Entrance examination. If you're reading this ❤️. What do you think about this problem? Hello My Friend ! Welcome to my channel. I really appreciate it! @higher_mathematics #maths #math
Write down the powers of 3: 1, 3, 9, 27, 81, 243, 729. From that you immediately see that all of the variables must be less than 7 and exactly one of x, y, z must be 6. Thus you reduced the problem to 3^u+3^v=837-729=108. Similarly you can notice that none of u, v can be can be greater than 4 and exactly one of them must be 4. So subtracting again 3^4=81 we reduce the problem to 3^s=108-81=27. Summing it up all the solutions are permutations of 3, 4 and 6.
Radix conversion, usually from base 10 to 2, and 2 to 10, used for all computer calculations, is one of the most common calculations on earth. In this case a radix conversion from 10 to 3 is required. Conversion of n to radix 3. The process is 1. Find the largest 3^(position - 1) < n. 2. Subtract 3. Repeat until remainder 0. Giving 3^6 (729) + 3^4 (81) + 3^3 (27) = n (837).
Did it a slightly differently, didnt factor out 3^z. Divided through by 3^3, so indices were x-3, y-3,z-3. Observed that indices must be positive, or wouldnt sum to an integer. ie x,y,z>=3. However, if they were all stricly greater than three, you could take a factor of 3 from the whole expression and divide through, which you cant because 31 is prime. Therefore one index must be precisely 0. say x-3=0, x=3. sub in, and subtract 1 from each side and you have the remaining terms sum to 30. But 30 has a factor of 3, so you can divide through to have the indices as y-4,z-4 and equalling 10. Trivially 9 and 1, which yield 5 and 13 respectively.
One of the terms must be greater than 1/3 of 837. Only 729 satisfies this, and leaves the sum of the other 2 terms having the last digit as 8. Powers of 3 can only end in digits 3, 7,1 and 9. Therefore perforce the reamaining terms must be 81 and 27.
Excellent exercise for understanding some the equations with multiple variables. I would not think on the multiple combinations of solutions, this is what I learned today. Thank you
At a quick glance I started by calculating powers of 3 and quickly find 3^6= 729 by adding 3^3 = 27 and 3^4 = 81 then 729 + 27 + 81 = 837 . One solution then is x = 3, y = 4 and z = 6.
Thinking in base 3, it's easy to compute that 837 (10) = 1011000 (3), and since base representation of an integer is unique, the only solution to the problem is 3^6 + 3^4 + 3^3. Any problem of the type "find x,y,z such that N^x + N^y + N^z = M", will have a solution iff the sum of the digits of the base N representation of M is equal to 3.
837 = 9*3*31 =(3^3) (27 + 3 +1) = 3^6 +3^4 +3^0 • digits of 837 add to a multiple of 9 so divide by 9 • 3 divides 93, 31 times • break down 31 in powers of 3 • done
I think the examinera are looking for the elegant answer: convert 837 to base 3 - 1011000 then by inspection x=6,y=4,z=3. I would score this solution 10/10! Trial and error 2/10 not scholarship material.
No need for math. The simple mechanical way is to keep dividing by 3 and up the factor by 1 from for each successful division. As soon as there's a remainder of 1 that's the smallest variable; if the remainder would be 2 we have two variables at once of same value. Subtract 1 from the factor as the remainder was not divided. Keep dividing the dividend ignoring this remainder (as it is already 'factored in') until the dividend is 0. 837/3 -> 279(0)/3 -> 93(0)/3 -> 31(0)/3 -> 10(1): after 4 divisions we have rest 1 which means the original size of the rest=1 * 3³ so z=3. Continuing with the dividend 10/3 -> 3(1) another rest one factor up; y=4. Continuing with 3/3 -> 1(0) -> 0(1) two factors up we have another rest of 1, so x =6. The remaining dividend is zero so we're done.
Write down the powers of 3: 1, 3, 9, 27, 81, 243, 729. From that you immediately see that all of the variables must be less than 7 and exactly one of x, y, z must be 6. Thus you reduced the problem to 3^u+3^v=837-729=108. Similarly you can notice that none of u, v can be can be greater than 4 and exactly one of them must be 4. So subtracting again 3^4=81 we reduce the problem to 3^s=108-81=27. Summing it up all the solutions are permutations of 3, 4 and 6.
Ok bat change 837-729 to 108.
@@Cagouille79 Fixed it, thanks!
This takes about 10 seconds by just thinking about the powers of 3.
Maybe more than 10.
Radix conversion, usually from base 10 to 2, and 2 to 10, used for all computer calculations, is one of the most common calculations on earth. In this case a radix conversion from 10 to 3 is required.
Conversion of n to radix 3. The process is
1. Find the largest 3^(position - 1) < n.
2. Subtract
3. Repeat until remainder 0.
Giving 3^6 (729) + 3^4 (81) + 3^3 (27) = n (837).
Really good solution. Very quick to compute without much algebra required.
@@emilegiesler9272
Thank you for your compliment : I first wrote a radix conversion program over 50 years ago.
Did it a slightly differently, didnt factor out 3^z. Divided through by 3^3, so indices were x-3, y-3,z-3. Observed that indices must be positive, or wouldnt sum to an integer. ie x,y,z>=3. However, if they were all stricly greater than three, you could take a factor of 3 from the whole expression and divide through, which you cant because 31 is prime. Therefore one index must be precisely 0. say x-3=0, x=3. sub in, and subtract 1 from each side and you have the remaining terms sum to 30. But 30 has a factor of 3, so you can divide through to have the indices as y-4,z-4 and equalling 10. Trivially 9 and 1, which yield 5 and 13 respectively.
One of the terms must be greater than 1/3 of 837. Only 729 satisfies this, and leaves the sum of the other 2 terms having the last digit as 8. Powers of 3 can only end in digits 3, 7,1 and 9. Therefore perforce the reamaining terms must be 81 and 27.
Excellent exercise for understanding some the equations with multiple variables. I would not think on the multiple combinations of solutions, this is what I learned today. Thank you
At a quick glance I started by calculating powers of 3 and quickly find 3^6= 729 by adding 3^3 = 27 and 3^4 = 81 then 729 + 27 + 81 = 837 . One solution then is x = 3, y = 4 and z = 6.
That one was beautiful👏👏
Thinking in base 3, it's easy to compute that 837 (10) = 1011000 (3), and since base representation of an integer is unique, the only solution to the problem is 3^6 + 3^4 + 3^3.
Any problem of the type "find x,y,z such that N^x + N^y + N^z = M", will have a solution iff the sum of the digits of the base N representation of M is equal to 3.
837 = 9*3*31 =(3^3) (27 + 3 +1) = 3^6 +3^4 +3^0
• digits of 837 add to a multiple of 9 so divide by 9
• 3 divides 93, 31 times
• break down 31 in powers of 3
• done
Genial. Gracias-----
Such sigma math, so skibidi sus ohio vibes like that 10/10 fanum tax gyat
In Cambridge, UK,it's pronounced zed!
More faster for my opinion: 3^x is max and
Bonne solution .
The problem has infinitely many solutions : choose e.g. x= 0 , y=1 .Then z = ln833/ ln 3 . This way you can construct arbitrarly many solutions.
I think the examinera are looking for the elegant answer: convert 837 to base 3 - 1011000 then by inspection x=6,y=4,z=3. I would score this solution 10/10! Trial and error 2/10 not scholarship material.
For the exam: ¿Do you have to write down your deduction or is it sufficient to give the solution and the proof 729 + 81 + 27 = 837?
There is more better and simply solution
Got the solution
How is it concluded that 3^z = 3^3?
No Power of 3 is equal to 31
No need for math. The simple mechanical way is to keep dividing by 3 and up the factor by 1 from for each successful division.
As soon as there's a remainder of 1 that's the smallest variable; if the remainder would be 2 we have two variables at once of same value. Subtract 1 from the factor as the remainder was not divided. Keep dividing the dividend ignoring this remainder (as it is already 'factored in') until the dividend is 0.
837/3 -> 279(0)/3 -> 93(0)/3 -> 31(0)/3 -> 10(1): after 4 divisions we have rest 1 which means the original size of the rest=1 * 3³ so z=3.
Continuing with the dividend 10/3 -> 3(1) another rest one factor up; y=4.
Continuing with 3/3 -> 1(0) -> 0(1) two factors up we have another rest of 1, so x =6.
The remaining dividend is zero so we're done.
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