<prompt> Tafuta nambari ndogo kabisa chanya n kiasi kwamba kuna r in 0, tfrac11000 kiasi kwamba nambari nr3 ni nambari kamili.  <response>: Tunadai kuwa r kama hiyo ipo ikiwa tu frac3n21000 frac3n10002 frac110003 1.Kwanza , tuseme kwamba nr3 ni nambari kamili, kwa baadhi r in left0, tfrac11000right. Since nr3 n3 na n3 ni nambari kamili, lazima tuwe na nr3 ge n3 1,hivyo 3rn2 3nr2 r3 ge 1. Tangu r tfrac11000 na n0, tunapata tfrac3n21000 tfrac3n10002 tfrac1103 3rn2 3nr2 r3 ge 1, unavyotaka.Kinyume chake, tuseme kwamba tfrac3n21000 tfrac3n10002 tfrac1103 1. Define fx 3xn 2 3nx2 x3, ili tuwe na flefttfrac11000kulia 1. Since f0 0 1 na f inaendelea, lazima kuwe na r in left0, tfrac11000right vile fr 1. Kisha kwa thamani hii ya r, tuna anza iliyopangwa nr3 n3 3rn2 3nr2 r3 n3 fr n3 1, endalignedambayo ni nambari kamili, kama inavyotakiwa.Kwa hivyo, inatosha kupata nambari ndogo kabisa chanya n ya kuridhisha frac3n21000 frac3n10002 frac110003 1 .Muhula wa kwanza katika upande wa kushoto ni mkubwa zaidi kuliko maneno mengine mawili, kwa hivyo tunatafuta n satisfying tfrac3n21000 approx 1, au n takriban sqrttfrac10003 takriban 18. Tunapata kwamba n 18 haikidhi usawa, lakini n 19 inatimiza. 