Remainder? Part 3
Find the remainder when,
1^(2019) + 2^(2019) + 3^(2019) + ........ + 2019^(2019) + 2020^(2019)
Is divided by 2019 .
The answer is 1.
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4 solutions
We know that the expression can be factorized as follows
( 1 + 2 + . . . + 2 0 1 9 ) 2 0 1 9 ( A ) + 2 0 2 0 2 0 1 9 Here A is a large number
= ( 2 0 1 9 ) 2 0 1 9 â‹… ( 1 0 1 0 ) 2 0 1 9 ( A ) + 2 0 2 0 2 0 1 9
Thus the expression containing A is completely divisible by 2 0 1 9 and the last term is congruent to 1 mod 2 0 1 9 making the answer 1
@Pi Han Goh Sir, could you help me please?? How do I align the colored part of the solution to the right??
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Your solution is actually incorrect.
I assumed you meant that A is an integer as well.
Your first math line tells us that ( 1 + 2 + ⋯ + 2 0 1 9 ) 2 0 1 9 is a factor of 1 2 0 1 9 + 2 2 0 1 9 + ⋯ + 2 0 1 9 2 0 1 9 . However, it's obvious to see that the former term is much larger than the latter term.
Here's a smaller example to illustrate this:
Using your argument, we know that there's an integer B satisfying 1 4 + 2 4 + 3 4 + 4 4 = ( 1 + 2 + 3 + 4 ) 4 â‹… B . However, solving for B gives B = 0 . 0 3 5 4 , which is clearly not an integer.
There's an easy fix to your solution here.
You just have to demonstrate/state/prove that 1 2 n + 1 + 2 2 n + 1 + ⋯ + m 2 n + 1 divides 1 + 2 + ⋯ + m , for all non-negative integers n .
Can you work it out from here?
The given thing can be written as,
[{1^(2019) + 2^(2019) + ...... + 2018^(2019)} + {2019^(2019) + 2020^(2019)}] ÷ 2019
When we divide 1^(2019) by 2019 , Remainder = 1
When we divide 2018^(2019) by 2019 , Remainder = 2018
When we divide [1^(2019)+2018^(2019)] by 2019 , Remainder = 0
Similarly, we can pair up (1,2018) ; (2,2017) ; (3,2016) ; ..... ; (1009,1010). And each time we get remainder as 0 .
So, the first part i.e. [1^(2019) + 2^(2019) + .... + 2018^(2019)] when divided by 2019 leaves remainder 0 .
When we divide, 2019^(2019) by 2019 we get 0 as remainder.
Finally when we divide 2020^(2019) by 2019 we get 1 as the remainder.
So, total remainder is = 0 + 0 + 1 = 1 ​
Brute force: M o d [ P l u s @ @ T a b l e [ P o w e r M o d [ i , 2 0 1 9 , 2 0 1 9 ] , i , 2 0 2 0 ] , 2 0 1 9 ] ⇒ 1
Plzz elaborate
"Brute force" is an American idiom for "using unreasoning strength." In this case, using a calculator powerful enough to do the computation directly. As in, We used the brute force of a large earth moving machine to pull the motorcycle out of the ditch.
PowerMod is a Wolfram Mathematics function that computes the first argument to the power of the second and applies the modulus of the third argument whenever needed to minimize the product. I used that function to produce a table of the reduced terms, added the terms and then applied a final modulus in case the sum was over 2019. Another idiom to the same effect: I used a pile driver to crack a nut.
1 2 0 1 9 + 2 2 0 1 9 + ⋯ + 2 0 1 9 2 0 1 9 + 2 0 2 0 2 0 1 9 ≡ ( − 2 0 1 8 ) 2 0 1 9 + ( − 2 0 1 7 ) 2 0 1 9 + ⋯ + ( − 1 0 1 0 ) 2 0 1 9 + 1 0 1 0 2 0 1 9 + ⋯ + 2 0 1 8 2 0 1 9 + 0 2 0 1 9 + 1 2 0 1 9 ( m o d 2 0 1 9 ) 1 2 0 1 9 + 2 2 0 1 9 + ⋯ + 2 0 1 9 2 0 1 9 + 2 0 2 0 2 0 1 9 ≡ − 2 0 1 8 2 0 1 9 − 2 0 1 7 2 0 1 9 − ⋯ − 1 0 1 0 2 0 1 9 + 1 0 1 0 2 0 1 9 + ⋯ + 2 0 1 8 2 0 1 9 + 0 2 0 1 9 + 1 2 0 1 9 ( m o d 2 0 1 9 ) 1 2 0 1 9 + 2 2 0 1 9 + ⋯ + 2 0 1 9 2 0 1 9 + 2 0 2 0 2 0 1 9 ≡ 1 2 0 1 9 = 1 ​ ( m o d 2 0 1 9 )