KCC's Quizzes AQQ279 about smart computing

Let's calculate the sum of the cubes of the even numbers from 2 to 24:

Breaking it down:

• 23=82^3 = 8

• 43=644^3 = 64

• 63=2166^3 = 216

• 83=5128^3 = 512

• 103=100010^3 = 1000

• 123=172812^3 = 1728

• 143=274414^3 = 2744

• 163=409616^3 = 4096

• 183=583218^3 = 5832

• 203=800020^3 = 8000

• 223=1064822^3 = 10648

• 243=1382424^3 = 13824

Now, let's sum them all up:

Adding these together, the value is:

Therefore, the sum of 23+43+63+83+...+223+2432^3 + 4^3 + 6^3 + 8^3 + ... + 22^3 + 24^3 is 49,000.

Pretty neat, isn't it? 

erm, something didn't go quite well with copying the answer in here, but the sum should be correct.

First, you observe that the number of terms of Y is the same as X. Then, you can see that each term in Y is obtained by the respective term in X multiplied by 2³. So Y=8*X= 48672.

and iof i ask Co-pilot the same question a second time, it gives me a different answer.  Can't trust anyone these days...

This ends up being (2*1)^3 + (2*2)^3 + (2*3)^3 ... = 8*1^3 + 8*2^3 ... = 8(X) = 48,672

Stupid Co-pilot: when asked a third time I got *another* wrong answer...   Asking Co-pilot if it was sure, it then confirms 48672 as being correct. Sigh! 

Thanks NorthBus and thanks gpiarino!!!

Hi,

A bit of math:

2^3+4^3+6^3+\cdots+22^3+24^3=

(2\cdot1)^3+(2\cdot2)^3+(2\cdot3)^3+\cdots+(2\cdot11)^3+(2\cdot12)^3=

2^3\cdot1^3+2^3\cdot2^3+2^3\cdot3^3+\cdots+2^3\cdot11^3+2^3\cdot12^3=

2^3\cdot(1^3+2^3+3^3+\cdots+11^3+12^3)

X = 1³ + 2³ + 3³ + 4³ +… + 11³ + 12³ = 6084 was given, so

Y=8\cdot X=8\cdot6084=48672

Given:

X = 1³ + 2³ + 3³ + 4³ +… + 11³ + 12³ = 6084

Y = 2³ + 4³ + 6³… + 22³ + 24³

The equation for Y can be rewritten as...

Y = (2*1)³ + (2*2) ³ + (2*3) ³ + ... + (2*12³)= 6084

Knowing that (a * b)^x = a ^x b ^x then...

Y = 2³* 1³ + 2³ * 2³ + ...+ 2³ * 12³

Y = 2³ (1³  + 2³ + 3³ + … + 12³ ), which is 2³ (X)

Y  = 8 * 6084 = 48,672

assuming the first sum is true, Im not checking then the sum is 6084X 2^3=48672