Using Remainder Th.
Since 2890 = 289*10, and 289 is square of 17. Thus 2890 is perfectly divisible by 17.
Now
2891 will leave 1 as remainder.
2892 will leave 2 as remainder.
2893 will leave 3 as remainder.
2894 will leave 4 as remainder.
2895 will leave 5 as remainder.
2896 will leave 6 as remainder.
2897 will leave 7 as remainder.
2898 will leave 8 as remainder.
2899 will leave 9 as remainder.
2990 will leave 10 as remainder.
Now\(\frac{1*2*3*4*5*6*7*8*9*10}{17}\)
Using remainder th. we can used smaller multiplication like
\(\frac{9*2}{17}\) gives 1 as remainder,
\(\frac{6*3}{17}\)gives 1 as remainder,
\(\frac{5*7}{17}\)gives 1 as remainder,
\(\frac{4*10}{17}\)gives 6 as remainder,
and 8 is remaining,
thus ultimate remainder is
1*1*1*6*8 = 48 which when divided by 17 gives 14 as remainder.
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Abhishek Parikh
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