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How many integers satisfy (sqrt n- sqrt 8836)^2<1? Solution) sqrt 8836 = 94 , let sqrt n=xthe equation becomes...(x-94)^2 < 1(x-94)^2 - 1 < 0(x-95)(x-93) < 0hence 938649 < n < 9025 the number of integers are 9024-8650+1=375
Hence the number of Integers are 375 (ANS).
Note on point of tangent
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ABCD is a parallelogram which AB and CD are divides by P and Q. Such that AP:PB=3:2 and CQ:QD=4:1. If PQ and AC are meet at R, show that AR=3/7AC.
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