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If [m](125^14)(48^8)[/m] is written out as an integer, how many consec
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06 Mar 2021, 05:03
06 Mar 2021, 05:03
00:00
Question Stats:
75% (01:02) correct
25% (02:14) wrong based on 4 sessions
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If \((125^{14})(48^8)\) is written out as an integer, how many consecutive zeroes will that integer have at the end?
A. 8
B. 14
C. 32
D. 42
E. 74
A. 8
B. 14
C. 32
D. 42
E. 74
Re: If [m](125^14)(48^8)[/m] is written out as an integer, how many consec
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06 Mar 2021, 05:15
06 Mar 2021, 05:15
2
\((125^14)(48^8)\) = \((5^42)(2^32)(3^8)\)
= \((5^10)(10^32)(3^8)\)
so there will be 32 consecutive zeros
option C is correct
= \((5^10)(10^32)(3^8)\)
so there will be 32 consecutive zeros
option C is correct
If [m](125^14)(48^8)[/m] is written out as an integer, how many consec
[#permalink]
06 Mar 2021, 05:33
06 Mar 2021, 05:33
2
Solution:
The number of zeros in a number is basically the number of 2s and 5s.
Here as the number of 2s in the above \(48^8\)= \((2^4)^8\) x \(3^8\)=\(2^32\) x\(3^8\)
Thus, there will be 32 consecutive zeros.
The number of zeros in a number is basically the number of 2s and 5s.
Here as the number of 2s in the above \(48^8\)= \((2^4)^8\) x \(3^8\)=\(2^32\) x\(3^8\)
Thus, there will be 32 consecutive zeros.
