Question Number 198435 by Abdullahrussell last updated on 20/Oct/23
Answered by Rasheed.Sindhi last updated on 20/Oct/23
![2 & 5 make 0 [2×5=10] There are more 2′s than 5′s ∴ Number of trailing zeros =Number of 5′s as a factor Counting 5′s: 5^(5!) ×10^(10!) ×15^(15!) ×20^(20!) ×25^(25!) ×...×40^(40!) ×45^(45!) 5^(5!) ×5^(10!) ×5^(15!) ×...5^(45!) ×5^(25!) 5^(5!+10!+15!+...+45!+25!) Number of trailing 0′s: =5!+10!+15!+...45!+25!](https://www.tinkutara.com/question/Q198444.png)
$$\mathrm{2}\:\&\:\mathrm{5}\:{make}\:\mathrm{0}\:\left[\mathrm{2}×\mathrm{5}=\mathrm{10}\right] \\ $$$$\mathcal{T}{here}\:{are}\:{more}\:\mathrm{2}'{s}\:{than}\:\mathrm{5}'{s} \\ $$$$\therefore\:{Number}\:{of}\:{trailing}\:{zeros} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:={Number}\:{of}\:\mathrm{5}'{s}\:{as}\:{a}\:{factor} \\ $$$${Counting}\:\mathrm{5}'{s}: \\ $$$$\mathrm{5}^{\mathrm{5}!} ×\mathrm{10}^{\mathrm{10}!} ×\mathrm{15}^{\mathrm{15}!} ×\mathrm{20}^{\mathrm{20}!} ×\mathrm{25}^{\mathrm{25}!} ×…×\mathrm{40}^{\mathrm{40}!} ×\mathrm{45}^{\mathrm{45}!} \\ $$$$\mathrm{5}^{\mathrm{5}!} ×\mathrm{5}^{\mathrm{10}!} ×\mathrm{5}^{\mathrm{15}!} ×…\mathrm{5}^{\mathrm{45}!} ×\mathrm{5}^{\mathrm{25}!} \\ $$$$\mathrm{5}^{\mathrm{5}!+\mathrm{10}!+\mathrm{15}!+…+\mathrm{45}!+\mathrm{25}!} \\ $$$${Number}\:{of}\:{trailing}\:\mathrm{0}'{s}: \\ $$$$=\mathrm{5}!+\mathrm{10}!+\mathrm{15}!+…\mathrm{45}!+\mathrm{25}! \\ $$