to set u = 10¹⁰⁰+3; then the expression becomes floor((u-3)²⁰⁰/u). Expand by the binomial theorem; the last term is then 3²⁰⁰/u =9¹⁰⁰/(10¹⁰⁰+3), which is clearly < 1 and gets ignored by the floor function. Since u ≡ 3 (mod 10); the remaining expression reduces to 3¹⁹⁹((1-1)²⁰⁰-1) This can be generalized to floor(10^(2m²)/(10^m+3)), whose last digit is ≡ 3^(m-1) (mod 10).
35=5×7, both are primes, so first I did it mod 5: 2023^2023 = 3^3 = 27= 2 (mod 5) Because 2023 = 3 (mod 5) and 2023 =3 ( mod 5-1) for the exponent using Fermat's little theorem. Next noticed that 7 divides 2023, so 2023^2023=0 (mod 7). Which number under 35 satisfies both conditions? Just starting with 2 and repeatedly add 5, until it is a multiple of 7. Well, the first try works, so the answer is 7.
Just clarifying your last step (because I'm not smart enough to directly see this): With 2023²⁰²³ = 5m + 2 = 7n, let m = 7k + q. So 5(7k + q) + 2 = 7n. Modulo 7 gives, 5q + 2 ≡ 0 (mod 7). So q=1 is a solution. Thus 2023²⁰²³ can be written as 5 (7k + 1) +2 = 35k + 7.
Puzzle, how can you get the Lorentz factor for any percentage of light speed using simple geometry? My solution, draw a 90 degree angle with the vertical side being 1 unit long and representing the speed of light, and the horizontal line extending as far as required when I describe the next steps. Draw a half circle from the top to the bottom of the vertical line, ie, center of the half circle being at the center of the line. Now choose whatever percentage of light speed you want to find the Lorentz factor for and mark a point on the vertical line that percentage of the distance up from the bottom end where it joins the horizontal line, ie, 0.5 light speed would be marked at 0.5 unit up. Next draw a quarter circle between that mark and the horizontal line, ie, centered at the intersection of the vertical and horizontal lines. Now you will have the half circle and quarter circle intersecting at a center point. Next draw a line from the top of the vertical line through the intersection point just described and extend it until it intersects the horizontal line. The length of that line will be the Lorentz factor for the chosen percentage of light speed. The procedure is ridiculously simple and entirely accurate. You would need to use a geometry drawing (CAD) program, of course, to get accurate results. The reason I said to draw the horizontal line as long as required when I describe the steps is because the higher the percentage of light speed you choose, the longer the horizontal line will need to be. The closer you get to 100% of light speed, the closer the horizontal line would get to being infinitely long, so this method is really only practical up to a certain point, like about 90% of light speed or less. I'm not suggesting that this method has any practical use, it's just to show that it can be done using simple geometry, not that you would actually want to do it instead of using the equation. If you do a search for a simple geometric method of obtaining Lorentz factors you won't be able to find one so I created one myself.
1st question is about cyclic groups. we show that C35 is isomorphic to C5 X C7. 2023^2023=3^3=27=2 in C5 2023^2023=0 in C7 extended euclidean algorithm shows that the answer is 7 in C35. very easy with euler's formula
For the first one there's also a binary way which is used in cryptography. 2023d=111 1110 0111b If the next bit is 1: rx=r(x-1)*fx%35, if the next bit is 0: rx=r(x-1), fx is in both cases: fx=f(x-1)²%35 f1=2023 r1=1*f1%35=28 f2=f1²%35=14 r2=r1*f2%35=7 f3=21 r3=7 f4=21 r4=r3=7 (bit is 0) f5=21 r5=7 (bit is 0) f6=21 r6=7 f7=21 r7=7 f8=21 r8=7 f9=21 r9=7 f10=21 r10=7 f11=21 r11=7 Result is 7
came here just to say this, it may seem complex but it's just fast exponentiation and modulus, maybe not the most elegant solution, but surely very easy
I did the first one slightly different by using the prime decomposition of 2023, noticing it's divisible by 7, and thus rewriting the thing as 2023^2022*289/5. Now you just need to check mod 5 (which means you can just check the last digit). So mod 10, that is: 3^2022*9 = 9^1011*9 = 9^1012 = 1^506 = 1 mod 10. So it's the same mod 5. Now we just multiply that by 7 (because we divided both quotient and remainder by 7 in our first step) and get a remainder of 7.
on the 2nd question I got the right answer even if how I did it might not be mathematically sound. I know if you divide 10 by 3 you get 3.3333(repeating) and the other numbers in the questions are factors of 10 so the last digit should also be a 3. I never learned any of that mod stuff lol.
For the Putnam problem, let's break down the expression 10^20000 / (10^100 + 3). We can represent it as the sum of two parts: (10^20000 - 3^200) / (10^100 + 3) + 9^100 / (10^100 + 3). It's important to note that the second part is less than 1. To demonstrate that the first part, obtained by subtracting 3^200 from 10^20000 and dividing by the sum of 10^100 and 3, is an integer, consider (10^100)^200 - 3^200. This is proportional to (10^100)^25 + 3^25, and this proportionality extends to 10^100 + 3. Therefore, the floor of 10^20000 / (10^100 + 3) is equal to the first part. Looking at the last digit of the numerator modulo 10, we find that 10^20000 - 3^200 is equivalent to 10 - 9^100, which is congruent to 9 modulo 10. Similarly, the last digit of the denominator, 10^100 + 3, is congruent to 3 modulo 10. Therefore, the last digit of the entire expression is 9/3, which equals 3, as this fraction results in a whole number.
About the last paragraph: why would the last digit of a ratio (assuming it is an integer) be the ratio of last digits of the numerator and denominator? The last digit of 125/25 is not 5/5=1
@@TaladrisKpop Quite straightforward! We can represent the fraction modulo 10 as 9 ≡ 3x (mod 10), where x is the unknown digit. Since 3 is coprime with 10, dividing both sides by 3 doesn't alter the equation, leading to x ≡ 3 (mod 10). However, in your example, 125 ≡ 25x (mod 10), 25 is not coprime with 10. Consequently, you can't naively divide both sides by 25."
@@TaladrisKpop Division modulo n does not work in general. But it works in the example above because 3 is coprime with 10. You need to have a multiplicative inverse for it to work. The multiplicative inverse of 3 mod 10 is 7 (because 3*7 = 1 mod 10). So 7*9= 3 (mod 10). Multiplying by 7 is equivalent of dividing by 3. In your example, 25 is not coprime with 10 and there is no multiplicative inverse k such 25*k =1 (mod 10).
Very clever solution. I assume you used the difference of squares to reduce (10^100)^200-3^200 and show it was divisible by (10^100)^25 + 3^25. Showing that (10^100)^25 + 3^25 was divisible by (10^100 + 3) was not initially obvious to me until I recalled that the sum of odd powers x^(2n+1)+y^(2n+1) have a factor of (x+y).
@@gregoryknapen9133 Yeah, I assumed that the sum of odd powers x^(2n+1) + y^(2n+1) has a factor of x+y because it's a remarkable fact that's easy to prove. Implicitly, as you correctly pointed out, I also employed the difference of squares for the initial reduction.
I just liked it, bro :). And it isn't that random. I think if I had thought enough about the second question, I could have solved it. It isn't that hard. I don't like it when it math gets very complicated as well. But the second question isn't that complicated. Probably thats the question you couldn't solve as well. (Sorry if i made any grammar mistakes.)
@@leif1075 You can guess what can happen when you use that serries. I mean you will have only 1 term to consider the remaining when it is divided to 10. And you can guess it.
Can you make a video about this interesting problem--> If z = x + iy is a complex number where x and y are integers. Then, the area of the rectangle whose vertices are the roots of the equation----> w z^3 + z w^3 = 350, where w=conjugate of z
For problem 1, we can also use sum of binomial coefficients sum of nCr = 2 to power n (2^n). Rewrite 7^2023 as (1 + 6) ^ 2023 and expand the binomial. Since the unit digit of 6^n is always 6, (1 + 6) ^ 2023 can be expanded as 1 + 6^1 x 2023 C 1 + 6^2 x 2023 C 2 + ... + 6^2023 x 2023 C 2023 = 1 + 10j + 6 (2023 C 1 + 2023 C 2 + ... + 2023 C 2023) = 1 + 10j + 6 (2023 C 0 + 2023 C 1 + 2023 C 2 + ... + 2023 C 2023) - 6 = 10j - 5 + 6 (2^2023) Since the unit digit of 2^n can be generalised as: n = 4j+1 => unit digit = 2 n = 4j+2 => unit digit = 4 n = 4j+3 => unit digit = 8 n = 4j+4 => unit digit = 6 => unit digit of 2^2023 = 8 Hence 10j - 5 + 6 (2^2023) can be rewritten as 10j - 5 + 6 (10m + 8) = 10j - 5 + 60m + 48 = 10(j + 6m + 4) + 8 - 5 = 10(j + 6m + 4) + 3 So we have 2023^2023 = 35k - 7^2023 = 35k - (10(j + 6m + 4) + 3) = 35k - 10(j + 6m + 5) + 10 - 3 = 35k - 10(j + 6m + 5) + 7
The first question could be solved this way: 2023 = 28 mod 35 2023^2 = 14 mod 35 2023^3 = 7 mod 35 2023^4 = 21 mod 35 2023^5 = 28 mod 35 ... Observe that the cycle repeats after every 4th power of 2023. Since 2023 = 3 mod 4, our answer is the 3rd element in the cycle, or 2023^3 = 7 mod 35. Therefore the remainder is 7.
First problem. The number is 0 mod 7 and 2 mod 5 (can use FLT and division algorithm on the exponent) 7 is a number with the above properties so 7 is a possible remainder mod 35. By chinese remainder, 7 is only possible choice. Hence answer is 7.
I used Chinese remainder theorem 6:04 after getting mod 5 and mod 7 values and then got 35k+7. By the way what books do you recommend to study combinatorics in detail for national olympiads
i did problem one like this since we are finding the last digit of 2023^2023 we can find the last digit of 3^2023 as the tens/hundreds and etc are useless when finding the last digit and we only have to study the behavior of the last digit of 3^2023 so keep multiplying until we have a repeat 1: 3, 2: 9, 3: (ignore 2) 7, 4: (ignore 2) 1, 5: 3 so we have 4 different numbers 3, 9, 7, 1. so we do 2023 mod 4 which gives 3, so we pick the third one which gives the digit 7, so 2023^2023 ends in 7
imagine if someone just brute forced the first question, goes through a stack of paper to multiply 2023 by itself 2023 times and then does long division on the resulting monstrosity
that is physically impossible for the human mind, and it will never be possible for a human to do that. in fact, no computer currently can even imagine to perform that computation. so unless the human has immortality and is extremely patient, good luck.
@@rohangeorge712 I wouldn't say physically impossible. After all, the actual computations can be done by any 5 year old. It's just that the human in question would probably not survive to finish the calculations. Computers could acutally do it reasonably easily, however. If we approximate 2023 by 2000 and 2^10 by 10^3, you would get that 2023^2023 ≈ 2000^2023 = (2^2023)(1000^2023). 2^2023 ≈ 10^674 and 1000^2023 = 10^6069, so their product will be around 6743 digits, which is quite manageable for any computer.
ur right about the computer part, but it is physically impossible because as you said, the human would not survive to finish the calculations. which is why its impossible.@@phoquenahol7245
First one is from jee mains. That doesn't even come close to jee advanced lol. Although I find these hard but I'm saying in general that jee advanced has much tougher ones
I've been meaning to ask this question for some time: Why do you use the phrase "the opposite of x" rather than "negative x"? I recognize that I may have an American English bias, but "opposite" seems ambiguous in a mathematical context. You're using it to mean "additive inverse" when it could just as logically mean "multiplicative inverse". So the "opposite of x" could be "1/x" just as easily as "negative x". Am I missing something? 🤔
Sir, will you please elaborate this? 0.9repeating =1(approaches 1) Does that mean 0.0repeating1=0(does this approach 0) so that I may better understand the concept of a/0=infinity (a>0). And, what about 0.9repeating+0.0repeating1?
If a person steals 100 ( any currency ) from a shopkeeper an then buys fruit worth 70 from that shopkeeper and gives him that 100 currency note , shopkeeper returns 30 Now total gain of that person will be ?
For the second one, you know 10 to the power of n is a multiple of 10, so you can surely ignore the powers. Knowing, simplified, one way or another the fraction, somewhere, will be 10/13, you do this equation, which gives you 0.7692307692.... This is a repeating decimal so surely knowing the last digit of this repeat is 3, it would give you your answer? Am I wrong? Was this complete luck?
I have literally no idea what maths you used in this video. Never seen this “mod” system before. I did manage to get the correct answer for question 2 however, using only my logic. Maybe I got lucky haha. Also, the thumbnail and question two are worded differently. What is the “units digit” versus what is the “last digit”. I don’t know what a “unit” is, in this instance, but I rationalized that a number ending in 0 and being divided by 3 would result in a 1/3 fractional remainder. So .3333333 repeating. You give the proofs, I bumble my way through to maybe get the right answer haha
I think the answer to the second question in the video is wrong. I got 7 as the units digit of the floor of 100^200/(100^100+3), not 3. Let's remove the word "floor" in the sentence above and replace it with the words "integer part". But since we are asked for "the units digit" of a number, it already implies we are getting it from the integer part of that number. So we can remove the words "of the floor" in the sentence above because it is not necessary. So now we can change the original question and simplifly it to: What is the units digit of 100^200/(100^100 + 3)? The answer is 7. Proof: 1. let 100^200/(100^100 + 3) = x 2. ((100^100 + 3)(100^100 - 3) + 9)/(100^100 + 3) = x 3. (100^100 + 3)(100^100 - 3)/(100^100 + 3)+9/(100^100 +3) = x 4. (100^100 - 3) + 9/(100^100 +3) = x 5. Notice in step 4 above that the fractional part of x is 9/(100^100 + 3). We can ignore the fractional part of x because it does not contain, or determine, the units digit of x which is what we are asked to find out. The integer part of x is 100^100 - 3. That is a 1 with 200 0s after it minus 3. This is the important part of x. This integer, 100^100 - 3, contains the units digit which is the last digit to the right. 6. The units digit of 100^100 - 3 is 7. If you have any integer which is a 1 with many 0s after it and then subtract 3 from it you get an integer which is many 9s with a 7 after it. For example: 10 - 3 = 7 100 - 3 =97 1000 - 3 = 997 10000 - 3 = 9997 100000 - 3 = 99997 1000000 - 3 = 999997 1000...000 - 3 = 999...997 Notice that there is one less 9 on the right side of the equal sign than the number of 0s on left side of the equal sign. Therefore the units digit of 100^200/(100^100 + 3) is 7.
Had a more prosaic (boring?) answer to Q1, as 2023 is divisible by 7 the sum boils down to 7^(6068)/5. 7^(6068)/5 is easy enough for the remainder as the pattern of remainders of powers of 7 divided by 5 is easily found to be 2,4,3,1,2,4,3,1... and 6068 is divisible by 4, so the remainder is 1. Multiplying back up to get the remainder on division by 35 gives 7. And THAT's the answer.
For the second answer, Here is my thinking so take the number 10/3; It will be a 3 Now do 1000/103; It'll Still be the same, 3 Do it again, 1000000000000000000000000000000000000000/10000000000003 it will be 3. This pattern goes on infinently. So there is your answer, 3! There is some exceptions ans as a kid, I wont tell you because I, Myself am confused :D.
6? It was seven😔. Reasoning: 2023= 19×19×7 35=7×5 Divide both by 7 19×19 5 19×19×19... always has one at end 5×5×5×5... always has 5 at end ....1-..5= 6
@@Padhai_Kar_Jake No i promise you can get a solution this way: 2023= 17×17×7 And (17×17)^x always ends in 9 or 3 And 5^y always ends in 5 And 9-7=2 And 12-5= 7 🤯😖🤔
@@blackholesun4942 it is wrong 😂, (17×17)^x deosn't always get 9 or something else rubbish you are talking about, But if you are interested in learning this search on youtube for problems on number theory, you will get a clear idea for which you are wrong
@@finnwilde - Post the alleged questions (or a certain sample) and/or Internet sites where the alleged problems are "lazily and invalid lyrics notated" to let others examine them. It may be that you are mistaken, and that they can be resolved.
@@robertveith6383I think the point is, the order of operations is often assumed rather than stated explicitly. In those cases, it’s perfectly reasonable to say that the question was poorly written, since it made an assumption that was not related. This is even bigger of an issue in exams that allow calculators - different brands and models have different order of operations, and it’s not something many people are taught in school, so tends to be an unpleasant surprise. As to whether PEMDAS, BODMAS etc. is technically ‘wrong’, no, they’re all perfectly correct. They’re just a component of notation if you think about it. You can always re-write an equation in any ordering you like, and if you use enough brackets, you can totally override any notational ordering. Notice that in all ordering schemes, brackets (parentheses) are always the highest priority!
you can solve the second one in an easier way x=10²⁰⁰⁰⁰ y=10¹⁰⁰+3 x mod (y) =( 10¹⁰⁰)²⁰⁰ mod y =(10¹⁰⁰-y)²⁰⁰ mod y =(-3)²⁰⁰ mod y =9¹⁰⁰ mod y = 9¹⁰⁰ ( since 9¹⁰⁰
Do you expect most pl to understand this notation..I got lost after the first line..what does x(mod y) even mean sorry..is this something even many math majors would recognize? Just wanted to understand sorry..
It depends on what you mean by "half" if you meant divide 50 by its half then the answer would be 2,so adding 20 will make it 22 If you mean the half of 50 ,which is 25 ,add 20 and you'll get 45
@@icecreamheh2131 well thats exactly what I mean because I actually found this on a short and the comments have different answers which starts the controversy (i think)
Yesterday, I saw one of ur videos about a puzzle in which we need to find out how many liars and truth tellers are there out of 100 people and I saw a mistake there: I do not think this is correct, see when u were comparing the answers of person 0 and people from 1 to 99 u took the possibility that either 0 shook hands with all liars or other person and 0 did not shook hands.....but here is a mistake, see by saying this u are assuming that person 0 is a truth teller already. There should have been one more possibility that 0 is a liar. Also due to this assumption only u got the final answer that only 0 is the truth teller!
It could be done either by dividing it with 10, or observing the repetition of the unit digit when 3 is squared, cubed, quadrupled, so on and so forth.
Hi Presh, making same comment I made in the last video with multiple problems once again. Please change the order of these videos where you show problem, solution, problem, solution instead of all problems up front and all solutions after. If a person has to skip around in your video to order it properly, the order is an impediment to the process of watching it.
You’re not making any sense. Both questions are stated at the beginning so that people can go away and try to solve them. Putting them in q, a, q, a order would totally ruin that. The video as it is does the job it’s creator intended perfectly IMHO.
My half-assed guesswork was (10 with infinite zeros) divided by (ten with huge number of zeros plus 3) is basically equivalent to 10^infinite divided by three, with gives an end of 3.33333... which gives a units digit of 3. Probably not logically valid but it had a 1 in 4 chance of working.
Hello pre, can i have a question. In question 2, you said that the sum of alternative series is less than the first term, What is the reason ,can anyone explain more deeply, thanks
The alternating series is a geometric series with first term a = (-3)^200/u and common ratio r = -3/u, where u = 10^100. Since |r| < 1, the series is convergent. Therefore, we can evaluate it by the formula S = a/(1-r). We have S = ((-3)^200/u)/(1+3/u) = 3^200/(u+3). But S < 3^200/u = 3^200/10^100 = (9/10)^100 < 1.
In question 1a, in the last step when we were doing calculation for -(7^2023)÷35, what if we do it as -(7^2022)÷5 (as 7 get cancelled) ? By doing this I am getting different answer. Can somebody please help me?
Yes the 7 gets cancelled now changing the look of the fraction to (289)(2023^2022)/5. It is the same final improper fraction number but just a new way of expressing it. Anyway, the answer is some integer +.2 decimal. 7/35=1/5=.2 decimal. A .2 decimal is equal to a remainder of 1. So if the 7 factor of 35 cancels out, leaving a division by 5, then the remainder is 1.
