Convert all '8's to 2^3. That gives a pattern of 2^15+2^12+2^9+2^6+2^3+1 or (in binary): 1001001001001001. Since all values are 16 bit, every CS student knows them by heart: 32768+4196+512+64+8+1. That's doable.
As a digital electronics designer in the field of Digital Signal Processing for 25+ years, I approve of this method. The powers of two have been drilled into my head through endless repetition and I can instantly tell you any power of two up to 2^32.
I would just put my 10 cents in by pointing out that you cna take advantage of the fact that 8 is 2 to the power of 3. Using this you can directly rewrite the answer in binary form as 11111000, This reduces the problem from an equation of powers to a simple mulitplication and addition problem. Bit 1::2 to the power of 0 is 1 Bit 2 : 2 to the power of 1 is 2 Bit 3: 2 to the power of 2 is 4 Bit 4: 2 to the power of 3 is 8 Bit 5: 2 to the power of 4 is 16 Bit 6: 2 to the power of 5 is 32 Bit 7: 2 to the power of 6 is 64 Bit 8: 2 to the power of 7 is 128 Bit 9: 2 5o the power of 8 is 256 Bit 10:2 to the power o f 9 is 512 Bit 11: 2 to the power of 10 is 1024 Bit 12: 2 to the power of 11 is 2048 Bit :13: 2 to the power of 12 is 4096 Bit 14 :2 to the power of 13 is 8192 Bit 15: 2 to the power of 14 is 16384 Bit 16: 2 to the power of 15 is 32768 Now: You just have to add the values for 2 to the power of 3*0, 2 to the power of 3*1, 2 to the power of 3*2, 2 to the power of 3*3, 2 to the power of 3*4, and 2 to the power of 3*5., or you can express it in binary form as 0100 1001 0010 0100. (This is easily converted to hexadesumal notation as well.)
Derivacija točke x0 iznad osi x je 1, a derivacija kad je točka na osi x, 0>0 je 0. Na to sam mislio. Moglo bi biti točno, ali i ne mora biti. Ne sjećam se što sam učio, ali mislim da sam učio da je 0>0 =1, makar mi to nema logike.
If we know value of Log8 then we can solve very easily without any calculator with simple multiplications. Every person who attends such exams will prepare to know Log1 to Log20 values, so that most of the problems can be solved.
Anyone attending any such exam who fails to just x8 and uses logs is failing as they can’t use an appropriate solution. My 10 year old niece could do this problem a year ago.
= 8《 8 { 8 [ 8 ( 8 + 1 ) + 1 ] + 1 } + 1 》+ 1 and you have only add 1 and multiply by 8 (i.e multiply 3 times per 2) several times, in fact only simple calculations.
The solution proposed looks like something a Harvard Grad would produce. Meanwhile...I just multiplied it out. Took about 30 sec. even without a calculator.
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37449 in 1 minute. 1 + 8 + 64 + 8^3 ( 1 + 8 + 64 ) = 73 + 8^3*73 = 73+73*64*8 = 73 ( 1+512 ) = 73 * 513. Take a sheet of paper = 37449.
Exactly.
If you add step by step from left to right you even don't need paper.
8^5 + 8^4 + 8^3 + 8^2 + 8^1 + 8^0 = (8 + 1) (4096 + 64 + 1) = 9 * 4161 = 41610 - 4161 = 37449
73 x 513 hanya kurang dari 60 detik? Jika sekelas anak SMA, brilian 😂😂😂
Not very elegant, but effective, and not long-winded.
How about just (8^6 - 1)/(8 - 1) = (262144 - 1)/7 = 262143/7 = 37449.
If you can't readily calculate 8^6, then (8^6 - 1)/7 = (8^3 - 1)(8^3 + 1)/7 = (512 - 1)(512 + 1)/7 = (511 × 513)/7 = 513 × (511/7) = 513 × 73 = 37449.
Convert all '8's to 2^3. That gives a pattern of 2^15+2^12+2^9+2^6+2^3+1 or (in binary): 1001001001001001. Since all values are 16 bit, every CS student knows them by heart: 32768+4196+512+64+8+1. That's doable.
We do indeed know at a glance that it is 4096 not 4196 😅
@BaldurNorddahl Touché - good catch!!
It's 1001001001001001 in base 2 Easy.
That was my strategy, but in base 8
11112
Sum of a G.P. : a(rⁿ - 1)/(r - 1)
a = First term = 1
r = common ratio = 8
n = number of terms
1.(8⁶ - 1)/(8 - 1)
= (2¹⁸ - 1)/7
= 37,449
@@guyhoghton399 how long do you calculate 2^18? 🤔🤔🤔
2^10=1024, it's a well known number.
Generally, Sn= 1+k+++k^n= (k^(n+1)-1)/(k-1), standard scholarship....
No idea why anyone competent in arithmetic would fail. Multiplying by 8.
1, 8,64,512,4096,32768. Add them up.
Absolutely no need for any factorising.
Sum of geometric sequence terms,
@@youben3468yes, but you don’t even need that!
1+8+64+512+4096+32768=37449
Can be done really fast without a calculator.
It can be done without calculation even, just write it in binary
0b1001001001001
As a digital electronics designer in the field of Digital Signal Processing for 25+ years, I approve of this method. The powers of two have been drilled into my head through endless repetition and I can instantly tell you any power of two up to 2^32.
I would just put my 10 cents in by pointing out that you cna take advantage of the fact that 8 is 2 to the power of 3. Using this you can directly rewrite the answer in binary form as 11111000, This reduces the problem from an equation of powers to a simple mulitplication and addition problem.
Bit 1::2 to the power of 0 is 1
Bit 2 : 2 to the power of 1 is 2
Bit 3: 2 to the power of 2 is 4
Bit 4: 2 to the power of 3 is 8
Bit 5: 2 to the power of 4 is 16
Bit 6: 2 to the power of 5 is 32
Bit 7: 2 to the power of 6 is 64
Bit 8: 2 to the power of 7 is 128
Bit 9: 2 5o the power of 8 is 256
Bit 10:2 to the power o f 9 is 512
Bit 11: 2 to the power of 10 is 1024
Bit 12: 2 to the power of 11 is 2048
Bit :13: 2 to the power of 12 is 4096
Bit 14 :2 to the power of 13 is 8192
Bit 15: 2 to the power of 14 is 16384
Bit 16: 2 to the power of 15 is 32768
Now: You just have to add the values for 2 to the power of 3*0, 2 to the power of 3*1, 2 to the power of 3*2, 2 to the power of 3*3, 2 to the power of 3*4, and
2 to the power of 3*5., or you can express it in binary form as 0100 1001 0010 0100. (This is easily converted to hexadesumal notation as well.)
64*65 = 32*2*13*5 = 32*13*10
11111(base8)=1/7(77777)=1/7(100000-1)
=1/7*((8^6)-1)
Geometric progression formule : Am I a joke to you?
64х65=65х65-65=4225-65=4160
Most easy answer: 1001001001001001 (binary)
64*65=65*65-65=4225-65=4160
Derivacija točke x0 iznad osi x je 1, a derivacija kad je točka na osi x, 0>0 je 0. Na to sam mislio. Moglo bi biti točno, ali i ne mora biti. Ne sjećam se što sam učio, ali mislim da sam učio da je 0>0 =1, makar mi to nema logike.
If we know value of Log8 then we can solve very easily without any calculator with simple multiplications. Every person who attends such exams will prepare to know Log1 to Log20 values, so that most of the problems can be solved.
Anyone attending any such exam who fails to just x8 and uses logs is failing as they can’t use an appropriate solution. My 10 year old niece could do this problem a year ago.
It is a sum of terms of geometric serie!!!!!
= 8《 8 { 8 [ 8 ( 8 + 1 ) + 1 ] + 1 } + 1 》+ 1 and you have only add 1 and multiply by 8 (i.e multiply 3 times per 2) several times, in fact only simple calculations.
Wow really good job 🎉🎉🎉
The solution proposed looks like something a Harvard Grad would produce. Meanwhile...I just multiplied it out. Took about 30 sec. even without a calculator.
8^3.73+73=73(512+1)=73×513.ans
等比数列求和都能算是奥林匹克题了？
Probably person is educated upto class 10th only and is unaware of geometric progression
A stupid problem
8^6 - 1
Or 2^18 -1
Harvard Entrance Exams: 8⁵ + 8⁴ + 8³ + 8² + 8¹ + 8⁰ =?
(8⁵ + 8⁴) + (8³ + 8²) + (8¹ + 8⁰) = 8⁴(8 + 1) + 8²(8 + 1) + (8 + 1) = (9)(8⁴ + 8²) + 9
8⁴ + 8² = 8²(8² + 1) = (64)(65) = (60 + 4)(60 + 5) = 3600 + 9(60) + 20 = 4160
8⁵ + 8⁴ + 8³ + 8² + 8¹ + 8⁰ = (9)(8⁴ + 8²) + 9 = (10 - 1)(4160) + 9 = 41600 - 4151 = 37449
16 bits
1001001001001001
That is just 111111 base 8..
37449 no calculator
God! Why not 64.65 = 4160 ?
बारबार विस्तार करना ठीक नही
Really? That seems really easy
8:03 (10-1)(4161)=
=41610
- 4161
=37449 😁
1438
37449
39448
Aise kaun karta hai
Show smart, it is(8^6-1)/7
Exactly, sum geometric sequence terms,
prolisso
Käse
I m from India , you can ask me some problems and solutions , pay me ten dollars per hour
Not smart
37448