Why go through such a long winded process? 1) First, note that 3^x grows much faster than a cubic function. Thus, for large ms, 3^m outstrips m^3. 2) When m = 3, 3^m and m^3 are equal. Beyond that, the two functions will never touch again. 3) Similarly, for negative integers, 3^m rapidly approaches zero, while m^3 gets to be a super negative number. 4) Thus, you only have to try integer values of 0,1,2 to find a solution. Takes all of 30 seconds.
(3^m)-m³=1 The trivial solutiin is m=0 For other possible silution, take modulo 2: mod(3^m,2)-mod(m³,2)=1 [mod(3,2)]^m-[mod(m,2)]³=1 (1^m)-1=[mod(m,2)]³ mod(m,2)=0 --> m=2k, k an integer 3^m=m³+1 --> 3^m=(m+1)(m²-n+1) (m+1) | 3^m (2k+1) | 9^k --> k=1 m=2 Therefore m={0, 2}
Hi SyberMath, big fan. I had a similar solution to you: 3^m = (m+1)(m^2-m+1) the terms on the RHS must be powers of 3, since they multiply to make a power of 3 this means that one of these terms must be a factor of the other term (m^2-m+1)/(m+1) = (m-2) + 3/(m+1) because (m+1) divides (m^2-m+1), 3/(m+1) has to be an integer hence (m+1) is a factor of 3 hence m= 0 or 2 (I am a high school student and just also wanted to thank you for improving my problem solving ability allowing me to qualify for a number of mathematics competitions through my school)
3^m - m^3 = 1 3^m = m^3 + 1 3^m = (m+1)(m^2 - m + 1) m+1 | 3^m Hence m+1 is a power of 3. Case 1: m+1 = 3^0 = 1 Hence m = 0. So, 3^0 - 0^3 = 1 - 0 = 1. Valid solution. Case 2: m+1 > 1. This means that m+1 is a power of 3, so m = 3k+2 But also, m^2 - m + 1 is a power of 3. And that's 9k^2 + 12k + 4 - 3k - 2 + 1 = 9k^2 + 9k + 3. Well, taking out a multiple of 3, we get 3(3k^2 + 3k + 1). So, 3k^2 + 3k + 1 is a power of 3. But it is clearly not divisible by 3, so it must be 1. Therefore, k = 0, and m = 3(0) + 2 = 2. Testing this, we get 3^2 - 2^3 = 9 - 8 = 1, so it's valid. Solutions are m=0 and m=2.
No factors in that case, but if the constant term was +9 (and not -9) you would be looking for two numbers with product (ac) = 2 x 9 = 18, and sum (b) = -9. -3 and -6 satisfy, therefore (x - 3).( 2x - 3) does the trick. In this case, it is nothing to do with b being equal to c, merely that 2 x 9 = 3 x 6. Always! Even early morning on a really crappy Friday.
@@dwm1943 someone else’s comment got deleted where it mentioned a “non monic factorization”. What is that and what did the person above ya mean by x base 12? Also if (x - 3)(2x - 3) gets foiled, u end up with +9 (due to the -3 • -3), not -9. Is this not factorable?
To me, at least, a mathematician is someone who works to develop entirely new ideas and results in mathematics, whereas a math educator is someone who takes existing ideas and results and teaches/explains them (and how to play around with them to problem-solve) to others. You can be a math educator without being a mathematician, or you can be a mathematician without being a math educator, or you can be both. I would say that SyberMath is a math educator. (As am I.)
@@scottcowan8036 Yes, and a good one. Some of the maths tutors on RUclips make it like an assault course. It's supposed to be fun, and SM makes it so. Which is why I keep coming back.
Answer 2 and 0 3^m - m^3 =1 3^ m= m^3 + 1 3^integer is always odd Hence, 3^ m is odd Hence, m^3 + 1 is odd Hence, m^3 is even Hence, m is even If m is even, then 3^ m unit digit is either 1 or 9 the difference between 3^ m and m^3 =1 3^ 0 = 1 0^3 =0 and 3^ 2 =9 and 2^3 =9 So 0 and 2 are two solutions since the difference between 3^m and m^3 is 1 in both cases. Let's try some more 3^4 and 4^3 81 and 64, a difference of 17 ' 3^6 and 6^ 3 729 and 216, a difference of 513 The difference becomes larger when m increase Hence, the solutions are 0 and 2 When m is negative, the difference 3^m is not an integer, but m^3 is
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1, 2 and 3 are small numbers so by little inspection we can say m=2
3^2_2^3=9_8=1 answer
Why go through such a long winded process?
1) First, note that 3^x grows much faster than a cubic function. Thus, for large ms, 3^m outstrips m^3.
2) When m = 3, 3^m and m^3 are equal. Beyond that, the two functions will never touch again.
3) Similarly, for negative integers, 3^m rapidly approaches zero, while m^3 gets to be a super negative number.
4) Thus, you only have to try integer values of 0,1,2 to find a solution.
Takes all of 30 seconds.
Nice!
it would have been better if he would have proved by induction that for m>=4; 3^m-m^3>1
(3^m)-m³=1
The trivial solutiin is m=0
For other possible silution, take modulo 2:
mod(3^m,2)-mod(m³,2)=1
[mod(3,2)]^m-[mod(m,2)]³=1
(1^m)-1=[mod(m,2)]³
mod(m,2)=0 --> m=2k, k an integer
3^m=m³+1 --> 3^m=(m+1)(m²-n+1)
(m+1) | 3^m
(2k+1) | 9^k --> k=1
m=2
Therefore m={0, 2}
Hi SyberMath, big fan.
I had a similar solution to you:
3^m = (m+1)(m^2-m+1)
the terms on the RHS must be powers of 3, since they multiply to make a power of 3
this means that one of these terms must be a factor of the other term
(m^2-m+1)/(m+1) = (m-2) + 3/(m+1)
because (m+1) divides (m^2-m+1), 3/(m+1) has to be an integer
hence (m+1) is a factor of 3
hence m= 0 or 2
(I am a high school student and just also wanted to thank you for improving my problem solving ability allowing me to qualify for a number of mathematics competitions through my school)
Wow! Thank you. I'm glad to hear that
Wow!
When m is 3 3^m=m^3. Starting from m=4 3^m will be much bigger than m^3->81>64 . Hence just try m= 0,1,2 and that's it
0 and 2 is ok. But 1 not good
good thinking!
from 6:40 3^k = 3^(2n) - 3^(n+1) + 3
if k = 0 , 3^(2n) - 3^(n+1) + 3 = 1 => (3^n - 1)(3^n - 2) = 0 => n = 0 => m =0
if k ≥ 1 , 3^(k - 1) = 3^(2n - 1) - 3^n + 1
=> 3^(k - 1) ≡ 1 (mod 3) => k - 1 = 0 => n =1 => m =2
3^m - m^3 = 1
3^m = m^3 + 1
3^m = (m+1)(m^2 - m + 1)
m+1 | 3^m
Hence m+1 is a power of 3.
Case 1: m+1 = 3^0 = 1
Hence m = 0. So, 3^0 - 0^3 = 1 - 0 = 1. Valid solution.
Case 2: m+1 > 1. This means that m+1 is a power of 3, so m = 3k+2
But also, m^2 - m + 1 is a power of 3. And that's 9k^2 + 12k + 4 - 3k - 2 + 1 = 9k^2 + 9k + 3. Well, taking out a multiple of 3, we get 3(3k^2 + 3k + 1). So, 3k^2 + 3k + 1 is a power of 3. But it is clearly not divisible by 3, so it must be 1. Therefore, k = 0, and m = 3(0) + 2 = 2. Testing this, we get 3^2 - 2^3 = 9 - 8 = 1, so it's valid.
Solutions are m=0 and m=2.
Very good!
I also got 0 and 2 as the only solutions.
m = 2
m = 0 or 2
3^m-m³=1
m=0 → ok.
when m>0,
-m³≡1 (mod3) → m=3k-1 (k>0)
3^m=m³+1=(m+1)(m²-m+1)
=3k(9k²-9k+3)=9k(3k²-3k+1)
∴3^p=9k ···①
3^(m-p)=3k²-3k+1 ···②
②→ m-p=0 → 3k(k-1)=0 → k=1
①→ 3^p=3^m=9 → m=2
∴m=0, 2
Nice solution
Thanks!
Good equation
Thanks!
Anyone know what to do when the b and c values in a factoring equation are the same, as in 2x^2 - 9x -9?
use quadratic formula
x₁₂ = -b ± √(b²-4ac) / 2a
No factors in that case, but if the constant term was +9 (and not -9) you would be looking for two numbers with product (ac) = 2 x 9 = 18, and sum (b) = -9.
-3 and -6 satisfy, therefore (x - 3).( 2x - 3) does the trick. In this case, it is nothing to do with b being equal to c, merely that 2 x 9 = 3 x 6. Always! Even early morning on a really crappy Friday.
@@dwm1943 someone else’s comment got deleted where it mentioned a “non monic factorization”. What is that and what did the person above ya mean by x base 12?
Also if (x - 3)(2x - 3) gets foiled, u end up with +9 (due to the -3 • -3), not -9. Is this not factorable?
What are you if not a mathematician?
Someone who likes math! 😜
@@SyberMath You gotta be at least engineer, right?
To me, at least, a mathematician is someone who works to develop entirely new ideas and results in mathematics, whereas a math educator is someone who takes existing ideas and results and teaches/explains them (and how to play around with them to problem-solve) to others. You can be a math educator without being a mathematician, or you can be a mathematician without being a math educator, or you can be both.
I would say that SyberMath is a math educator. (As am I.)
@@scottcowan8036 Yes, and a good one. Some of the maths tutors on RUclips make it like an assault course. It's supposed to be fun, and SM makes it so. Which is why I keep coming back.
Great solution !
Thanks!
Osm explanation sir
Let f(m) = 3^m/m^3
ln f = m ln 3 - 3 ln m
f'/f = ln 3 - 3/m
f' > 0, if m >= 3
Answer 2 and 0
3^m - m^3 =1
3^ m= m^3 + 1
3^integer is always odd
Hence, 3^ m is odd
Hence, m^3 + 1 is odd
Hence, m^3 is even
Hence, m is even
If m is even, then 3^ m unit digit is either 1 or 9
the difference between 3^ m and m^3 =1
3^ 0 = 1 0^3 =0
and 3^ 2 =9 and 2^3 =9
So 0 and 2 are two solutions since the difference between 3^m and m^3 is 1 in both cases.
Let's try some more
3^4 and 4^3 81 and 64, a difference of 17
'
3^6 and 6^ 3 729 and 216, a difference of 513
The difference becomes larger when m increase
Hence, the solutions are 0 and 2
When m is negative, the difference 3^m is not an integer, but m^3 is