Solving 2 questions from the world's hardest exams

Thanks to Sanket for suggesting problem 1! It is adapted from the JEE Mains 2022 test in India. Problem 2 is from the 1986 Putnam competition, A2.

0:00 problems
0:58 solution 1a
4:10 solution 1b
6:50 solution 2

Problem 1
Brainly
Sarthaks
StackExchange

Problem 2
John Scholes Putnam problems
StackExchange
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[johnnypoker]

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

[gregoryknapen]

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.

[khangphuc]

After this question, I just realized I need a time machine to go back to year 1 so the questions there are a lot easier (1^1 ÷ 35)😂

[almazantitiming]

Number 2 is actually pretty easy, if we divide 10^n by 3, 30, 300 etc we always get a number with infinite amount of 3

[mike.]

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!

[quasirandomname]

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.

[koenth]

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.

[bobzarnke]

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).

[shlokdubey]

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

[pierreabbat]

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.

[Nowmylifeis.swee.like.cinnamon]

1st question is of JEE MAINS, yes a nice question

[pauselab]

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

[RolanRoyce]

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.

[L]

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

[learningforeverofficial]

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?

[ozelot]

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

[zeroplays]

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

[philstubblefield]

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? 🤔

[hdthor]

11:59 You can instantly know that x^4 = 1 mod 10 because phi(10) = (2-1)(5-1) = 1*4 = 4. So (x^199 = x^(199 mod 4) = x^(-1 mod 4) = x^3) mod 10.

[albinwikgren]

I have a intresting problem from my Swedish math 4 final.
Find all values of z
1+Imz=(Rez)³
IzI=3
Sorry if my english is bad 😊