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heureka

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 #2
avatar+26396 
0

Find the ones digit of the largest power of 2 that divides into (2^4)!.

 

[...] integer Part

 

power of 2:

[2421]+[2422]+[2423]+[2424]=23+22+21+1=8+4+2+1=1524!=215

 

power of 3:

[1631]+[1632]=5+1=624!=21536

 

power of 5:

[1651]=324!=2153653

 

power of 7:

[1671]=224!=215365372

 

power of 11:

[16111]=124!=215365372111

 

power of 13:

[16131]=124!=215365372111131

 

16!=2153653721113

 

laugh

01.08.2016
 #2
avatar+26396 
+5

How would you end up solving, |2-|1-|x||| = 1,
i realize the basic idea but what do you do with the negative after the 2, do you distribute it?
edit: through the powers of trial and error i found
that you dont need to distribute but rather just divide the -1 and the answers are, +-4, +-2, and 0.

 

| 2| 1|x| | |=1|square both sides( 2| 1|x| | )2=1244| 1|x| |+(1|x|)2=144| 1|x| |+12|x|+x2=144| 1|x| |2|x|+x2=0x22|x|+4=4| 1|x| ||square both sides(x22|x|+4)2=16( 1|x| )2(x22|x|+4)(x22|x|+4)=16( 12|x|+x2 )x44|x|x2+12x216x+16=1632|x|+16x2x44|x|x2+12x216x=32|x|+16x2x44|x|x2+16|x|4x2=04|x|(4x2)=4x2x44|x|(4x2)=x2(4x2)|square both sides16x2(4x2)2=x4(4x2)216x2(4x2)2x4(4x2)2=0(4x2)2(16x2x4)=0(4x2)2x2(16x2)=0(4x2)(4x2)xx(16x2)=0

 

1.

4x2=0x2=4x=±4x1=2x2=2

 

2.

x=0x3=0

 

3.

16x2=0x2=16x=±16x4=4x5=4

 

laugh

26.07.2016