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heureka

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Questions 17
Answers 5678

 #1
avatar+26396 
+3

What is the remainder when the sum

12+22+32+...+20162+20172+20182+20192+20202 

is divided by 17?

 

12(mod17)=122(mod17)=432(mod17)=942(mod17)=1652(mod17)=862(mod17)=272(mod17)=1582(mod17)=1392(mod17)=13102(mod17)=15112(mod17)=2122(mod17)=8132(mod17)=16142(mod17)=9152(mod17)=4162(mod17)=1172(mod17)=0

cycle mod 17: (1,4,9,16,8,2,15,13,13,15,2,8,16,9,4,1,0)

 

circle mod 17

1+4+9+16+8+2+15+13+13+15+2+8+16+9+4+1+0(mod17)=136(mod17)=0(mod17)

 

2020/17 = 118 cycles remainder 14

 

remainder 14, this is the sum in the cycle from 1 to 14  mod 17

1+4+9+16+8+2+15+13+13+15+2+8+16+9(mod17)=131(mod17)=12(mod17)

 

12+22+32+...+20162+20172+20182+20192+20202(mod17)=0118(mod17)118 cycles+12(mod17)remainder 14=12(mod17)

 

The remainder when the sum 12+22+32+...+20162+20172+20182+20192+20202 is divided by 17 is 12

 

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May 27, 2022
 #1
avatar+26396 
+2

Find all complex solutions such that z^4 = -4
Note: All solutions should be expressed in the form a+bi,
where a and b are real numbers help!

 

z2=±2i

 

z2=±2i|z=a+bi(a+bi)2=±2ia2+2abi+b2i2=±2i|i2=1a2+2abib2=±2ia2b2+2abi=±2ia2b2+2abi=+2ia2b2+2abi=0+2i|compare2ab=2b=1aanda2b2=0a2=b2a2=(1a)2a4=1a=±1a1=1a2=1b1=1a1b1=1b2=1a2b2=1ora2b2+2abi=2ia2b2+2abi=02i|compare2ab=2b=1aanda2b2=0a2=b2a2=(1a)2a2=(1a)2a4=1a=±1a3=1a4=1b3=1a3b3=1b4=1a4b4=1

 

z=a1+b1i=1+iz=a2+b2i=1iz=a3+b3i=1iz=a4+b4i=1+i

 

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May 26, 2022
 #1
avatar+26396 
+2

Compute

 

1232+1333+1434++110310

 

s=1232+1333+1434++110310=12(221)+13(321)+14(421)++110(10219=1(21)2(2+1)+1(31)3(3+1)+1(41)4(4+1)++1(101)10(10+1)++1(n1)n(n+1)s=10n=21(n1)n(n+1)1(n1)n=1n11n1(n+1)n=1n1n+11(n1)n(n+1)=12(1n12n+1n+1)s=1210n=2(1n12n+1n+1)=12(10n=21n110n=22n+10n=21n+1)=12(9n=11n10n=22n+11n=31n)9n=11n=11+12+9n=31n10n=22n=2221029n=31n11n=31n=110+111+9n=31ns=12(11+12+9n=31n2221029n=31n+110+111+9n=31n)=12(11+1222210+110+111)=12(12210+110+111)=12(12110+111)=14120+122=520120+122=420+122=15+122=22+5522s=27110

 

1232+1333+1434++110310=27110

 

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May 22, 2022
 #1
avatar+26396 
+2
May 20, 2022