3 difficult questions of Mental Calculations World Championship

  1. What is (5 1/7)^5 to the nearest integer?

  2. Calculate the exact 17th root of a given 45-digit number

  3. How many years in the century 1801-1900 were leap years with 53 Sundays?

Source: https://worldmentalcalculation.com/2019/08/24/mind-sports-olympiad-results-2019/


note that I am not a mental mathlete, I just like math.

The first one is not too hard, but it takes a while to do with brute force.

First Question

Basically you have (36^5) / (7^5) so lets start easy.

7x7 is 49, we can ll do that one. 49x49 is still easy, as you can just grab 50x50 (2500) and subtract 99 (once 50 and once 49) to get 2401. Next, 2401 times 7 would be 16807.

36^5 is tougher, and I am sure there are easier ways to do it than I do, but I can figure this out. 36*36 is 1296. 1296x1296 is pretty easy to do as a grid, 1679616 is what I end up with. Using another grid to do that times 36 lets me end up with 60466176.

So, what we want to know is the integer closest to 60466176 / 16807, for as far as I can see, this is irreductible, so we end up with a fun division. This is the easiest bit. Basically I wrote the first 9 multiplications of 16807 and went on with a long division, which was quick to give me 3597 with a leftover of 11397, which is more than half of 16807.

That means the nearest integer is 3598

The second is magic to me, I have no clue how to do this. But the third I can again do with a semi-brute force approach, rather than fully mathematically.

Third Question

The leap year rule is that every 4 years there is a leap year. Every 100 years there is no leap year, unless that year is also divisible by 400. That means that in this question, 1900 is not a leap year. That leaves us with 100/4-1=24 leap years in total.

Now, what parameters need to be met for a year to have 53 sundays? You need at least 52x7+1 days to get 53 of any day, which is exactly 365. That is important, because that means that as long as a leap year starts on a saturday or sunday, there will be exactly 53 sundays in a leap year.

The thing making this easier is that there is a pattern to the days a year starts with. If you let monday be 0, then climb up to sunday being 6, you can find the day any year starts with by adding 1 for every year you add, and adding an additional 1 for every 4 years. meaning you get n+y+⌊y/4⌋ (mod 7) where y is the number years you add, and n is the day on which 1800 started (you could take 1801 as well, but you will see why I pick 1800)

january 1st 2019 was a tuesday, thus a 1. Reversing the formula gets us to 1-219-(⌊219/4⌋)=1-219-54 = -272. Though we also counted 1900 there, that was not a leap year, so lets add one back to get -271. We need a positive number between 0 and 6, so using modulo 7 we get 2, that is a wednesday.

Now we have our equation, 2+y+⌊y/4⌋ (mod 7), and since we only assume leap years, this can be simplified to 2+y/4*5, so our equation adds 5 points for every jump of 4 years. That is a very pretty equation, as it allows us to work from te answer rather than to it.

from our numbered days, we found that we want a 5 mod 7 or 6 mod 7 to be our answer, and we now have an equation that just turns into a +5 per y/4. The chain we get, counting from y/4=1 ( thus year 4, aka 1804) is:

0, 5, 3, 1, 6, 4, 2, 0…

It repeats. That means that every second and every fifth leap year in a cycle of seven is a year with 53 sundays.

4 times 7 is 28, meaning that it repeats every 28 years. That is already 3*2=6 leap years with 53 sundays in 84 years, with 16 years left. that lets us stuff one more leap year into it.

So my answer here is 7 leap years with 53 sundays between 1801 and 1900


Alternatively, if you do calendar calculation events and you got the year codes memorized just count the number of leap years in the relevant memory palaces.

Obviously, this only works if you got those seven memory palaces set up; however, if you have them you’ll be super fast. Also if pressed for time during the event, you’ll most likely guess 7 as the answer anyway as there are either 3 or 4 leap years per memory palace. If you don’t have the memory palaces set up…

Do basically what @Mayarra explained:

  1. 100 / 4 = 25 (every four years you got a leap year)
  2. 25 - 1 = 24 (remove 1800 because it’s not evenly divisible by 400)
  3. 7 x 52 = 364 (normal year 365 and leap year 366)

…or step 3. without math: if this year starts on a Monday then next year starts on a Tuesday, so this year ends on the same day it started on. If it’s a leap year, next year starts on a Wednesday because you got an extra day this year, so you could also start one day earlier and still have the aforementioned day included.

  1. 24 / 7 = 3.42 (rounding up or down you get 3 or 4, respectively)
  2. 3 x 2 = 6 and 4 x 2 = 8, so just guess 7 (doing x 2 because of the extra day)

This question is similar to when will XYZ’s birthday be on a Monday again or what date is Thanksgiving any given year (falls on the last Thursday in November that’s a full week).

I kinda feel that 1900 - 2000 for the above question would have been harder because now you need to consider that 1900 is not a leap year but 2000 is… maybe I do too much calendar math, but the statistical guess described above basically gives you 7, which happens to be the correct answer in most xx01 - xx00 scenarios.

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  1. What do you mean by grid? I think I am going to learn something good here :slight_smile:

I would have done my 1296 x 1296 by saying 1300 less 4

\begin{align} x &=1296^{-4} * 1296^{-4}\\ &= 1300 * 1292 + 16\\ &= (1300 * 1200) + (1300 * 90) + (1300 * 2) +16\\ &= 1560000 + 117000 + 2600 + 16\\ &= 1,679,616\\ \end{align}

but now I need to go 1,679,616 * 36 and you say grid and I say what what what???


Okay, bear with me, this is the steps for the grid for 1296 times 1296. I will explain below.

So, starting out. One of the numbers is horizontal from left to right, the other is vertical from top to bottom. (first image) That creates a grid, and in that grid we will fill the multiplications of the numbers. (second image). For example, the top left one is 1 and 1, so 1 times 1. Going one to the right we find 2 and 1, so 2 times one. Third on the bottom is 9 and 6, so 9 times 6.

Having done that, we will add some numbers diagonally. The order here is important, this has to do with the fact that the 1 times 1 actually represents a million in the answer, which is way more than the 6 times 6 which just represents 36. The blue lines show what to add. (third image)

after adding the numbers, we get 7 numbers, there we are going to add as well, in the stair shape as shown (image four).

This same trick works with the 1.679.616 times 36, but rather than a 4x4 grid, we will work in a 2x7 grid. The idea is the same though.

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Ok, I have seen this before but not linked it to the term “grid” and not considered it as a mental multiplication calculation method so much as a nice way of remembering my criss-cross when I have an uneven number of terms.

Are you suggesting that you are actually using this grid mentally in a 4x4 multiplication and that you can visualize this table?

I really like the visual representation but am not clear how I would effectively represent a table such as this in my head. … Please continue…

I suspect there is a learning opportunity for me here. How do you mentally build this table so that you can use it?

In my practice, I follow the same steps but I simply keep a running subtotal and add the next term. This is great but not nearly as nice as having the grid that you have presented available mentally. If I was able to keep this grid in my head I could cross-check my work trivially and expanding multiplication to higher numbers of terms would be a much easier path. I suspect I could peg the grid in an awkward fashion with the major system but it would “probably” take me more time than the calculation.

This is really making me go hmmmm.


This method works best with a piece of paper, as you could imagine, but there are ways to do it mentally

If I had to do it 100% mentally, I would use a peg system. Using consonants, you can actually build a grid like this.

Say you have a 4x4 grid, you could use any consonants you like, or even involve vowels, but that is up to you. I would get B, C, D and F

The Grid would then be:
BB - BC - BD - BF
CB - CC - CD - CF
DB - DC - DD - DF
FB - FC - FD - FF

The images then could be
Baby (BB) - Bacon (BC) - Bed (BD) - Beef (BF)
Cab (CB) - Fuel tank (CC) - CD (CD) - Coffee (CF)
Dab (DB) - Batman (DC) - Breasts (DD) - Deaf (DF)
Facebook (FB) - Face (FC) - FedEx (FD) - Fortisimo (FF)

Then use a number system to stick the images to the pegs. After some practice you can know which ones to add.

I was just playing with the grid in my head with simple 3 digit squares and it is quite doable without mnemonics. I will have to play about. a^2 + 2ab + b^2 has generally been my go to method for 3 and 4 digit squares but the grid has real potential for mental visualization. definitely hm. thanks again.

… additionally 3x3 mental multiplication has a very low mental effort using this visualization. It feels like running an abacus rather than doing multiplication. I am going to play with this for a few weeks and see where it fits in. I suspect that this will be a nice way of getting over the hump for of multi digit mental calculation. At a minimum it is another way to think about things and that always helps.


I think the @Mayarra technique (which look like Lattice multiplication) can be useful with 5 digit by 5 digit multiplication.

I says this because with criss-cross multiplication method I reached my limit at 4 digit by 4 digit multiplication.


I based it on the line method, which became annoying at anything above 3x3. But the Lattice method seems indeed very similar.


@Mayarra it seems a good technique BUT to master it it would take at least 1 to 1.5 years. Basically I should memorize something else?

I think it depends on what it is you are looking for. Doing it fully mentally is a challenge, but with pen and paper, it creates an solution for multiplications with large numbers of which the steps remain the same, no matter how many digits. It is far from perfect, but with pen and paper, it works for me :slight_smile:


@Mayarra Can you help me with logarithms?

@benjamin1990 what is it you need help with on that area?

Im struggling to calculate log (43)

since it is that specific, I think you are better off making a new topic for it. I can sort-of do it, but as I said, I am not a mental mathlete, so there will be many people who can both do and explain something that specific better than I can

Ok. Can you teach me with your method how would be 345 × 127?

That one is actually very easy.

35 x 127 = 4445

That means that 350 x 127 = 44450

Subtract the 5x we counted too many and we get 43815

3 digit numbers are often easy to modify

Another method would be to make the numbers easier to work with.

Dividing the 345 by 5 gets us 69. This transformation gets us 635 instead of 127.
7 x 635 = 4445
So 70 times is 44450

Remove the 635 we counted too many and we get 43815 again.

Good but I was asking about your method. I do my calculations WITHOUT shortcuts. I apply brute-force calculation so to do this one I would applied criss-cross multiplication.

This is easier than it sounds.

First of all allthough we cannot see the given number, we know the logarithm of that same number is between 44 and 45.
The very fact that it is a number of 45 digits means the answer is between 388 and 443.
44 / 17 \approx 2.588
45 / 17 \approx 2.647
10^{2.588} \approx 388
10^{2.647} \approx 443

(Use the posts I wrote about calculating logarithms and antilogs on how to do this mentally)

So we now have 55 numbers that - raised to the 17th power - leads to a 45 digit number.

Let’s assume the number we are trying to find is 399.

399^{17} = 164641466832372473541255851374408235498246799

First take the log of the first couple of digits. The more accurate the better.
The log of 1.6464 \approx 0.216 and therefore the
log(164641466832372473541255851374408235498246799) \approx 44.216

44.216 / 17 \approx 2.6009...

10^{2.6009} \approx 398 or 399

If you just use the post about calculating antilogs and even forget the correction, you will find that 10^{2.6} \approx 400.

Next step.
Only numbers ending in 99 will also end in 99 when raised to the 17th power.
If you multiply a number ending in 99 with itself, the result ends in 01. If you multiply that with a number ending in 99, then the result ends in 99. So you get a sequence 99, 01, 99, 01, etc. After 17 times the result ends in 99.

In general; if you know the last 2 digits of a number that is raised to the 17th power, you can find the last 2 digits of the base number.

So in this case the answer has to be 399, since we are looking for a number close to 400 that ends in 99.