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If 125^1448^8 were expressed as an integer, how many : Multiple-choice Questions — Select One Answer Choice

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If 125^1448^8 were expressed as an integer, how many [#permalink]
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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?

To find the number of consecutive numbers we need to find what is the power of 10 in the given number.
And to find the power of 10 we need to find out how many 5's and 2's are there which can get multiplied to give us 10

\(125^{14}*48^8\)
125 can be written as \(5^3\)
=> \((5^3)^{14}*48^8\) (and 48 can be written as 16*3 )
=> \(5^{42}*(16*3)^8\) (16 can be written as \(2^4\))
=> \(5^{42}*(2^4*3)^8\)
=> \(5^{42}*2^{4*8}*3^8\)
=> \(5^{42}*2^{32}*3^8\)
=> \(5^{10}*5^{32}*2^{32}*3^8\)
=> \(5^{10}*10^{32}*3^8\)

So, we can have 32 consecutive 0's in the end as we have \(10^{32}\) as the maximum power of 10 in the number.

So, Answer will be B
Hope it helps!

Watch the following video to MASTER Exponents

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Re: If 125^1448^8 were expressed as an integer, how many [#permalink]
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Re: If 125^1448^8 were expressed as an integer, how many [#permalink]
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