April has 30 days.

=> Zeus needs to throw: 30.12=360 (lightning bolts) in April.

In the first week of April, he has thrown: 15.7=105 (lightning bolts)

In the next 30-7=23 days, Zeus needs to throw: 360-105=255 (lightning bolts)

So the average is: 255:23=11.08≈

11 (lightning bolts per day).

The difference between the loud rock concert and the normal conversation is 120-60=60 dB. It's 3 times of 20 dB. Because every increase of 20 dB corresponds to sound becoming 10 times as loud, the rock concert must be 10 × 10 × 10 = 1000 times as loud as the normal conversation.

a + b = 273

a - b = 23

⇒

 (a + b) - (a - b) = 250

     2b = 250 

       b = 125​ ​​ ​​ ​​​ ​​ ​​ ​​ ​​ ​​ ​​​ ​​ ​​ ​

​⇒

​ a = 273 - 125 = 158

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We have: |7x−5y|≥0; |2z−3x|≥0 and |xy+yz+zx−2000|≥0

=> P≥0+0+0≥0

P=0

 only when 7x−5y=2z−3x=xy+yz+zx−2000=0

=> 7x=5y;2z=3x;xy+yz+zx=2000

=> x5=y7

; z3=x2 and xy+yz+zx=2000

=> x10=y14=z15

 and xy+yz+zx=2000

Put x10=y14=z15=k

We have: k2=x10⋅y14=y14⋅z15=z15⋅x10=xy+yz+zx10⋅14+14⋅15+15⋅10=2000140+210+150

=2000500=4

=> k=±2

=> (x;y;z)=(20;28;30)

 and  (−20;−28;−30)

So minP=0

 only when (x;y;z) = (20;28;30) and (-20;-28;-30)

(2x1−5y1)2018≥0;(2x2−5y2)2018≥0;.......;(2x2019−5y2019)2018≥0

⇒(2x1−5y1)2018+(2x2−5y2)2018+......+(2x2019−5y2019)2018≥0

Because (2x1−5y1)2018+(2x2−5y2)2018+..........+(2x2019−5y2019)2018≤0

⇒(2x1−5y1)2018+(2x2−5y2)2018+......+(2x2019−5y2019)2018=0

⇒2x1=5y1;2x2=5y2;..........;2x2019=5y2019

⇒x1y1=x2y2=..........=x2019y2019=52=x1+x2+.......+x2019y1+y2+.......+y2019

i think the title is wrong. It is: 34(AB+AC+BC)<AD+BE+CF<AB+AC+BC

We have :

x2−5x+1=0

⇒x2+1=5x

 ( 1 )

x2+1x2=(x2+1x2+2.1x.x)−2.1x.x=(x+1x)2−2

                =(x2+1x)2−2

 ( 2 )

( 1 )( 2 )⇒x2+1x2=(5xx)2−2=25−2=23

Ans : 23.

0≤a,b≤9⇒−9≤a−b≤9;a+b≤18

We have 53ab98¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯⋮99

,so :

+ 53ab98¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯⋮9⇒25+a+b⋮9⇒a+b=2;11

+53ab98¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯⋮11⇒|5+a+9−3−b−8|=|3+a−b|⋮11

⇒a−b=−3;8

If a + b = 2,then a + b is even but 2b is even

=> a + b - 2b = a - b is even => a - b = 8 > a + b (absurd)

=>{a+b=11a−b=−3⇒{2a=8b=11−a

⇒{a=4b=7

We have  [(n-1)n(n+1)=(n^2-n)(n+1)=(n^2-n)n+n^2-n=n^3-n^2+n^2-n]

          [=n^3-n]

[\Rightarrow n^3=(n-1)n(n+1)+n]   [(1)]

Apply (1) to the expression, we have : 13 + 23 + 33 + ... + 20083 

= (1 - 1)1(1 + 1) + 1 + (2 - 2)2(2 + 2) + 2 + ....... + (2008 - 2008) 2008 (2008 - 2008) + 2008

= 1 + 2 + 1.2.3 + 3 + 2.3.4 + ........ + 2008 + 2007.2008.2009

= (1 + 2 + 3 + ...... + 2008) + (1.2.3 + 2.3.4 + ...... + 2007.2008.2009)

= 2017036 + 2007.2008.2009.2010 / 4 

= The number to mik not count

a724b¯¯¯¯¯¯¯¯¯¯¯¯⋮12

.So we have :

+ a724b¯¯¯¯¯¯¯¯¯¯¯¯⋮4⇒4b¯¯¯¯¯⋮4⇒b=0;4;8

+a724b¯¯¯¯¯¯¯¯¯¯¯¯⋮3⇒13+a+b⋮3

If b = 0,then ab = 0

If b = 4,then 17+a⋮3⇒a=1;4;7⇒ab=4;16;28

If b = 8,then 21+a⋮3⇒a=0;3;6;9⇒ab=0;24;48;72

Hence,the maximum value of ab is 72 (only when a = 9 ; b = 8)

m1725n¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯⋮4⇒5n¯¯¯¯¯¯⋮4⇒n=2;6

If n = 2,then m17252¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯

divided by 11 with remainder 5

⇒m17247¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯⋮11

⇒|m+7+4−1−2−7|=|m+1|⋮11⇒m∈∅

If n = 6,then m17256¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯

 divided by 11 with remainder 5

⇒m17251¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯⋮11

⇒|m+7+5−1−2−1|=|m+8|⋮11⇒m=3

 Hence,(m ; n) = (3 ; 6),so m + n = 9

1962mn6¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯⋮72

 .So we have :

+ 1962mn6¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯

 is divisible by 8 

⇒mn6¯¯¯¯¯¯¯¯¯¯⋮8⇒mn6¯¯¯¯¯¯¯¯¯¯⋮4⇒n∈{1;3;5;7;9}

+ 1962mn6¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯⋮9⇒24+m+n⋮9

- If n = 1,then 25+m⋮9⇒m=2

.

Because 216⋮8

,we choose (m ; n) = (2 ; 1)

- If n = 3,then 27+m⋮9⇒m=0;9

Because36⋮̸ 8;936⋮8

,we choose (m ; n) = (9 ; 3)

- If n = 5,then 29+m⋮9⇒m=7

Because 756⋮8̸ 

,we don't choose this case

- If n = 7,then 31+m⋮9⇒m=5

Because 576⋮8

,we choose (m ; n) = (5 ; 7)

- If n = 9,then 33+m⋮9⇒m=3

Because 396⋮̸ 8=49,5=99/2

,we don't choose this case

Hence,(m ; n) = (2 ; 1) ; (9 ; 3) ; (5 ; 7)

¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯

divided by 8 with remainder 2

⇒n78¯¯¯¯¯¯¯¯

divided by 8 with remainder 2

=> n = 1 ; 3 ; 5 ; 7 ; 9

506mn78¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯

⋮9⇒26 + m + n⋮9

If n = 1,then 27 + m⋮9⇒ m = 0 ; 9

If n = 3.then 29 + m⋮9⇒ m = 7

If n = 5,then 31 + m⋮9⇒ m = 5

If n = 7.then 33 + m⋮9⇒ m = 3

If n = 9,then 35 + m⋮9⇒ m = 1

Hence,(m ; n) = (0 ; 1) ; (9 ; 1) ; (7 ; 3) ; (5 ; 5) ; (3 ; 7) ; (1 ; 9)

506mn78¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯

divided by 8 with remainder 2

⇒n78¯¯¯¯¯¯¯¯

divided by 8 with remainder 2

=> n = 1 ; 3 ; 5 ; 7 ; 9

506mn78¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯⋮9⇒26+m+n⋮9

If n = 1,then27+m⋮9⇒m=0;9

If n = 3.then 29+m⋮9⇒m=7

If n = 5,then 31+m⋮9⇒m=5

If n = 7.then 33+m⋮9⇒m=3

If n = 9,then 35+m⋮9⇒m=1

Hence,(m ; n) = (0 ; 1) ; (9 ; 1) ; (7 ; 3) ; (5 ; 5) ; (3 ; 7) ; (1 ; 9)

3a+5b⋮13⇒2(3a+5b)⋮13or6a+10b⋮13

We have :13a+13b⋮13⇒(13a+13b)−(6a+10b)⋮13

⇒7a+3b⋮13

Trịnh Đức Phát 22/03/2017 at 12:43

We have: 

⇒a=2017n−2016b

We have: 2a+2015b=2(2017n−2016b)+2015b=2.2017n−2017b⋮2017

Same as above. We hava: ⎧⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎨⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎩3a+2014b=3.2017n−2.2017b⋮20174a+2013b=4.2017n−3.2017b⋮2017...2015a+2b=2015.2017n−2014.2017b

So that: A⋮2017.2017...2017

⇒A⋮20172014

(*) 1+N=.....=x2/(x-2)2+4

We have x2 ≥

 0 , (x-2)2+4 ≥

 4 > 0 

So 1+N ≥

 0 => N ≥

 -1 ;equality : x=0 

(*)1-N=....=(x-4)2/(x-2)2+4 

....... -> N ≤

 1 , equality : x=4

We have :

x2−2x+5=x2−2x+1+4

                      =(x−1)2+4≥4

⇒H=1x2−2x+5≤14

⇒MaxH=14

⇔(x−1)2=0

⇔x=1

Use Cauchy's inequality for positive numbes a,b,c.

a+b≥2ab−−√,b+c≥2bc−−√,c+a≥2ca−−√

⇒(a+b)(b+c)(c+a)≥8ab−−√⋅bc−−√⋅ca−−√=8abc

⇒A=abc(a+b)(b+c)(c+a)≤18

MaxA=18⇔a=b=c>0

There is something is wrong here 

Change to :

We have a2+b2≥5

=> 9−(a2+b2)≤4

=> 2ab≤4

=> ab≤2

<=> a2b2≥4

=> 4a2b2≥16