let m is a set of even no between 0 to 23 and odd numbers between 23 to 100. if we multiply elements of m with each other. how many zeros we get in their product..

• any zero can be made only mutiplying any even number with 5 so here count the number of 5s b/w 0 to 23(only even) and i e 10 and 20 , two 5 are here
  and b/w 23 to 100(only odd)
  the number of 5s are 25, 35, 45,......95 that is 8 5s
  
  so total 5s are 8+2=10
  so total zeros are 10
• 9 years agoHelpfull: Yes(8) No(11)
• any zero can be made only by multiplying any even number with 5 so count the number of 5s b/w 0 to 23(only even) and i e 10 and 20 , 2 5s are there
  and b/w 23 to 100(only odd)
  the number of 5s are 25, 35, 45,......95 that is 10 5s.
  Total no. of 5s=12
  hence,no. of zeroes will b=12
• 9 years agoHelpfull: Yes(5) No(4)
• after multiplication we get power of 5 is 12, so number of zeroes sud be 12.
• 9 years agoHelpfull: Yes(3) No(1)
• For a single zero . we need one 2 ands one 5 ... from 0 to 23 we have 10 and 20 as no. which contain 5 ... and from 23 to 100 we have 8 numbers which contain 5 ... so no. of zeros in the product is 10
• 9 years agoHelpfull: Yes(1) No(0)
• i think your question is wrong it may be as ....even no between 0 to 23 and odd numbers between 1 to 100.
• 9 years agoHelpfull: Yes(0) No(7)
• I only have one thing to explain i.e the no. of 5s in 25,35...95 =10 nd not 8 which most of people have misunderstood in thier solution as 25 nd 75 have two 5
• 7 years agoHelpfull: Yes(0) No(0)