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Lê Quốc Trần Anh Coordinator
10/06/2018 at 06:32
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Compare: \(8\sqrt{3}\) with \(5\sqrt{7}\)

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    Nguyễn Mạnh Hùng 14/06/2018 at 02:22

    We have:

    \(8\sqrt{3}=\sqrt{8^2.3}=\sqrt{64.3}=\sqrt{192}\)

    \(5\sqrt{7}=\sqrt{5^2.7}=\sqrt{25.7}=\sqrt{175}\)

    Because 192>175 so \(\sqrt{192}>\sqrt{175}\)

    or \(8\sqrt{3}>5\sqrt{7}\) 

    Lê Quốc Trần Anh selected this answer.

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Lê Quốc Trần Anh Coordinator
11/06/2018 at 02:08
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Given a,b,c,d satisfys \(ab=cd\). Prove that: \(a^5+b^5+c^5+d^5\) isn't a prime number.


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Lê Quốc Trần Anh Coordinator
11/06/2018 at 02:11
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Prove that the product of four consecutive integers is not a square number.


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Lê Quốc Trần Anh Coordinator
11/06/2018 at 02:14
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Find \(n\in N\) so that \(n^5-n+2\) is a square number

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    Dao Trong Luan Coordinator 11/06/2018 at 12:10

    \(n^5-n+2=n\left(n^4-1\right)+2=n\left(n^2+1\right)\left(n^2-1\right)+2\)

    \(=n\left(n-1\right)\left(n+1\right)\left(n^2+1\right)+2\)

    We see: \(\left(n-1\right)\cdot n\cdot\left(n+1\right)⋮3\) (is product of 3 consecutive natural numbers) 

    \(\Rightarrow n^5-n+2\equiv2\left(mod3\right)\)

    But no square numbers are in the form 3k + 2

    So \(n\notin\varnothing\)

    Lê Quốc Trần Anh selected this answer.

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Lê Quốc Trần Anh Coordinator
23/05/2018 at 09:34
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If p, q and r are prime numbers such that pq + r = 73, what is the least possible value of p + q + r? 

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    quanpham04106 24/05/2018 at 02:24

    because p,q,r is a prime number,pq+r=73 , ''the odd number'' multiply ''the odd number'' equal ''the odd number'' and ''the odd number'' plus ''the even number'' equal ''the odd number''. number.

    We have 2 case:

    case 1:r is even. so r = 2

    pq+2=7

    pq=71=1*71(eliminate)

    case 2:p or q is even

    example:p is even.so p=2

    first:r=3

    2*q+3=73

    2*q=70

    q=35

    q isn't the prime number(eliminate)

    second:r=5

    2*q+5=73

    2*q=68

    q=34(eliminate)

    third:r=7

    2*q+7=73

    2*q=66

    q=33

    fourth:r=9

    2*q+9=73

    2*q=64

    q=32(eliminate)

    fifth:r=11

    2*q+11=73

    2*q=62

    q=31(possible)

    so p+q+r=2+31+11=44

    Lê Quốc Trần Anh selected this answer.
  • ...
    Nguyễn Thành Long 26/05/2018 at 08:21

    vì p, q, r là số nguyên tố, pq + r = 73, '' số lẻ '' nhân '' số lẻ '' bằng '' số lẻ '' và '' số lẻ '' cộng ' 'số chẵn' 'bằng' 'số lẻ' '. con số.

    Chúng ta có 2 trường hợp:

    trường hợp 1: r là chẵn. vậy r = 2

    pq + 2 = 7

    pq = 71 = 1 * 71 (loại bỏ)

    trường hợp 2: p hoặc q là chẵn

    ví dụ: p là even.so p = 2

    đầu tiên: r = 3

    2 * q + 3 = 73

    2 * q = 70

    q = 35

    q không phải là số nguyên tố (loại bỏ)

    thứ hai: r = 5

    2 * q + 5 = 73

    2 * q = 68

    q = 34 (loại bỏ)

    thứ ba: r = 7

    2 * q + 7 = 73

    2 * q = 66

    q = 33

    thứ tư: r = 9

    2 * q + 9 = 73

    2 * q = 64

    q = 32 (loại bỏ)

    thứ năm: r = 11

    2 * q + 11 = 73

    2 * q = 62

    q = 31 (có thể)

    do đó p + q + r = 2 + 31 + 11 = 44


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Lê Quốc Trần Anh Coordinator
23/05/2018 at 09:40
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Francisco is born at 1:00 a.m. on a Tuesday and gets married exactly 218 hours later. On what day of the week does Francisco get married?

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    Kaya Renger Coordinator 24/05/2018 at 13:21

    We have : \(2^{18}=262144=262121+23\)

    Total time we can calculate since John was born until he was married is : 

    \(S=2^{18}\left(h\right)+1\left(h\right)=262144\left(h\right)+1\left(h\right)=262128\left(h\right)+17\left(h\right)\)  

        \(=10920\left(days\right)+17\left(h\right)=10922\left(days\right)+2\left(days\right)+17\left(h\right)\)

        \(=1560\left(weeks\right)+2\left(days\right)+17\left(h\right)\)

    Because after a week , the day that Fransisco was born was unchange so it still on Tuesday 

    After Tuesday 2 days is Thurday 

    So , exactly the day that Francisco was married is at 17:00 on Thursday 

    Lê Quốc Trần Anh selected this answer.
  • ...
    huynh anh phuong 23/05/2018 at 10:28

    218=131072 (hours)

    Francisco get married at:131072+1=131073 hours

    131073 :24 = 5461,375 days

    The percent of one day is:24:100= 0,24%

    Because 0,5 day = 12 hours so 6 hours=0,25 day.

    And 0,375-0,25=0,125(day) so 0,125 day=7 hours.

    Answer:5461 days 7 hours


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Lê Quốc Trần Anh Coordinator
18/05/2018 at 12:44
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Prove that: \(abc\ge\left(b+c-a\right)\left(a+c-b\right)\left(a+b-c\right)\)

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    Dao Trong Luan Coordinator 18/05/2018 at 13:36

    Apply inequality cauchy, we have:

    \(\left\{{}\begin{matrix}\left(b+c-a\right)+\left(a+c-b\right)\ge2\sqrt{\left(b+c-a\right)\left(a+c-b\right)}\\\left(a+c-b\right)+\left(a+b-c\right)\ge2\sqrt{\left(a+c-b\right)\left(a+b-c\right)}\\\left(a+b-c\right)+\left(b+c-a\right)\ge2\sqrt{\left(a+b-c\right)\left(b+c-a\right)}\end{matrix}\right.\)

    \(\Rightarrow\left\{{}\begin{matrix}2c\ge2\sqrt{\left(b+c-a\right)\left(a+c-b\right)}\\2a\ge2\sqrt{\left(a+c-b\right)\left(a+b-c\right)}\\2b\ge2\sqrt{\left(a+b-c\right)\left(b+c-a\right)}\end{matrix}\right.\)

    \(\Rightarrow\left\{{}\begin{matrix}c^2\ge\left(b+c-a\right)\left(a+c-b\right)\\a^2\ge\left(a+c-b\right)\left(a+b-c\right)\\b^2\ge\left(a+b-c\right)\left(b+c-a\right)\end{matrix}\right.\)

    \(\Rightarrow\left(abc\right)^2\ge\left[\left(b+c-a\right)\left(a+c-b\right)\left(a+b-c\right)\right]^2\)

    \(\Rightarrow abc\ge\left(b+c-a\right)\left(a+c-b\right)\left(a+b-c\right)\)

    Lê Quốc Trần Anh selected this answer.
  • ...
    Fc Alan Walker 18/05/2018 at 13:41

    Apply inequality cauchy, we have:

    ⎧⎩⎨⎪⎪⎪⎪(b+c−a)+(a+c−b)≥2(b+c−a)(a+c−b)−−−−−−−−−−−−−−−−−√(a+c−b)+(a+b−c)≥2(a+c−b)(a+b−c)−−−−−−−−−−−−−−−−−√(a+b−c)+(b+c−a)≥2(a+b−c)(b+c−a)−−−−−−−−−−−−−−−−−√

    ⇒⎧⎩⎨⎪⎪⎪⎪2c≥2(b+c−a)(a+c−b)−−−−−−−−−−−−−−−−−√2a≥2(a+c−b)(a+b−c)−−−−−−−−−−−−−−−−−√2b≥2(a+b−c)(b+c−a)−−−−−−−−−−−−−−−−−√

    ⇒⎧⎩⎨⎪⎪c2≥(b+c−a)(a+c−b)a2≥(a+c−b)(a+b−c)b2≥(a+b−c)(b+c−a)

    ⇒(abc)2≥[(b+c−a)(a+c−b)(a+b−c)]2

    ⇒abc≥(b+c−a)(a+c−b)(a+b−c)


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Lê Quốc Trần Anh Coordinator
20/05/2018 at 10:31
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Given a,b,c,d are 4 consecutive positive integers. Prove that: \(abcd+1\) is a square number

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    Dao Trong Luan Coordinator 20/05/2018 at 11:06

    Choose the smallest number in this list is a, and b = a+1, c = a+2, d = a+3

    So abcd + 1 = a(a+1)(a+2)(a+3) + 1 = a(a+3).(n+1)(n+2) + 1 

    = (a2 + 3a)(n2 + 3n +2) + 1 

    = [a2 + 3a][(n2 + 3n) + 2] + 1

    = (a2 + 3a)2 + 2.(a2 + 3a).1 + 12 = (a2 + 3a + 1)2 is a square number

    Lê Quốc Trần Anh selected this answer.

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Lê Quốc Trần Anh Coordinator
20/05/2018 at 10:39
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Prove that: \(n^4-10n^2+9⋮384\) with all even natural n.

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    Dao Trong Luan Coordinator 21/05/2018 at 01:26

    I think you are wrong in the number of even and odd numbers in English

    even numbers: số chẵn                                              odd numbers: số lẻ

    So in this post must be n is a odd number.

    \(n^4-10n^2+9=n^4-n^2-9n^2+9=n^2\left(n^2-1\right)-9\left(n^2-1\right)=\left(n^2-1\right)\left(n^2-9\right)=\left(n-1\right)\left(n+1\right)\left(n-3\right)\left(n+3\right)\)

    Because n is a odd number so we can write n at the form n = 2k+1

    So \(\left(n-1\right)\left(n+1\right)\left(n-3\right)\left(n+3\right)=\left(2k+1-1\right)\left(2k+1+1\right)\left(2k+1-3\right)\left(2k+1+3\right)\)

    \(=2k\left(2k+2\right)\left(2k-2\right)\left(2k+4\right)=2k\cdot2\left(k+1\right)\cdot2\left(k-1\right)\cdot2\left(k+2\right)=16\left(k-1\right)k\left(k+1\right)\left(k+2\right)⋮16\)

    But (k-1)k(k+1)(k+2) is the product of four consecutive integers so has multiples of 2,3,4 so \(⋮\left(2\cdot3\cdot4\right)=24\)

    So \(n^4-10n^2+9⋮\left(16.24\right)=384\)

    Lê Quốc Trần Anh selected this answer.

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Nguyễn Thị Linh
21/05/2018 at 02:01
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 B = 1.2.3 + 2.3.4 + ... + (n - 1)n(n + 1)

B=?

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    Huy Toàn 8A (TL) 21/05/2018 at 02:06

    B = 1.2.3 + 2.3.4 + ... + (n - 1)n(n + 1)

    We have:

    4B = 1.2.3.4 + 2.3.4.4 + ... + (n - 1)n(n + 1).4

    = 1.2.3.4 - 0.1.2.3 + 2.3.4.5 - 1.2.3.4 + ... + (n - 1)n(n + 1)(n + 2) - [(n - 2)(n - 1)n(n + 1)]

    = (n - 1)n(n + 1)(n + 2) - 0.1.2.3 = (n - 1)n(n + 1)(n + 2)

    =>B = \(\dfrac{\left(n-1\right)n\left(n+1\right)\left(n+2\right)}{4}\)

    Nguyễn Thị Linh selected this answer.
  • ...
    Nguyễn Thành Long 26/05/2018 at 08:34

    Ta có :           

    \(B=1.2.3+2.3.4+...+\left(n-1\right).n.\left(n+1\right)\)

    \(4B=1.2.3.4+2.3.4.4+...+\left(n-1\right).n.\left(n+1\right).4\)

    \(4B=1.2.3.4+2.3.4\left(5-1\right)+...+\left(n-1\right).n.\left(n+1\right).\left[\left(n+2\right)-\left(n-2\right)\right]\)

    \(4B=1.2.3.4+2.3.4.5-1.2.3.4+...+n.\left(n+1\right).\left(n+2\right)-\left(n-1\right).n.\left(n+1\right)\)

    \(4B=\left[1.2.3.4+2.3.4.5+...+n.\left(n+1\right).\left(n+2\right)\right]\)\(-\left[1.2.3.4+2.3.4.5+...+\left(n-1\right).n.\left(n+1\right)\right]\)

    \(4B=n.\left(n+1\right).\left(n+2\right)\)

    \(B=\dfrac{n.\left(n+1\right).\left(n+2\right)}{4}\)


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