#1**+1 **

**If I understand your question.........!!!.**

**I wrote a short computer code and it found 2 sequences of positive consecutive 4-digit integers whose product divides 2010^2 evenly, and both are 12 consecutive terms as follows: n=1000;p=0;a=productfor(m, n, n+p, (m));if(a%4040100==0, goto loop, goto next);printp;loop:printn,", ",;next:p++;if(p<=12, goto2,0);p=0;n++;if(n<9999, goto2, 0)**

**(4478, 4479, 4480, 4481, 4482, 4483, 4484, 4485, 4486, 4487, 4488, 4489)=12 terms mod 2010^2 =0 (8967, 8968, 8969, 8970, 8971, 8972, 8973, 8974, 8975, 8976, 8977, 8978) Total = 12 terms mod 2010^2=0**

Guest Nov 10, 2019

#2**+1 **

Help Melody or Cphill or anybody smart!

Is there a mathematical way to solve this?

CalculatorUser
Nov 10, 2019

#3**+3 **

I know the answer I believe

I started by breaking 2010^2 into ist prime factors, 2*3*5*67 *2*3*5*67

Now 67*67 probably needs to be in one of the numbers 67*67=4489

I might as well knock over as many factors as possible with the first number so I will times that by 2

So my first number is 8978 that takes care of 3 of the factors

Now I need 5 and 5 to be factors so I need 8990 or better still 8975 that will take care of 5*5 both at the same time

So i have

8975, to 8978 That is 4 numbers.

How many factors will that take care of?

What factors still need to be included,

What is the best way of doing it.

What is the answer?

Melody Nov 11, 2019

#4**+2 **

I think you are one short Melody!

A numerical check shows that (8975*8976*8977*8978) mod 2010^2 = 1346700,

but (8975*8976*8977*8978*8979) mod 2010^2 = 0, so a minimum of 5 is required.

Alan
Nov 11, 2019