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# What is the remainder when 43^43+ 33^33 is divided by 10

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CEO
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What is the remainder when 43^43+ 33^33 is divided by 10 [#permalink]  04 Jan 2008, 13:17
What is the remainder when 43^43+ 33^33 is divided by 10?
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Re: exponents - remainders [#permalink]  04 Jan 2008, 13:37
(43^4)^10 * 43^3 + (33^4)^8 * 33

unit digit 1 * unit digit 7 + unit digit 1 * unit digit 3 = unit digit 0

So divisible by 10.
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Re: exponents - remainders [#permalink]  04 Jan 2008, 15:16
bmwhype2 wrote:
What is the remainder when 43^43+ 33^33 is divided by 10?

3^1=3
3^2=9
3^3=27
3^4=81
3^5=243
...........
3^33=...3
3^43=...7

remainder is 0
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Re: exponents - remainders [#permalink]  04 Jan 2008, 18:59
remainder 0
good qn
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Re: exponents - remainders [#permalink]  05 Jan 2008, 05:20
ashkrs wrote:
(43^4)^10 * 43^3 + (33^4)^8 * 33

unit digit 1 * unit digit 7 + unit digit 1 * unit digit 3 = unit digit 0

So divisible by 10.

what is the purpose of breaking it down like you did? i don't quite follow
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Re: exponents - remainders [#permalink]  05 Jan 2008, 05:31
bmwhype2 wrote:
What is the remainder when 43^43+ 33^33 is divided by 10?

All we care about is the units digit of 43^43 and the units digit of 33^33. Both are dependent on powers of 3.
Units digit of powers of 3 are:
3^1 = 3
3^2 = 9
3^3 = 7
3^4 = 1

3^43 = (3^4)^10 * 3^3 = 1*7 = 7
3^33 = (3^4)^7 * 3^5 = 1*3 = 3

Remainder of (7+3)/10 is 0. That's the answer.
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Re: exponents - remainders [#permalink]  05 Jan 2008, 20:30
why do we only look at the units digit?
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Re: exponents - remainders [#permalink]  05 Jan 2008, 23:16
bmwhype2 wrote:
why do we only look at the units digit?

because all multiples of 10 have their units digit as 0, and no other.
so once we prove that the units digit is 0,we crack the ans.
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Re: exponents - remainders   [#permalink] 05 Jan 2008, 23:16
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