Tricky Division

Number Theory Level 2

Find the remainder when $2012 \times 2013 \times 2014$ is divided by $2011$ .

The answer is 6.

6 solutions

Math Man
Sep 16, 2014

2012 x 2013 x 2014 is congurent to 1 x 2 x 3 (mod 2011)= 6 (mod 2011)

Can you tell me from which site I can learn mod that you have used in your anwer

tarun mali - 6 years, 8 months ago

You can search this topic in "Congruence Theory".

Asama Zaldy Jr. - 6 years, 8 months ago

Rocel Polintang
Sep 30, 2014

Let n=2011 [(n+1)(n+2)(n+3)]/n simply +1+2+3=6 :)

Tarun Mali
Sep 17, 2014

Let 2011=x , than we have to find remainder when (x+1)×(x+2)×(x+3) is divided by x ,when me multiply it each term will contain x and hence divisible by 2011 but the last term 6 is not divisible and the remainder. No knowledge of mod required in this way

Trevor Arashiro
Sep 16, 2014

We have $(a+1)(a+2)(a+3)\equiv b (mod a)$ . Upon expansion, every term on the LHS is clearly will be divisible by a. Thus we only need the last term: 6. And that is our remainder.

Vishal S
Dec 15, 2014

When we divide 2012 with 2011, the remainder is 1.when 2013 is divided by 2011, the remainder is 2.And when 2011 divides 2014, the remainder is 3.

Therefore the remainder when 2011 divides 2012 x 2013 x 2014 is 1+2+3=6.

No .. that clearly is the wrong approach, because if you keep going, you would get that the remainder of 2012x2013x2014x2015 divided by 2011 is 10, when the correct answer is 24. 2012=2011+1, so let 2011=x and write (x+1)(x+2)(x+3)(x+4) .. if you cross multiply, the only term that will not be divisible by x is the last one, which will be 1 2 3*4=24.

David Moore - 6 years, 1 month ago

Rohit Sharma
Dec 1, 2014

add 1,2,3 to 2011 to get 2012,2013,2014 and by adding 1,2,3 we get 6 which is the remainder

Tricky Division

The answer is 6.

6 solutions

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