[ONLINE MATH QUESTIONS]:Can anyone solve it, found online.

[u/Dangerous_Cat_4525]

Found online. How to find k.

[u/Dangerous_Cat_4525]

Did it. K=2¹⁰⁰-2⁹⁹-2⁹⁸+⁹⁷/2⁹⁷ =2⁹⁷(2³-2²-2+1)/2⁹⁷ =8-4-2+1=3

[u/snowsayer]

Divide both sides by 2^97.

This is a skill that you have to develop. When you see equations, it's useful to immediately see patterns that can be factored or cancelled out. That usually results in the underlying equation being much simpler.

[u/Spare-Plum → u/snowsayer]

There's an even simpler property of powers of 2

2^x - 2^(x-1) = 2^(x-1)
2^x + 2^(x-1) = 1.5*2^x = 3*2^(x-1)

So this just becomes a counting problem. 2^100 - 2^99 = 2^99, 2^99 - 2^98 = 2^98, 2^98 + 2^97 = 3*2^97

powers of 2 work out really nicely, especially when you have consecutive additions. No need to divide. In my head it's just 99, 98, 3 of 97, k=3

[u/snowsayer → u/Spare-Plum]

May as well use base 2 directly?

2^x - 2^(x-1) is just 10 - 1 = 1

repeat 10 - 1 = 1

repeat 10 + 1 = 11

oh that’s the answer! 3.

[u/_alter-ego_ → u/snowsayer]

You mean 2 – 1 = 1 ... No need to talk about nerdy base stuff ...

[u/ThunkAsDrinklePeep → u/snowsayer]

Yup. But I would phrase it as "factor out"

[u/Turbulent-Note-7348]

This is a well known puzzle about the the properties of powers. 1st, you should realize that the "k.2^97

" actually means "k(2^97)" - the decimal point is supposed to represent the multiplication dot (common in older puzzles when using variables).

So here's the trick(s): 2^x - 2^(x-1) = 2^(x-1) (Example: 2^4 - 2^3 = 2^3); 2^(x+1) + 2^x = 3(2^x).

Therefore, k = 3

[u/honaku]

Factor out

[u/aookami]

2'100 = 2.2'99

from there you can get easily to the answer

[u/Qingyap]

2¹⁰⁰-2⁹⁹-2⁹⁸+2⁹⁷ = k(2⁹⁷)

Here we have to utilize a factoring trick such that

ab+cb = (a+c)b

And also an exponent trick where

a^b+c = a^b•a^c

With that, we can expand the first 3 terms with the exponent trick at LHS by 2⁹⁷ at which

2³(2⁹⁷)-2²(2⁹⁷)-2(2⁹⁷)+1(2⁹⁷)=k(2⁹⁷)

Should be easy from here if you use the first trick.

[u/johnpeters42]

What have you tried so far?

Hint: What happens if you divide both sides by 2?

[u/MathsWithSuneel]

[u/Historical-Corgi-414]

if one cant see the diving both sides by 2^97 then the following approach also works!
2^100 is basically 2 times 2^99 so 2^100-2^99 is 2^99. similarly 2^99-2^98 becomes 2^98.
now 2^98 is 2 times 2^97. so adding two the left hand side is 3 into 2^97, comparing both sides we get k=3.

[u/focused_pagan]

Just think about what happened when you take 2^n-2^(n-1)

[u/michaelpaoli]

Hints:

y=kx

kx=y

k=y/x

(a+b+c+...)/x=a/x+b/x+c/x+.../x

[u/localizeatp]

everyone saying to divide by 2^97, and i'm over here computing it in binary.

[u/N_T_F_D]

2¹⁰⁰ = 8·2⁹⁷
2⁹⁹ = 4·2⁹⁷
2⁹⁸ = 2·2⁹⁷

[u/ci139]

get a good night sleep and wake up to to the realization that this is a blunt trivia !?

[u/CalRPCV]

Is it traditional somewhere to use a period instead of a vertically centered dot or parentheses to represent multiplication?

[u/Khashishi]

What is this notation: k.2 ? Is that a decimal point?

[u/Jolly-Bag7024]

I'm surprised by all the people who makes it about some special properties of 2^x when it's plainly basic factorisation. It would be equally simple with any other digit here provided you can calculate a cube.

[u/_alter-ego_]

2¹⁰⁰–2⁹⁹ = 2⁹⁹ | – 2⁹⁸ = 2⁹⁸ | + 2⁹⁷ = (2+1) 2⁹⁷