number

I won't take credit for this solution, as it goes to this website: https://www.cut-the-knot.org/proofs/LinesDividePlane.shtml
Applying 2016 to the formula, we'll get 2033137 regions, which has the sum of digit be 19

Thank you!!!!! :D

3a2 = 2a3

=> a3a2

 = 32 => a = 32

3a² = 2a³

=> \(\dfrac{a^3}{a^2}\) = \(\dfrac{3}{2}\) => a = \(\dfrac{3}{2}\)

my answer is a)708

my answer is 708, too

please check

The total of the cards are: \(9+8+7+6+5+4+3+2+1=45\)

The total of the cards that can removed 1 card is: \(\left(45-1\right)to\left(45-9\right)\)

From this operation and the question, we see only \(C.54\) satisfy these.

So the answer is C

The total of the cards are: 9+8+7+6+5+4+3+2+1=45

The total of the cards that can removed 1 card is: (45−1)to(45−9)

From this operation and the question, we see only C.54

 satisfy these.

So the answer is C

The total of the cards are: 9+8+7+6+5+4+3+2+1=459+8+7+6+5+4+3+2+1=45

The total of the cards that can removed 1 card is: (45−1)to(45−9)(45−1)to(45−9)

From this operation and the question, we see only C.54C.54 satisfy these.

So the answer is C

Because this topic is not to say that the faces should be different so that their sum is the smallest, then the remaining 3 [can be said to be x, y, z] smallest and satisfy:
  - 12*x = 9*y = 6*z
=> x = y = z = 0 is the smallest
So the sum is: 0 + 0 + 0 + 6 + 9 + 12 = 27
Answer: 27

Because this topic is not to say that the faces should be different so that their sum is the smallest, then the remaining 3 [can be said to be x, y, z] smallest and satisfy:
  - 12*x = 9*y = 6*z
=> x = y = z = 0 is the smallest
So the sum is: 0 + 0 + 0 + 6 + 9 + 12 = 27
Answer: 27

The answer is 901.

From 1 to 9 we have sum : 45.

From 10 to 19 we have : 1 + 0 + 1 + 1 + 1 + 2 + 1 + 3 +...+ 1 + 9 we have sum 55.

Next from 20 to 29 we have sum 65.

................. 30 to 39 we have sum 75.

................. 40 to 49 we have sum 85.

................. 50 to 59 we have sum 95.

................. 60 to 69 we have sum 105.

................. 70 to 79 we have sum 115.

................. 80 to 89 we have sum 125.

................. 90 to 99 we have sum 135.

And the last 100 we have sum 1 + 0 + 0 = 1.

So the sum she obtained is : 65 + 75 + 85 + 95 +105 +115 +125 +135 + 1 = 901.

The answer is 901.

From 1 to 9 we have sum : 45.

From 10 to 19 we have : 1 + 0 + 1 + 1 + 1 + 2 + 1 + 3 +...+ 1 + 9 we have sum 55.

Next from 20 to 29 we have sum 65.

................. 30 to 39 we have sum 75.

................. 40 to 49 we have sum 85.

................. 50 to 59 we have sum 95.

................. 60 to 69 we have sum 105.

................. 70 to 79 we have sum 115.

................. 80 to 89 we have sum 125.

................. 90 to 99 we have sum 135.

And the last 100 we have sum 1 + 0 + 0 = 1.

So the sum she obtained is : 65 + 75 + 85 + 95 +105 +115 +125 +135 + 1 = 901.

From 1 to 9 the sum of the individual digits are: \(1+2+3+4+5+6+7+8+9=45\)

From 10 to 19: the tens digit all have the same digit, the ones digit have the rules from 1 to 9.

So as the same with 20-99.

The number 100 has the individual digits sum: \(1+0+0=1\)

She has obtained the sum: \(\left[\left(1+2+3+4+5+6+7+8+9\right).10\right]+\left[\left(1+2+3+4+5+6+7+8+9\right).10\right]+1\)

\(=450+450+1=901\)

So she obtained the sum: \(901\)

That number is 322

\(That\) \(number\) \(is\) \(322\)

i think is 322

Without loss of generality, we may assume gcd(a,b,c) = 1.

(otherwise, if d=gcd(a,b,c) the for a'=a/d, b'=b/d, c'=c/d, the equation still holds for a', b', c' and a'b'c' is still a cube if only if abc is a cube).

We multiply equation by abc, we have:

  \(a^2c+b^2a+c^2b=3abc\)(*)

if \(abc=\pm1\), the problem is solved.

Otherwise, let p be a prime divisor of abc. Since gcd(a,b,c)=1, the (*) implies that p divides exactly two of a, b,c. By symetry, we may assume p divides a, b but not c. Suppose that the lagest powers of p dividing a, b are m, n, respecively.

If n < 2m, then \(n+1\le2m\)  and \(p^{n+1}\)| \(a^2c,b^2c,3abc\). Hence \(p^{n+1}\)|\(c^2b\), forcing \(\)\(p\)|\(c\) (a contradiction). If n > 2m, then \(n\ge2m+1\) and \(p^{2m+1}\)|\(c^2b,b^2a,3abc\). Hence \(p^{2m+1}\)|\(a^2c\), forcing \(p\)|\(c\) (a contradicton). Therefore n = 2m and \(abc=\Pi p^{3m}\), \(p\)|\(abc\), is a cube.