You, me, and my first International Math Olympiad problem

I remember when I was 14 and my friend, who was also into math, took some IMO problem to school and showed me. We were both able to solve it, which really boosted our confidence back then. Later the friend won a gold medal at IMO (1,5 year later) and I also did have decent results on the national stage. It is good to have some easier tasks at IMO, just to give kids joy and a feeling they can work hard to reach for their dreams.

mans got the supreme hoodie

this video is pretty cool IMO

Tinha de ser japones.

When you compute the upper bound on the number of digits using log base ten, you take the whole number upper bound of the log function as equal to the number of digits — but when the log function equals an integer n, the input actually has n + 1 digits. For example, log base 10 of 100 = 2 but 100 has three digits and log base 10 of 1000 = 3 and 1000 has four digits. Thus when you use the log rule to find the number of digits of 4444^4444 and get 4444*4 = 17776, you should really have done 4444*5 = 22220. This is because 4444 has less than 5 digits because it is less than 10000. This makes sense because it has 4 digits so it should be less than 5, not less than 4. 

This doesn't change the answer because in the next step it reduces to 47 rather than 46 but then when you replace 46 with 49, you still get 13. However, I thought I would note it because it confused me when I was watching.

I took number theory and still had no idea how to get to the last answer…

This video just shows how fabulous calculators really are. Would be able to solve this in 1 min.

If we see carefully, 4,444 when divided by 9 leaves a remainder of 7, and 7 raised to any power will give remainders in cyclic patterns of 7, 4, 1, and then 7, 4, 1, and then 7, 4, 1, and so on....

So, I'd see it as if its digit sum is similar to the quantity 7^7, which, at the end, lands up to be 7.

This is a very rough way to deal with the problem. But, sometimes, even very rough means to solve a problem can give quick results.

Hello!
You should definitely check out 2nd question of IMO 1995 related to inequalities. Its a proving question. You can solve it by using simple concepts!!

i really liked the part when foufoufoufoufoufoufoufoufoufoufoufoufoufoufoufoufoufoufoufoufoufoufoufoufoufoufoufou

I am not on this level yet but I was able to follow it with you ☺️✨️💛 your joy for solving this was adorable, thank you for sharing

Saw the question and thought the answer would be remainder when divided by 9. Although solution is incomplete without proving it is less than 13. I didn't know how to prove that

I somehow solved it on my own and got the correct answer

You're an excellent teacher. Entertaining and educating. Thank you

The question makes my brain hurt

It actually seems pretty easy to prove that
n = d(n) (mod 9)

If x is a digit of n,
x * 10^k = x * 1 (mod 9)

To calculate 4444^4444 mod 9. You can mod the exponent by phi(9)=6. End up having 7^4 = 4^2 = 7 mod 9

This basically should mean that d(d(d(x))) is a very nasty way to replace "x mod 9" for any non-astronomical number. (4444^4444 has an astronomical size, but he is still on the brink.)

My way of solving this is to first bound the number with 4000^4444 and 4500^4444. Then, compute the approximate log_10 for the lower and upper logarithm bounds: 4444 * (3 + 2 * 0.301) and 4444 * (3 + 2 * 0.4771 - 0.301). You will get the rough bounds (16007.2, 16234.9). So, 4444^4444 has at most 16235 digits. The first sum of digits (SOD) is at most 9 * 16235 = 146115. So, the second sum of digits is at most max(SOD(149999), SOD(99999)) = 45. So, the third sum of digits is at most max(SOD(45), SOD(39)) = 12. Now, we constrain the answer to be a non-negative integer <= 12.

Observe that SOD(x) mod 9 = x mod 9 as a power of ten mod 9 is 1. So, the answer mod 9 = 4444^4444 mod 9 = (-2)^4444 mod 9 = (-2)^(3 * 1481 + 1) mod 9 = (1^1481 * (-2)) mod 9 = 7. With the above two constraints, the answer can only be 7.

It is a snow day with no school (I am a high school math teacher) and here I am watching this. Time well spent!

broooo , he have infinite IQ

Epic

I miss that feeeling

The idea of combining the reduced value space with the exact modulo 9 digit sum calculation is astonishing. When he started to speak about divisibility by 9 i first thought: What the heck is he thinking to approach with that stuff? It has all but nothing to do with the task at hand, does it?! But when he ended up applying this principle to the digitsum of digitsum recursively, the realization suddenly flared up.

I understood everything and it was very clear, however, I wouldn't be able to solve that yet, too much reasoning I am not atuned with

i bet its 7

Hi,
D(b*9)=9
D(d+9)=d
and D(A)=Mod9(A)
Thank
Today I view this video, firt time,
Exelent😅

Really quite a tough problem. I have a PhD in maths but had no idea how to tackle this problem. Think this was a very elegant, clear solution and presented with great enthusiasm. Think you need slight larger whiteboards!

I'm over here checking 8*5=40 with my calculator and BPRP is multiplying 9*17776 off the top of his head

genius! when you got the congruency I couldn't believe what I was seeing

It is easy to show that 4444^3 is 1(mod9) and that is easy to compute the higher powers of 4444.

I cheated and looked at the digit sums for the first few exponentials for 4444, and realized that 4444^2 started with a digit sum of 40, and that it increased by 15 each time. So 40 + (15 * 4442) = 66,670. d(66,670) = 25, and d(25) = 7

>>> import sys
>>> 4444**4444
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
ValueError: Exceeds the limit (4300 digits) for integer string conversion; use sys.set_int_max_str_digits() to increase the limit
>>> sys.set_int_max_str_digits(9999999)
>>> def digitsum(x):
... s = 0
... for a in str(x):
... s += int(a)
... return s
...
>>> print(digitsum(digitsum(digitsum(4444**4444))))
7

I guess you're right. I feel just calculating it is much simpler though.

wow

Mashalaah

I have got easy method take mod 9 and if that is possible then sum of digits if the number woukd be divisible by 9 or guve a remainder less then it just try it

The actual solution: The sum of the digits of 4444^4444 = 8 times 4 = 32, the sum of the digits of 32 = 5. Easy!

Why did he keep the first digit and change all the rest to 9

CHHOTA DIMAAG KAA GHULAAM...😂😂😂

So most of you guys focused on the math, which I appreciate, so thank you. But how are you doing?