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Re: What is the units digit of 13^4*17^2*29^3? [#permalink]
17 Oct 2010, 13:32

3

This post received KUDOS

I think that the fastest and more important, the safest way to solve this is: First brake the problem to three multiplication problems: 1. 13*13*13*13 2. 17*17 3. 29*29*29 In all of the problems we don't have to multiply by the whole number, but only by the last digit (because that is what we are asked about). Second, simply multiply: 13*13*13*13. It is actually 3*3*3*3 (we are interested in the last digit). Thus - 3*3=9, 9*3=7(the last digit), 7*3=1 (the last digit). 17*17 is actually 7*7 = 9 (the last digit). 29*29*29 is actually -9*9*9- that can be interpreted into 9*9=1 (the last digit) and 1*9=9. So eventually we will have- 9 (29^3) * 9 (17^2) * 1 (13^4). Therefore - 9*9=1(the last digit), 1*1=1

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