Exponents Problem (m04q20) : Retired Discussions [Locked] - Page 2
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# Exponents Problem (m04q20)

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15 Oct 2013, 21:39
An easy question. Product of 2 and 5 has the last digit is 0 so E is the correct answer.
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16 Oct 2013, 06:12

Solution: (2^15 * 3^15 * 5^15) * (2^7 * 7^1 * 5^1)= (2^15 * 3^15 * 5^15) * (5*7*128)=> no need to find the whole product of 2nd part = (2^15 * 3^15 * 5^15) * (XXXX0) => considering that the number ends with zero product will end with zero.
Re: Exponents Problem (m04q20)   [#permalink] 16 Oct 2013, 06:12

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# Exponents Problem (m04q20)

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