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# #21MT

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7 How do you do this?

Nov 10, 2019

#1
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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
+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.

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
#5
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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
+2