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Question Number 147661 by mathdanisur last updated on 22/Jul/21

$${41}^3+{42}^3+{43}^3+...+{59}^3$$$$Findthelastthreedigitsofthe$$$$number$$

Answered by nimnim last updated on 22/Jul/21

$$S=(1^3+2^3+3^3+....+{59}^3)-(1^3+2^3+3^3+....+{40}^3)$$$$={\left(\frac{59\times 60}{2}\right)}^2-{\left(\frac{40\times 41}{2}\right)}^2$$$$={\left(\cancel{1770}\right)}^2-{\left(\cancel{820}\right)}^2$$$$=(..70-..20)(..70+..20)\to lasttwodigits$$$$=..50\times ..90$$$$=....500\to lastthreedigits$$

Commented by mathdanisur last updated on 23/Jul/21

$$Sir,\left(\frac{59\cdot 60}{2}\right)-...canyouexplainthis$$$$partplease.?howdidyougetit.?$$

Commented by mathdanisur last updated on 22/Jul/21

$$thanksSir$$

Commented by mathdanisur last updated on 23/Jul/21

$${\left(\frac{59\cdot 60}{2}\right)}^2-...$$

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