It took me like 10 seconds to figure out the 32 + 16 + 8 = 56 so we have 3,4,5 I guess the main question is, is there some analytic way to figure out problems like this when the numbers are much larger... Since two to the power of 6 is 64 it is a fairly straightforward matter to check and guess numbers under six. Try setting up the same type of problem but replace 56 with a 12-digit number
@@SchoolClassMath Is it stated in the problem what kind of numbers x, y and z are? It should be, otherwise the problem is poorly formulated. If x, y and z are arbitrary real numbers, there are infinitely many solutions. If they are positive integers, there is only one solution (which one should perhaps prove).
No complicated math is needed to solve the problem if the solution are entirely integers. As well x, y and z has to be 5 or less (as 2^6 = 64) Let’s try setting z to 5: 2^x + 2^y + 2^5 = 56 This means 2^x + 2^y + 32 = 56 leading to 2^x +2^y = 24. Both x and y must be less than 5 (as 2^5 = 32). Let’s try setting y to 4: 2^x + 2^4 = 24. This means 2^x + 16 = 24 leading to. 2^x = 8. It is now easy to see that x must be 3 (as 2^3 = 8). The three exponents (x,y,z) are 3, 4 and 5. It is not possible to say “which is which”, any combination will work. The method might seem complicated, but is done by head is seconds. 😁
I think there must be a general way of solving problems....whatever the value of x, y, z be it whole numbers or decimal fractions, even complex numbers
@@SchoolClassMath There probably is. However to determine the value of three variables you would need three equations. In this case using a general way seems a bit like "crossing the river to get water". 😊
Just work out the equation.I hope you dont teach young students because you are the type of teachers who make working out mathematical problems a nightmare. Talking too much explaining nothing.
"quickly and effectively" you say. Rubbish I say. Even ignoring the infinite number of solutions where the variables are not integer such as x=0, y=0, z=log to base 2 of 54. There are 6 different integer solutions, so your long-winded method 'proving' x=3 is more probably wrong as there are twice as many alternative answers, namely x=4 or x=5. x, y and z are indeed 3, 4 and 5, but the order cannot be determined with the information provided. You fail.
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56 in Boolean 111000. Ans 5,4 3.
Good observation
It took me like 10 seconds to figure out the 32 + 16 + 8 = 56 so we have 3,4,5
I guess the main question is, is there some analytic way to figure out problems like this when the numbers are much larger... Since two to the power of 6 is 64 it is a fairly straightforward matter to check and guess numbers under six. Try setting up the same type of problem but replace 56 with a 12-digit number
Sir...I really appreciate your observation
Honestly speaking, this is helpful
56=32+16+8
=2^×+2^y+2^z
=>(×=5>y=4>z=3)
An assimulation only
Basis is all needed in every work of life..... before even mentioning an alternative. This is where we all begin from. Thanks, nice 👍 observation
Finally everything is correct but 34+32=66 and 56 is correct
👍🤔😉
nice observation
Good observation...it shows that it is followed line by line
Assuming that x is a positive integer, it takes about 10s to realize that a solution is (x, y, z) = (3, 4, 5).
What if x is not.....how long is it likely to take ?
@@SchoolClassMath Is it stated in the problem what kind of numbers x, y and z are? It should be, otherwise the problem is poorly formulated.
If x, y and z are arbitrary real numbers, there are infinitely many solutions. If they are positive integers, there is only one solution (which one should perhaps prove).
@@asterixx6878 Actually, there are 3! = 6 answers depending on the order of x, y, and z.
No complicated math is needed to solve the problem if the solution are entirely integers.
As well x, y and z has to be 5 or less (as 2^6 = 64) Let’s try setting z to 5: 2^x + 2^y + 2^5 = 56 This means 2^x + 2^y + 32 = 56 leading to 2^x +2^y = 24.
Both x and y must be less than 5 (as 2^5 = 32). Let’s try setting y to 4: 2^x + 2^4 = 24. This means 2^x + 16 = 24 leading to. 2^x = 8.
It is now easy to see that x must be 3 (as 2^3 = 8).
The three exponents (x,y,z) are 3, 4 and 5. It is not possible to say “which is which”, any combination will work.
The method might seem complicated, but is done by head is seconds. 😁
I think there must be a general way of solving problems....whatever the value of x, y, z be it whole numbers or decimal fractions, even complex numbers
@@SchoolClassMath There probably is. However to determine the value of three variables you would need three equations. In this case using a general way seems a bit like "crossing the river to get water". 😊
Just work out the equation.I hope you dont teach young students because you are the type of teachers who make working out mathematical problems a nightmare. Talking too much explaining nothing.
@@lusigijustus7274
For 3 unknown variables
With one eqn ???.
Thanks!!!!
The actual problem is, students are in different degrees, not just assimilating at the same rate....this calls for full explanation
"quickly and effectively" you say. Rubbish I say. Even ignoring the infinite number of solutions where the variables are not integer such as x=0, y=0, z=log to base 2 of 54. There are 6 different integer solutions, so your long-winded method 'proving' x=3 is more probably wrong as there are twice as many alternative answers, namely x=4 or x=5. x, y and z are indeed 3, 4 and 5, but the order cannot be determined with the information provided. You fail.
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