number

Tôn Thất Khắc Trịnh 23/07/2018 at 16:04
I won't take credit for this solution, as it goes to this website: https://www.cuttheknot.org/proofs/LinesDividePlane.shtml
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Applying 2016 to the formula, we'll get 2033137 regions, which has the sum of digit be 19 
Mr Puppy 24/07/2018 at 01:44
Thank you!!!!! :D
Uchiha Sasuke
14/06/2018 at 11:24Uchiha Sasuke
22/04/2018 at 12:40

Lê Anh Duy 22/04/2018 at 15:48
3a^{2} = 2a^{3}
=> \(\dfrac{a^3}{a^2}\) = \(\dfrac{3}{2}\) => a = \(\dfrac{3}{2}\)
huynh anh phuong
13/04/2018 at 12:02Đỗ Phi Phi
26/07/2017 at 23:12Alex placed 9 number cards and 8 addition symbol cards on the table as shown.
9  +  8  +  7  +  6  +  5  +  4  +  3  +  2  +  1 
Keeping the cards in the same order he decided to remove one of the addition cards to form a 2digit number. If his new total was 99, which 2digit number did he form?
A.32 B.43 C.54 D.65 E.76

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: \(\left(451\right)to\left(459\right)\)
From this operation and the question, we see only \(C.54\) satisfy these.
So the answer is C
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FA KAKALOTS 08/02/2018 at 22:04
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

Duy Trần Đức 11/02/2018 at 08:30
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
Đỗ Phi Phi
26/07/2017 at 22:44
FA KAKALOTS 08/02/2018 at 22:01
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 
Dao Trong Luan 28/07/2017 at 19:04
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
Đỗ Phi Phi
26/07/2017 at 22:28
Searching4You 27/07/2017 at 09:29
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.
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FA KAKALOTS 08/02/2018 at 22:01
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 2099.
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: \(901\)
Đỗ Duy Mạnh
11/04/2017 at 11:15

Đỗ Duy Mạnh 11/04/2017 at 20:08
\(That\) \(number\) \(is\) \(322\)

Vũ Hà Vy Anh 11/04/2017 at 17:49
i think is 322

John 10/04/2017 at 15:24
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.
Carter selected this answer.