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The number of zeros at the end of number 28! + 29! is _____
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- so here ans is
we get 6 zeros at the end
28! + 29! = 28! + 29*28! = 28!( 1 + 29 ) = 28! * 30 = 30.28.27.26.25.24.23.22.21.20.19.18.17.16.15.14.13.12.11.10.9.8.7.6.5.4.3.2.1
now 30/5 = 6 we get 6 zeros at the end .
for example
( 5*2 = 10) whenever there is 5 and 2 present in the number we get the zeros at how many 5 present in the largest no
(1*2*3*4*5*6*7*8*9*10) we have to divide 10/5 = 2. so we get zeros - 4 years agoHelpfull: Yes(6) No(3)
- number of zeroes in 28! is 6 i.e 304888344611713860501504000000
number of zeroes in 29! is 6 i.e 8841761993739701954543616000000
after adding them we will get 7 zeroes at the end...because look at the end 4000000+6000000...we get 10000000 - 4 years agoHelpfull: Yes(3) No(1)
- great observation by Ashutosh Kumar.
Slight correction in ans though.
we have two 5's in number 25,
so we 7 zero's - 4 years agoHelpfull: Yes(2) No(1)
- 28! + 29! = 28! (1+29) = 28! * 30
No. of zeroes in 28! = 28/5 +28/(5*5) = 5+1 = 6
Therefore no. of zeroes in 28! + 29! = 28! * 30 = 6+1=7
hence 7 is the answer - 3 years agoHelpfull: Yes(1) No(0)
- 28! => 6 zeros. (28/5 = 5, 5/5 = 1, total = 5+1 = 6)
28! + 29! = 28!(1 + 29)
= 28! * 30
=> 6zeroes + 1(for 30)
=> 7 zeroes(ans) - 7 Months agoHelpfull: Yes(0) No(0)
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