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How do you do this?

 Nov 10, 2019
 #1
avatar
+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

 Nov 10, 2019
 #2
avatar+2557 
+1

Help Melody or Cphill or anybody smart!
 

Is there a mathematical way to solve this?

CalculatorUser  Nov 10, 2019
 #3
avatar+106993 
+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?

 Nov 11, 2019
 #4
avatar+28357 
+1

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
 #5
avatar+106993 
+2

Hi Alan,

I was not short. I had left it to the asker to find the last one

 

My words:

"What factors still need to be included,

What is the best way of doing it. "

 

 

But yes I agree the answer is 5.

 

8975 throught to 8979

Melody  Nov 11, 2019
 #6
avatar+28357 
+2

My apologies Melody, I'll read your answers more carefully in future!
 

Alan  Nov 11, 2019
 #7
avatar+106993 
0

It is not a problem Alan, it is good to chat to you :)

Melody  Nov 11, 2019

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