Find the number of zeros in the expression 15*32*25*22*40*75*98*112*125.
A. 9
B. 7
C. 14
D. 12

• (3*5)*(2^5)*(5^2)*(11*2)*(5*2^3)*(3*5^2)*(49*2)*(7*2^4)*(5^3)
  so no of 5 = 9
  & no of 2=14
  as 5*2=10
  that's why no of zero=9
• 11 years agoHelpfull: Yes(67) No(13)
• we will get a zero for one (5*2)
  15=5*3
  32=2*2*2*2*2
  25=5*5
  22=2*11
  40=2*2*2*5
  75=3*5*5
  98=2*7*7
  112=2*2*2*2*7
  125=5*5*5
  therefore no. of pairs [5*2] are 9 . therefore total no. of zeros are 9
• 11 years agoHelpfull: Yes(57) No(3)
• Number of zeros is depend on (5*2) ...
  here no. of 2 > no. of 5..
  so no. of zeros depend on no. of 5..
  Checking no. of 5..
  Number No. of 5
  15 1
  25 2
  40 1
  75 2
  125 3
  So no. of zeros will be 9.. :)
  
• 11 years agoHelpfull: Yes(29) No(4)
• in the above numbers the 5 are obtained as
  
  15-one 5
  25-two 5's
  40-one 5
  75-two 5's
  125-three 5's
  
  here the total number of 5's is 9
  so total number of 0's can be formed as 9.
• 10 years agoHelpfull: Yes(6) No(0)
• here factorize no in term of 5 ther have total no of 5 is 9 then the answer is5
  A is answer
• 11 years agoHelpfull: Yes(2) No(2)
• [3*2]*[2*2*2*2*2]*[5*5]*[11*2]*[2*2*5]*[3*5*5]*[2*7*7]*[7*2*2*2*2]*[5*5*5]
  
  total number of pairs of (2,5) equals total number of zeros
  
  = 8
• 9 years agoHelpfull: Yes(2) No(4)
• We have to count only the numbers of 5 in each digit!
  so 15=5*3-->1
  25=5*5-->2
  40=5*8-->1
  75=5*5*3-->2
  125=5*5*5-->3
  so total number of 5=(1+2+1+2+3)=9--->number of zeroes=9!
• 9 years agoHelpfull: Yes(2) No(0)
• no. of zeros in any expn. depends on the no. of 5*2 pairs......nd we knw the no. of 2's is always > the no. of 5's thus no. of zeros depends on the no. of 5's in the expn. thus from the above expn. we get 9 5's thus no. of zeros is 9
• 11 years agoHelpfull: Yes(1) No(2)
• a will b the answer
  
• 11 years agoHelpfull: Yes(1) No(0)
• ans is 9. the possible combination of 2 and 5 is only 9 which produces zeros.
• 11 years agoHelpfull: Yes(1) No(0)
• 9
  no of zeros=no of pairs of (2*5)
  
• 11 years agoHelpfull: Yes(1) No(1)
• Factorize in terms of multiple of 2 and 5 as far as possible and now count the number of 2 and 5. The number of 2 and 5 will form a zero. Here 9 is the answer
• 11 years agoHelpfull: Yes(0) No(1)
• answew is9
• 11 years agoHelpfull: Yes(0) No(2)
• no. of 2>no. of 5
  14>9
  and 2*5=10
  so, ans is 9.
• 11 years agoHelpfull: Yes(0) No(0)
• ans-A
  because we are getting 9 combinations of 5*2 on breaking the above equation.
• 11 years agoHelpfull: Yes(0) No(0)
• as there are total nos of 5 is 9 so 5*2 gives 0 that is why with the help of 5 we can say probability of 0 is 9....
  9
• 11 years agoHelpfull: Yes(0) No(0)
• no. of zeros=no. of 5's....
  ie. 9
• 11 years agoHelpfull: Yes(0) No(2)
• 15*32*25*22*40*75*98*112*125
  =(15*32)*(25*22)*(40)*(75*98)*(112*125)
  =480 * 550 * 40 * 7350 * 14000
  = 264000 * 40 * 102900000
  = 264* 100 * 4 * 10 * 1029 * 100000
  = 264*4*1029* 10^8
  
  So total 8 zeroes.
  
• 9 years agoHelpfull: Yes(0) No(0)
• multi of 5's and 2's is
  5's=9
  2's=14
  5*2=10
  total how many 5 is there in the above eq is 9
  A is the ans
• 9 years agoHelpfull: Yes(0) No(0)
• option a
  5^9* 2^14, so we get 9 zeros
• 9 years agoHelpfull: Yes(0) No(0)
• 5*3*4*8*5*5*2*11*2*10*5*5*52*49*56*2*5*5*5
  
  cobinations give 0
  5*4,8*5 ,5*2,10,2*5,5*52,56*5,2*5 =9 zeros
  
• 9 years agoHelpfull: Yes(0) No(1)
• multiply in the calculator and you will get the below number
  1086624000000000
  so, there are totally 10 zeros
  (why everyone answered "9" i don't understand,please enlighten me)
  
• 8 years agoHelpfull: Yes(0) No(0)
• A:9
  multiply last two digits of number
• 8 years agoHelpfull: Yes(0) No(0)
• See, the number of zeroes at the end of any number/product depend on the number of times 1010 occurs in its factorization.
  
  Therefore, to find zeroes at the any product we can simply find 1010 in its product.
  
  Since 10=2∗510=2∗5, we can further simply our task to finding number of pairs of 22 and 55 in our product.
  
  That being said, our product can be written as 15∗32∗25∗40∗75∗98∗112∗12515∗32∗25∗40∗75∗98∗112∗125
  =213∗32∗59∗73=213∗32∗59∗73
  
  We clearly have 1313 number of 22 and 99 of 55.
  
  Given that, we can make maximum 99 pairs of 22 and 55. This means, we have 99 times 1010 in our multiplaction. Which means we will have 99 zeroes at the end of the given product.
• 7 years agoHelpfull: Yes(0) No(0)
• A. 9
  
  There are 9 pairs of (5*2) in the given expression.
  e.g. 75=5*15
• 6 years agoHelpfull: Yes(0) No(0)
• No of fives in the abv expression=9.
• 5 years agoHelpfull: Yes(0) No(0)