I am stuck at the question:
How many zeros are there when numbers between 1 and 100 are multiplied including 1 and 100, devise some technique for this .
Regards
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Do you mean trailing zeroes? If so, it's fairly easy. If not, it's hard work $-$ you have to do the arithmetic. – TonyK Jul 01 '14 at 14:51
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yes the trailing zeros – user2170497 Jul 01 '14 at 14:52
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2This is a duplicate by any account: Here for the non-trailing, and here for trailing zeros – apnorton Jul 01 '14 at 14:52
2 Answers
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This is asking, what is the highest power of 10, that divides 100! . This is same as the highest power of 5 that divides 100! (why?) , which is given by: [100/5] + [100/5^2] + [100/5^3] + ... = 20 + 4 = 24. This is the de Polignac's Formula.
Somabha Mukherjee
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A number ends with an additionnal 0 iff it is multiplied by 5 and 2.
In your product, there are many more times '$\times$ 2' than '$\times$ 5', hence you can only counts the 5s.
Do not forget to count 2 5s for the multiples of $5\times 5$ (there are no higher powers since 5$\times$5$\times$5>100).
It gives: 20 + 4 = 24 zeros.
mookid
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