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Find the remainder when 1^2 + 2^2 + 3^2 + ... + \(60^2\) is divided by 15.

 Feb 5, 2022
 #1
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sumfor(n,1, 60, n^2)  mod 15 ==10 - the remainder

 Feb 6, 2022
 #2
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+1

 

 

1^2 + 2^2 + 3^2 + ... +60^2    is divided by 15.

 

Im going to split this up into

 

\((1^2+2^2+ \dots +14^2)+(15^2\dots+29^2)+(30^2\dots+44^2)+(45^2\dots+56^2)+60^2 \mod(15)\\ =4(1^2+2^2+ \dots +14^2)+0 \mod(15)\\ =4(1^2+2^2+ 3^2+4^2+5^2+6^2+7^2+8^2+9^2+10^2+11^2+12^2+13^2+14^2) \mod(15)\\ =4(1^2+2^2+ 3^2+4^2+5^2+6^2+7^2+(-7)^2+(-6)^2+(-5)^2+(-4)^2+(-3)^2+(-2)^2+(-1)^2) \mod(15)\\ =8(1^2+2^2+ 3^2+4^2+5^2+6^2+7^2) \mod(15)\\ =8(1+4 -6+1-5+6+4) \mod(15)\\ =8( 5) \mod(15)\\ =40 \mod(15)\\ =30+10 \mod(15)\\ =10 \mod(15)\)

 

 

 

 

 

You need to check for careless errors

 

 

 

LaTex

(1^2+2^2+ \dots +14^2)+(15^2\dots+29^2)+(30^2\dots+44^2)+(45^2\dots+56^2)+60^2 \mod(15)\\
=4(1^2+2^2+ \dots +14^2)+0 \mod(15)\\
=4(1^2+2^2+ 3^2+4^2+5^2+6^2+7^2+8^2+9^2+10^2+11^2+12^2+13^2+14^2) \mod(15)\\
=4(1^2+2^2+ 3^2+4^2+5^2+6^2+7^2+(-7)^2+(-6)^2+(-5)^2+(-4)^2+(-3)^2+(-2)^2+(-1)^2) \mod(15)\\
=8(1^2+2^2+ 3^2+4^2+5^2+6^2+7^2) \mod(15)\\
=8(1+4 -6+1-5+6+4) \mod(15)\\
=8( 5) \mod(15)\\
=40 \mod(15)\\
=30+10 \mod(15)\\
=10 \mod(15)

 Feb 6, 2022
edited by Melody  Feb 6, 2022

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