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Re: Find the unit's digit in the product (2467)^153 * (341)^72 [#permalink]

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28 Oct 2010, 21:33

Hence D[/quote]

+1 Thanks.. Your answer is correct. I am little confused with the solution here

7^1 =7 7^2 = 49 7^3 = 343 7^4 = ...1 7^5 = ...7

what does it mean ?[/quote]

Please refer to the section "LAST DIGIT OF A PRODUCT" in the link math-number-theory-88376.html . This is an exhaustive post on Number theory by Bunuel.

Re: Find the unit's digit in the product (2467)^153 * (341)^72 [#permalink]

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01 Nov 2010, 05:29

We only have to look at the first part of a equation, as the second part will always end in 1, having no affect on the first part. If we know that 7 has a cyclicity of 4 (what gurpeet showed), we can simply divided

153/4

which gives us the remainder 1.

7^1 ends in 7 (as does 7^5), so the product also ends in 7.

Re: Find the unit's digit in the product (2467)^153 * (341)^72 [#permalink]

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24 Mar 2015, 03:29

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Re: Find the unit's digit in the product (2467)^153 * (341)^72 [#permalink]

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24 May 2016, 13:20

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Re: Find the unit's digit in the product (2467)^153 * (341)^72 [#permalink]

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17 Aug 2017, 17:20

1

This post was BOOKMARKED

here 7^2=49 7^3=343 7^4=...1(unit digit) 7^152=...1(unit digit) 152 is the multiple of 4 (38*4). The question is 152 is also the multiple of of 2(76*2)..then why 4 is taken?