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exponent problem

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ash_maverick Senior | Next Rank: 100 Posts Default Avatar
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exponent problem

Post Sun Jan 21, 2007 4:43 am
What is the remainder when 43^43+ 33^33 is divided by 10?

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telugupilla Newbie | Next Rank: 10 Posts Default Avatar
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Post Fri Feb 02, 2007 2:00 pm
Thanks for this cool tip!

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aim-wsc Legendary Member
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Post Fri Feb 02, 2007 9:14 pm
Sad thats my suicide note Laughing none but this problem is responsible for it. Twisted Evil

haha i know such big numbers are justbubbles. i dealt with units 5 6 1 they are of course easy. but this with 3 as a unit must belong to ''tough'' category. Smile

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telugupilla Newbie | Next Rank: 10 Posts Default Avatar
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Post Fri Feb 02, 2007 2:00 pm
Thanks for this cool tip!

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aim-wsc Legendary Member
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Post Fri Feb 02, 2007 9:14 pm
Sad thats my suicide note Laughing none but this problem is responsible for it. Twisted Evil

haha i know such big numbers are justbubbles. i dealt with units 5 6 1 they are of course easy. but this with 3 as a unit must belong to ''tough'' category. Smile

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limits660 Senior | Next Rank: 100 Posts
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Post Sun Feb 04, 2007 6:36 am
Stacey Koprince wrote:
Whenever they ask you a "units digit" question with really high exponents, there will be a pattern and you will only have to follow the units digit through the problem (b/c it is all multiplication).

43^43:

3^1 = 3^2 = 3^3 = 23^4 = 83^5 = etc. so the units digit pattern is 3-9-7-1. The pattern repeats every 4th term. So 3^4, 3^8, 3^12, etc, will all have the units digit 1. 3^40 will be 1, 3^41 will be 3, 3^42 will be 9, 3^43 will be 7. Same pattern as above. 3^32 will be 1, 3^33 will be 3. units digit 7 + units digit 3 will equal units digit 0. Anything that ends in 0 will have a remainder of 0 when divided by 10.
WOW, beautiful trick. I totally missed that one 8) Thanks Laughing

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Post Mon Jan 22, 2007 6:13 pm
Whenever they ask you a "units digit" question with really high exponents, there will be a pattern and you will only have to follow the units digit through the problem (b/c it is all multiplication).

43^43:

3^1 = 3^2 = 3^3 = 23^4 = 83^5 = etc. so the units digit pattern is 3-9-7-1. The pattern repeats every 4th term. So 3^4, 3^8, 3^12, etc, will all have the units digit 1. 3^40 will be 1, 3^41 will be 3, 3^42 will be 9, 3^43 will be 7. Same pattern as above. 3^32 will be 1, 3^33 will be 3. units digit 7 + units digit 3 will equal units digit 0. Anything that ends in 0 will have a remainder of 0 when divided by 10.

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maxim730 Senior | Next Rank: 100 Posts Default Avatar
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Post Sun Jan 21, 2007 5:20 pm
chandra_adesh wrote:
Last digit in the expansion of 43^43=7
Last digit in the expansion of 33^33=3

Last digit of 43^43+33^33=0

Hence remainder=0
how did you get the last digit? 8)

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