Question 1207599
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<pre>
Such an integer must be 3 more than a multiple of 47. They form an arithmetic
sequence, with common difference d=47

To find the first term, we first find the first multiple of 47 above 1000.

1000/47 = 21.27659574

So we know from that, that since the whole part is 21, that 21x47 is the last
multiple of 47 less than 1000, and that the first multiple of 47 above 1000 will
be 22x47=1034.

Add 3 to that and we get the first term, 

{{{a[1]=1034+3=1037}}}

{{{a[n]=a[1]+(n-1)*d}}}

{{{a[n]=1037+(n-1)*47}}}

We require the terms to be less than or equal to 2000.

{{{1037+(n-1)*47<=2000}}}

Solve that inequality and get

{{{n<=1010/47}}}

{{{n<=21.4893617}}}

So the answer is the largest integer less than 21.4893617, which is 21.

Edwin</pre>