11Abhirup, y 13? Very unlucky [17][3]
2.9523279903960414084761860964352e+38
Can anyone convert this to normal numbers and pls also tell me how to do it???? When calcultors return this as the answer, I get so angry...
Pls reply
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19 Answers
106yes but u have not counted 25 and 50 which themselves have 2 multiples of 5 so total multiples is 12 + 2 (1 each for 25 and 50 more)
1look........10=5 into 2
so no. of 2 in 60! is...[60/2]+[60/4]+[60/8]+[60/16]+[60/32]
=30+15+7+3+1
= 56
and no. of 5 in 60! is..[60/5]+[60/25]
=12+2
= 14
so HCF of 14 and 56 is....14
so ther are 14 zeroes
13but there is one 25....that gives u an extra 5
so 13
11Since there are 12 multiples of 5 in 60 and since everytime u multiply by a 0, an xtra 0 gets added...so 12 0's should be there....
PLS TELL WHAT SHOULD BE MODIFIED IN THIS LOGIC?? ABHIRUP, ALSO WHY DID U EDIT UR REPLY FROM 12 TO 13?
106at the end of 60! there are 14 zeros..
60! = 256.514.(I) where I is an integer with no powers of 2 or 5
hence 60! = 1014.(I)
so there are 14 zeros.
11Pls also tell how many 0's there are in 60!........i thought 12.......but it's given ** [3]
11I was multiplying by e and adding it to that number!!!!
That is actually 34! [3]
13how many zeros are at the end of 60!?
13 0s at the end
112.9523279903960414084761860964352 * 10^38
Pink me my answer is rite
138.32098711 × 1081
does not mean that there r 81 0s