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Mr Puppy
23/07/2018 at 10:34
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2
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Let N be the largest number of region that can be formed by drawing 2016 straight lines on a plane. Find the sum of all digits of N.

number

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    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.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

    Selected by MathYouLike
  • ...
    Mr Puppy 24/07/2018 at 01:44

    Thank you!!!!! :D


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LOL
01/07/2018 at 13:57
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0
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The operation @ on two numbers produces a number equal to their sum minus 2. The value of  (1 @ 2 @ 3 @ ... @ 2017)

number


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Uchiha Sasuke
14/06/2018 at 11:24
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0
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Let N be the largest number of region that can be formed by drawing 2016 straight lines on a plane. Find the sum of all digits of N.

number


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Uchiha Sasuke
22/04/2018 at 12:40
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2
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Find the number not equal to 0 such that triple of its square is equal to twice its cube.

number

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    FA NAKROTH 25/04/2018 at 14:39

    3a2 = 2a3

    => a3a2

     = 32 => a = 32

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    Lê Anh Duy 22/04/2018 at 15:48

    3a2 = 2a3

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


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huynh anh phuong
13/04/2018 at 12:02
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3
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Choose the correct answer:

The number lighter than 709 and greater than 707 was:

a)708   b)710   c)712   d)714

number

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    Fixida 13/04/2018 at 13:32

    my answer is a)708

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    Anh Kiều 14/04/2018 at 06:53

    my answer is 708, too

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    Fixida 13/04/2018 at 13:32

    please check


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Đỗ Phi Phi
26/07/2017 at 23:12
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3
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Alex 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 2-digit number. If his new total was 99, which 2-digit number did he form?

   A.32                 B.43                  C.54                D.65            E.76

number

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    Lê Quốc Trần Anh Coordinator 27/07/2017 at 09:13

    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

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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

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    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 Cundefined


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Đỗ Phi Phi
26/07/2017 at 22:44
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2
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6 9 12 This cube has a different whole number on each face, and has the property that whichever pair of opposite faces is chosen, the two numbers multiply to give the same result.What is the smallest possible total of all 6 numbers on the cube?

numbergrade 5Grade 6

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    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

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    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


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Đỗ Phi Phi
26/07/2017 at 22:28
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3
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Rani wrote down the numbers from 1 to 100 on a piece of paper and then correctly added up all the individual digits of the numbers. What sum did she obtain?

grade 5Grade 6number

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    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.

    Selected by MathYouLike
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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.

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    Lê Quốc Trần Anh Coordinator 27/07/2017 at 09:24

    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: \(901\)


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Đỗ Duy Mạnh
11/04/2017 at 11:15
Answers
10
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Between the numbers : 321 and 323 is how much ?

number

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    Siêu nhân họ Đỗ 11/04/2017 at 20:15

    That number is 322

    Đỗ Duy Mạnh selected this answer.
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    Đỗ Duy Mạnh 11/04/2017 at 20:08

    \(That\) \(number\) \(is\) \(322\)

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    Vũ Hà Vy Anh 11/04/2017 at 17:49

    i think is 322


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Carter
10/04/2017 at 08:39
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Let a;b;c be integers such that \(\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a}=3\). Prove that abc is the cube of an integer.

intergernumber

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    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.

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