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Question Number 155816 by cortano last updated on 05/Oct/21

 How many of the number formed   by using all the digits 1,2,3,4,5,6   only once divisible by 25.

$$\:\mathrm{How}\:\mathrm{many}\:\mathrm{of}\:\mathrm{the}\:\mathrm{number}\:\mathrm{formed} \\ $$$$\:\mathrm{by}\:\mathrm{using}\:\mathrm{all}\:\mathrm{the}\:\mathrm{digits}\:\mathrm{1},\mathrm{2},\mathrm{3},\mathrm{4},\mathrm{5},\mathrm{6} \\ $$$$\:\mathrm{only}\:\mathrm{once}\:\mathrm{divisible}\:\mathrm{by}\:\mathrm{25}. \\ $$

Answered by talminator2856791 last updated on 05/Oct/21

 4! = 24

$$\:\mathrm{4}!\:=\:\mathrm{24} \\ $$

Commented by prakash jain last updated on 05/Oct/21

Should subtract numbers divisble  by 625 from 24.

$$\mathrm{Should}\:\mathrm{subtract}\:\mathrm{numbers}\:\mathrm{divisble} \\ $$$$\mathrm{by}\:\mathrm{625}\:\mathrm{from}\:\mathrm{24}. \\ $$

Commented by Rasheed.Sindhi last updated on 05/Oct/21

Should also subtract numbers divisible  by 25^3  and perhaps by 25^4 ...

$${Should}\:{also}\:{subtract}\:{numbers}\:{divisible} \\ $$$${by}\:\mathrm{25}^{\mathrm{3}} \:{and}\:{perhaps}\:{by}\:\mathrm{25}^{\mathrm{4}} ... \\ $$

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