Almost solved 49^9!

I calculated 49^7 correct which was 678.233.072.849 but then I made a mistake at 49^8. As you can see in the screenshot, my answer was very close to 49^10, which is 79792266297612001. I overcalculated haha I think I will calculate a 2 digit number to the 10+ power very soon. Very proud that I got 49^7 correct.

Wow, Johnny!

Really amazing.

I can go to 49^4. After that, things become really slow and difficult to do for me.

Iāll take the readers to 49^4. Letās first square 49.

49^2=2401. This is easy. There are two ways of doing this

1 49^2 = (50-1)^2 = 50^2 - 2*50 + 1 = 2500 - 100 + 1

2 49^2 = 49*(50-1)= 2450-49=2401.

Mentally, multiplying by 50 is easy.

First multiply by 100, or just add two zeroās.

Then just have the result.

Letās do this with 49.

49 => 4900 => 2450.

All that is left to do is subtract 49:

2450-49=2401

Squaring 2401 is also not very difficult.

24 | 01

squared is:

24^2 | 2*24 | 01^2 = 576 | 48 | 01 = 5,765,801

If you know that 12 squared is 144, then 24 squared is 4 * 144 = 576.

Why is it difficult for you to calculate 49^5?

Does the method you use not work with larger numbers?

All my life I calculated things that I was and am interested in.

For most practical purposes, I am generally fine with 3, max 4 digit precision.

I calculated 49^4 as 5,675,801.

Calculating 5,675,801 * 49 with full precision means I have to juggle 9 digit numbers in my head.

This is beyond my abilities.

You already know that what your have - the ability to use numbers with many more digits than most people are able to do - is extremely rare.

If I had to calculate 49^5 I would think:

49 = 100, divided by 2, minus 2%.

So I would go 100^5 = 10^{10} or 10 billion.

We divide this by 2^5 or 32.

(Intermediate answer 312 million).

Subtracting 5 times 2% is a little less than 10% in total. 10% of 312 is 31.2. For āa little lessā I would go with 30. 312-30 = 282.

So 282 million would be my answer.

So yes, irl I have a lot of shortcuts to āguesstimateā.

I have no experience with calculating higher powers, however I have lots of experience with being terrible with communication and being generally social

What I have done is memorize the single digit numbers to single digit powers, for example, 6 to the 9th power = 10077696 etc.

Does anyone consider using memory place to store numbers and them manipulate them for large mental multiplications? Or is it inefficient?

I have considered it. Iām not sure how efficient it is; Iām not an expert. Although it does seem like it has potential.

User Nodas has a nice explanation

It is inefficient. Mental math techniques like the cross method make large calculation so much easier to calculate, you donāt even need to train your memory or be born with a special one. Everyone can do it.

But this gets a little bit tricky with power calculations because you are forced to hold a very large amount of digits during the calculation.

Anyone who wants to calculate a large power has to memorize a crazy amount of smaller powers or numbers to use as a shortcut or indeed try to use something like a memory palace or other memory techniques. However, I donāt think using them will make it any easier because converting the numbers with a memory technique while calculating is, I suspect, pretty difficult.

You would need to convert your subresults to images hundreds of times without making a single mistake AND your calculation needs to be accurate as well just so you can calculate something as small as 56^18.

Rudiger gamm is known for large 2 digit power calculations. But he memorized hundreds of digits so he could calculate for example 56^18 very fast. And for something like 87^100 he memorized 300k+ digits.

I donāt think he could actually calculate beyond 5th power without those memorized digits.

Fortunately I was born with a good memory so I donāt need to memorize anything and I donāt need to convert the numbers with a memory technique to remember them. This saves me time and reduces the risk for mistakes.

I donāt have any experience with memory techniques in mental calculation but from my perspective, the way I calculate powers, it would seem pretty exhausting to convert the numbers at every step of the calculation.

Of course, I donāt know how it would actually feel to do that. When you are calculating powers, how do you use your memory techniques?

Could you explain the steps with an example like 13^6?

Might not be a good example just because of the way the numbers fallā¦

13^6=169^3

ā¦since 169 is super close to 170 itās easy to get toā¦

169^2=28,561

All that is left is to multiply by 169 one more timeā¦

ā¦so now you āstoreā 28,561 as P-02 / A-85 / O-61 if you use PAO, or O-028 / O-561if you prefer an OO 3-digit system. Store that in the first location. Now just multiply by 169 and here Iād make use of the image you stored.

- Double the whole thing to get to whatever x200 would be (add two zeros of course)
- Subtract that same number decimal point shifted to get to x180 (i.e., 200-20)
- Subtract one more 28,561 to get to x179

It depends a bit on your calculation speed how many of those you need to storeā¦ I donāt think storing double of your PAO image is necessary, because even if you forgetā¦ doubling again (i.e., redoing the calculation is not going to kill you).

I might put another image for the x180 result just because itās after subtracting the same from the same (decimal point shifted) which was just double of my stored PAO imageā¦ however, for the last step I might want to have a ārestoreā point just in case.

This one is pretty simple because 169 is so close to 170 and 169 = 200 - 20 -1 so a lot can be reused and times one and doubling is the only ācalculationā you have to do along the way.

169^2=170^2-170*2+1=28,561

ā¦just seems a bit overkill to go all crisscross on that 169 thereā¦ and if you know 69 squared, Iām sure you know 17 squared. Dunno, just sayingā¦ seems easier and faster to me.

I think 12^5 might actually end up being more work than 13^6 despite the result being lower.

Yeah, ditto probablyā¦ my problem is that I find x^7 pretty annoying because of the extra step compared to x^6 or x^8. In fact, I wouldnāt mind the larger number with 8, just because you can simply square the squared square.

Define ārelatively easyāā¦ as already mentioned, taking something to the eighth power would just be squaring the squared squareā¦ so in terms of crisscrossing and storing intermediate result with mnemonicsā¦ x^8 is easier than x^7.

37 is also past the 32 mark, which is where your squares turn 4-digits. So generally complexity increases past 31^2ā¦you could argue that 35 is still ok because it ends in a 5 and you can easily shortcut the squaring of the square.

17^2=289 for example, Iād consider similar to your 13^6 example because because itās just one removed from 290 and you can do what I did before using 29^2 to quickly get the square of the squareā¦ 19, 21, 23, 27, 29, 31 all the same kind of categoryā¦ and then of course 25 is easy.

Generally, any of the above taken to either 4 or 8 would be easy; even though, 8 would require an additional time consuming crisscross. Any other base taken to 5 or 7 would as previously mentionedā¦ be annoying.

I donāt think thatās what @anon3561952 is sayingā¦ like I mentioned before

ā¦not all 2-digit numbers are equal in terms of calculation complexity if you donāt brute force the whole thing like you do.

Eye of the beholder. For me, the end result is irrelevant. The fact that itās to the seventh power is super annoying because 37^8 is easier to do than 37^7. For you thatād be an additional step because you just multiply by 37 over and over and over. I, on the other hand, square 37, then I square that result, and then I square that result and done, so for meā¦ taking it to the eighth power is much easier than taking it to the seventh, even though the result will be a larger number.