What?!?! I figured out that it is 34 in about 2 seconds, in my head. 30 x 30 = 900.....40 x 40 = 1600. So the number must fall between 30 and 40. Only 4 squared and 6 squared end in 6, so it has to be one of those two numbers in the singles position. Since 1156 is much closer to 900 than 1600, the singles number must be 4. Therefore 34 is the answer. Multiplied 34 x 34 to verify--also in my head. I am not a mathematician or an engineer, and I am 74 years old. Your method is far too complicated. I stopped watching after about a minute.
My approach is simply to find the prime factors. It's obviously divisible twice by 2, leaving us only to find the square root of 289. It's also obviously not divisible by 3. 5 can be rules out as 5^4 is 625 which means we must hop that there's an exact square root of 289, or it gets very messy. It's clearly greater than 13 squared (169) which most people know, and the next prime is then 17, the square of which is 289. So, we have sqrt(2*2*17*17) = 34. nb. I don't think merits being an olympiad question.
I used the "by hand" method, which only took two steps = less than a minute. Grade 4 students used to be taught this method, which is similar to long division.
1156 is between 900 and 1600 and squared numbers that end in 6 are either X4 or X6 so the answer is either 34 or 36 and after that you just need to check these
I tried beginning to solve it before opening the video, my method was to try to find the prime factors of 1156 to if possible remove most of the work from the square root, found that you could divide it by 2 two times wich gives 2√289, then I had to think for a bit longer until I reached 17 wich fits perfectly, giving √1156 = 34, ngl I was not expecting the answer to be an integer
1156 is a very long way from being a "huge number". As others have pointed out, this answer takes only a few seconds to work out in your head (divide by 2, divide by 2, left with 289 which is 17^2, answer is 2*17). Also, you didn't need to multipy by 4/4, what you had was (10^2 + 2) ^ 2, giving the result of 102/3 after the square root which is 34. Also, also, what's with the unrelated #-tags? The only one relevant to this video is #squareroot...
To verify you can just multiply with the standard algorythm: 34^2 = 30*34+4*34 = 1020+136 = 1156 . There's no need of looking for algebraic identities since it's an arithmetic problem that doesn't require further generalisation.
@@lucho2868 of course you can, but if you know rules for nearby perfect squares than my way is quicker. You might as well say that you didn't need to use your knowledge that 32^2 = 1024 in the first place!
@@kicorse You also used 1024^.5=32. In fact you wrote a proof around the fact that that 32^2=1024. And your answer is formally incorrect since you didn't even proove that 32^2=1024. Your solution is useless because it requires differences, elementary algebra and (over all that) a long mutiplication whilst mine only require the long multiplication to check at the end. Making it shorter, more rigurous, and more generally applycable than yours.
@@kicorse for example, if you wanted approximate the value of any square root by hand the long multiplication's result of a nearby approximation could be used as an input in a Bakhshali-kind formula to get a good approximation result whilst the method you suggest would work less generally.
What?!?! I figured out that it is 34 in about 2 seconds, in my head.
30 x 30 = 900.....40 x 40 = 1600. So the number must fall between 30 and 40.
Only 4 squared and 6 squared end in 6, so it has to be one of those two numbers in the singles position.
Since 1156 is much closer to 900 than 1600, the singles number must be 4.
Therefore 34 is the answer. Multiplied 34 x 34 to verify--also in my head.
I am not a mathematician or an engineer, and I am 74 years old. Your method is far too complicated. I stopped watching after about a minute.
I did too. I used 32^2=1024; well-known if you're familiar with powers of 2. So it's higher than 32, but ends in 6, so the rest is as you stated.
That's nice! Thank you!!!
My approach is simply to find the prime factors. It's obviously divisible twice by 2, leaving us only to find the square root of 289. It's also obviously not divisible by 3. 5 can be rules out as 5^4 is 625 which means we must hop that there's an exact square root of 289, or it gets very messy. It's clearly greater than 13 squared (169) which most people know, and the next prime is then 17, the square of which is 289.
So, we have sqrt(2*2*17*17) = 34.
nb. I don't think merits being an olympiad question.
Nice! Thank you. I was just trying to show another way of solving it.
34
That's rigth!
I used the "by hand" method, which only took two steps = less than a minute. Grade 4 students used to be taught this method, which is similar to long division.
Thats great! Thank you!
I used the method I learned in freshman year of high school. I had the answer in less than a minute.
Wow! that's nice!
They should do one of these where it isn't conveniently an integer.
I'll solve one of this soon! thanks!
prime factorization gets answer very fast 1156 = 2x2x17x17 =34^2
Yes!!!!
1156 is between 900 and 1600 and squared numbers that end in 6 are either X4 or X6 so the answer is either 34 or 36 and after that you just need to check these
Thank you! That's a nice way too!
I tried beginning to solve it before opening the video, my method was to try to find the prime factors of 1156 to if possible remove most of the work from the square root, found that you could divide it by 2 two times wich gives 2√289, then I had to think for a bit longer until I reached 17 wich fits perfectly, giving √1156 = 34, ngl I was not expecting the answer to be an integer
Nice! Thank you!
Took me a whole 5 seconds to solve. 9mins and 7 secs of my life saved not following the solution
Nice!😉
i did it in my head in like 20 seconds when I saw the thumbnail. It's really fun!
That's nice Thank you!
1156 is a very long way from being a "huge number". As others have pointed out, this answer takes only a few seconds to work out in your head (divide by 2, divide by 2, left with 289 which is 17^2, answer is 2*17). Also, you didn't need to multipy by 4/4, what you had was (10^2 + 2) ^ 2, giving the result of 102/3 after the square root which is 34. Also, also, what's with the unrelated #-tags? The only one relevant to this video is #squareroot...
Thanks! the other hashtags are intended for the video to reach more people.
32^2=1024 and 33^2 is odd so (asuming the answer is an integer number) 34^2=1156
I had the same reasoning but verified it by noting that 1024 + 4*33= 1156. In general, x^2+4(x+1) = (x+2)^2.
To verify you can just multiply with the standard algorythm: 34^2 = 30*34+4*34 = 1020+136 = 1156 . There's no need of looking for algebraic identities since it's an arithmetic problem that doesn't require further generalisation.
@@lucho2868 of course you can, but if you know rules for nearby perfect squares than my way is quicker. You might as well say that you didn't need to use your knowledge that 32^2 = 1024 in the first place!
@@kicorse You also used 1024^.5=32. In fact you wrote a proof around the fact that that 32^2=1024. And your answer is formally incorrect since you didn't even proove that 32^2=1024. Your solution is useless because it requires differences, elementary algebra and (over all that) a long mutiplication whilst mine only require the long multiplication to check at the end. Making it shorter, more rigurous, and more generally applycable than yours.
@@kicorse for example, if you wanted approximate the value of any square root by hand the long multiplication's result of a nearby approximation could be used as an input in a Bakhshali-kind formula to get a good approximation result whilst the method you suggest would work less generally.
This is mental. So slow! Anyone doing a math olympiad would just look at it and know the answer, surely?
Probably yes!