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64 is 8 squared, and 63 is 7 times 9. In general, n^2-1 is (n+1)*(n-1). If I subtract 1 from 4444488889, I get 4444488888, and that has obvious factors. It's 100002 times 44444. 100002 has digits that add to 3, divide by 3, get 33334. So the number minus one is equal to 33334 times 3 times 11111 times 4. 33334 and 33333 differ by 1; we want numbers that differ by 2, so take one of the twos out of 4. The answer seems to be 66667*sqrt(2).
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The square root of 4444488889 is 66667. It is a perfect power (a square), and thus also a powerful number. 4444488889 is digitally balanced in base 3, because in such base it contains all the possibile digits an equal number of times.
It can be written as a sum of positive squares in only one way, i.e., 4061895289 + 382593600 = 63733^2 + 19560^2 . It is not a de Polignac number, because 4444488889 - 211 = 4444486841 is a prime. It is a Duffinian number. It is a plaindrome in base 10. It is not an unprimeable number, because it can be changed into a prime (4444488883) by changing a digit. It is a polite number, since it can be written in 8 ways as a sum of consecutive naturals, for example, 10866517 + ... + 10866925.
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Not a good job, because that's plain wrong! For starters: sqrt(y) = x => only one solution (with x >= 0 for non-negative real y), it's called the principal square root. y = x² => two solutions, + sqrt(y) and - sqrt(y). Still sqrt(y) is only the positive root here, the sign is on the outer side!
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Or sqrt of 49 = 7 and sqrt of 4489 = 67 and sqrt of 444889 = 667 and sqrt of 44448889 = 6667 so sqrt of 4444488889 = 66667.
nice job
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Nice algebraic trick. But i doubt if this was actually a Harvard entrance exam question.
thank you so much!
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math olympiad question
@@AsadInternationalAcademy that's more like it. Its lots of puzzle with little maths
Harvard doesn’t have an ‘entrance exam’ 😂
Such a beautiful solution!
Glad you think so!
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I dont get the step where you omit the 12b
The Babylonian method is easier and quicker. Nice video just to illustrate a different way to solve.
Cool, thanks
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Easy, answer is X.
Yes, it's easy
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64 is 8 squared, and 63 is 7 times 9. In general, n^2-1 is (n+1)*(n-1). If I subtract 1 from 4444488889, I get 4444488888, and that has obvious factors. It's 100002 times 44444. 100002 has digits that add to 3, divide by 3, get 33334. So the number minus one is equal to 33334 times 3 times 11111 times 4. 33334 and 33333 differ by 1; we want numbers that differ by 2, so take one of the twos out of 4. The answer seems to be 66667*sqrt(2).
okay thanks
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@quadibloc2 - you seem to have picked up a spurious factor of sqrt(2)?
Calculator on your phone says 66667.
Nice. But how would one know that this is the way to solve it?
Thanks, there are more methods to solve this, not only this. But, I'm sure, this is the easiest method than others
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I just did it the basic long way. Took a minute but i got there
Wonderful job, thanks
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I would take my phone (calculator)out of my pocket. 😮
ok
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How do you in starting that it is a perfect Square
The square root of 4444488889 is 66667. It is a perfect power (a square), and thus also a powerful number. 4444488889 is digitally balanced in base 3, because in such base it contains all the possibile digits an equal number of times.
It can be written as a sum of positive squares in only one way, i.e., 4061895289 + 382593600 = 63733^2 + 19560^2 .
It is not a de Polignac number, because 4444488889 - 211 = 4444486841 is a prime.
It is a Duffinian number.
It is a plaindrome in base 10.
It is not an unprimeable number, because it can be changed into a prime (4444488883) by changing a digit.
It is a polite number, since it can be written in 8 ways as a sum of consecutive naturals, for example, 10866517 + ... + 10866925.
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Please write slower, I cant follow that fast....
no problem, you can slow the speed, see options
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What about 12B ?
very simple, follow the steps
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Do it by number my number right going left
okay but this method is very simple and easy to understand
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I was tought in school how to calculate square roots by hand, don't know how difficult would be to get this solution.
this is easy problem
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I tried, it took me about 3 minutes.
❤
thank you so much!
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you should share this lesson
AcademicaLy pretty Not practicaL to daiLy Life
thanks
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Well the actual answer is 66667 or -66667 so where is the working for that
Good job! thanks
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Not a good job, because that's plain wrong!
For starters: sqrt(y) = x => only one solution (with x >= 0 for non-negative real y), it's called the principal square root.
y = x² => two solutions, + sqrt(y) and - sqrt(y). Still sqrt(y) is only the positive root here, the sign is on the outer side!
For the latter: if y = 0, there's only one solution, and that's x = 0. For y < 0, there's no real solution. But that's just a disclaimer!
@rainerzufall42 where does it say that it has to be positive only, the squareroot of 4 is 2 or -2, same here but 66667 and -66667
try v(2004848)
nice idea, thanks
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