Remainder of Dividing By 5
Calculate the remainder of dividing N = 1 2 0 0 7 + 2 2 0 0 7 + 3 2 0 0 7 + 4 2 0 0 7 + 5 2 0 0 7 + 6 2 0 0 7 + ⋯ + 2 0 0 5 2 0 0 7 + 2 0 0 6 2 0 0 7 + 2 0 0 7 2 0 0 7 By 5
The answer is 4.
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5 solutions
wrong procedure bro.......there are 401 numbers which are divisible by 5....so u cant take this cubic series upto 2007......it is equal to= 401(1^3 + 2^3 + 3^3+ 4^3)+ 1^3+2^3 as n=1,2,3,4(mod5), as (1^3 + 2^3 + 3^3+ 4^3)=100 and 1^3+2^3=9, so the required remainder =4
sorry it is okk as it remains 0 always
We know that x^n+y^n is divisible by x+y for all odd values of n. Here 1^2007+2^2007+3^2007+..........+2007^2007 , notice that 1^2007+2^2007+3^2007+4^2007 is divisible by 5 . And the next number in the series 5^2007 is already a multiple of 5. Similarly all the number till 2005^2007 will leave no remainder when divide by 5. Now we are left with 2006^2007+2007^2007, now 2006^2007 leaves remainder 1 when divided by 5 and 2007^2007 leaves remainder 3 when divided by 5.Therefore when N is divided by 5 the remainder is 3+1=4.
Its simple logic I first found the first terms of 1^2007 to 10^2007 then added it all up...This turns out to be 34 .A little deduction reveals that this pattern continues so 34 when divided by 5 leaves remainder 4 so 4 is the answer .... please comment on its logic
EVERY NO. CAN BE EXPRESSED AS (5n+1),(5n+2),(5n-2),(5n-1) or 5n so each and every term upto the 2005 th term if expanded cancels each other other out and becomes a multiple of 5 only the 2006 th and 2007th term remains........................(2006)^{2007}=(5 401+1)^{2007} = M5+1......................... (2007)^{2007}=(5 401+2)^{2007} = M5+3 [using modulo5]......................... so the remainder is 3+1=4
1^2007 = 1;// reminder 2^2007 = 3; 3^2007 = 2; 4^2007 = 4; 5^2007 = 0; (1+5)^2007 = 1; (2+5)^2007 = 3; (3+5)^2007 = 2; etc. for 5^2005, we got the reminder is 0; and for the 2006 & 2007's reminder it is = 1+3 = 4 .
How about this:
i = 1 ∑ 2 0 0 7 i 2 0 0 7 ≡ i = 1 ∑ 2 0 0 7 i 3 = 2 2 2 0 0 7 2 ( 2 0 0 8 ) 2 = ( 2 0 0 7 ) 2 ( 1 0 0 4 ) 2 ≡ ( 2 ) 2 ( − 1 ) 2 ≡ 4 ( m o d 5 )
Just use FLT.
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beautiful technique ..
but u should mention when n is divisible by 5 it is equivalent to 0 and hence u can take any power of it........there is no use of FLT
What theorem floor ??
1^2007 + 2^2007 + ... + 2007^2007
the unit digit is 1^3 + 2^3 + 3^3 + ... 2007^3
sum of cubes = [(2007 × 2008)/2]² = unit digit in this is 8² = 4
4 ÷ 5 remainder is 4