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If n = (33)^43 + (43)^33 what is the units digit of n?

0 2 4 6 8

pls suggest approach for such probs

In such type of exponents problem, first find the cyclicity of unit digit of 3. which is (3,9,7,1)

Now convert exponent value in to multiple of 4. here 43=4(10)+3 so take 3^3 as unit value of first term which is 27 ===> unit digit is 7 (3,9,7,1) Now 33=4(8)+1,so take 3^1 as unit value of second term which is 3^1====> unit digit is 3 (3,9,7,1)

Now unit digit of n = unit digit of first term + unit digit of 2nd term = 7+3=10 = unit digit is 0

Re: If n = (33)^43 + (43)^33 what is the units digit of n? [#permalink]

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04 Oct 2012, 04:27

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Expert's post

Archit143 wrote:

If n = (33)^43 + (43)^33 what is the units digit of n?

A. 0 B. 2 C. 4 D. 6 E. 8

Topic moved to PS forum.

Check Number Theory chapter of our Math Book to know how to deal with such problems: math-number-theory-88376.html (LAST DIGIT OF A POWER part) _________________

Re: If n = (33)^43 + (43)^33 what is the units digit of n? [#permalink]

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If n = (33)^43 + (43)^33 what is the units digit of n?

0 2 4 6 8

pls suggest approach for such probs

In such type of exponents problem, first find the cyclicity of unit digit of 3. which is (3,9,7,1)

Now convert exponent value in to multiple of 4. here 43=4(10)+3 so take 3^3 as unit value of first term which is 27 ===> unit digit is 7 (3,9,7,1) Now 33=4(8)+1,so take 3^1 as unit value of second term which is 3^1====> unit digit is 3 (3,9,7,1)

Now unit digit of n = unit digit of first term + unit digit of 2nd term = 7+3=10 = unit digit is 0

I can't understand why it has to be a multiple of 4. I mean, 43 can also be expressed as (2)(21)+1 but this would be wrong. Generally in such exercises it has to be a multiple of 4, or I don't get something?

If n = (33)^43 + (43)^33 what is the units digit of n?

0 2 4 6 8

pls suggest approach for such probs

In such type of exponents problem, first find the cyclicity of unit digit of 3. which is (3,9,7,1)

Now convert exponent value in to multiple of 4. here 43=4(10)+3 so take 3^3 as unit value of first term which is 27 ===> unit digit is 7 (3,9,7,1) Now 33=4(8)+1,so take 3^1 as unit value of second term which is 3^1====> unit digit is 3 (3,9,7,1)

Now unit digit of n = unit digit of first term + unit digit of 2nd term = 7+3=10 = unit digit is 0

I can't understand why it has to be a multiple of 4. I mean, 43 can also be expressed as (2)(21)+1 but this would be wrong. Generally in such exercises it has to be a multiple of 4, or I don't get something?

If n = (33)^43 + (43)^33 what is the units digit of n?

A. 0 B. 2 C. 4 D. 6 E. 8

First of all, the units digit of (33)^43 is the same as that of 3^43 and the units digit of (43)^33 is the same as that of 3^33. So, we need to find the units digit of 3^43 + 3^33.

Next, the units digit of 3 in positive integer power repeats in blocks of four {3, 9, 7, 1}: 3^1=3 (the units digit is 3) 3^2=9 (the units digit is 9) 3^3=27 (the units digit is 7) 3^4=81 (the units digit is 1) 3^5=243 (the units digit is 3 again!) ...

Thus: The units digit of 3^43 is the same as the units digit of 3^3, so 7 (43 divided by the cyclicity of 4 gives the remainder of 3). The units digit of 3^33 is the same as the units digit of 3^1, so 3 (33 divided by the cyclicity of 4 gives the remainder of 1).

Therefore the units digit of (33)^43 + (43)^33 is 7 + 3 = 0.

If n = (33)^43 + (43)^33 what is the units digit of n?

A. 0 B. 2 C. 4 D. 6 E. 8

First of all, the units digit of (33)^43 is the same as that of 3^43 and the units digit of (43)^33 is the same as that of 3^33. So, we need to find the units digit of 3^43 + 3^33.

Next, the units digit of 3 in positive integer power repeats in blocks of four {3, 9, 7, 1}: 3^1=3 (the units digit is 3) 3^2=9 (the units digit is 9) 3^3=27 (the units digit is 7) 3^4=81 (the units digit is 1) 3^5=243 (the units digit is 3 again!) ...

Thus: The units digit of 3^43 is the same as the units digit of 3^3, so 7 (43 divided by the cyclicity of 4 gives the remainder of 3). The units digit of 3^33 is the same as the units digit of 3^1, so 3 (33 divided by the cyclicity of 4 gives the remainder of 1).

Therefore the units digit of (33)^43 + (43)^33 is 7 + 3 = 0.

Answer: A.

For more on this check theory and problems listed below:

Re: If n = (33)^43 + (43)^33 what is the units digit of n? [#permalink]

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09 Sep 2014, 07:13

Expert's post

honchos wrote:

Bunuel,

I found mistake in the solution given by you-

The Cycle here is

9, 7, 1, 3 Not 3, 9, 7 , 1 (as you stated)

Multiply 33 x 33 Once, what do we get 1089, Unit digit is 9. So the cycle should start at 9

By that logic we should add

1 + 9 = 10 Answer is still A.

But lets tweak it-

(33)^43 + (43)^34

Bunuel you will do this with your chosen cycle- 7+9 = 16, Your answer will be 6

With my chosen Cycle- 1+7 = 8.

That could be a trap if they play on the starting point of the cycle.

Please re-read the solution and follow the links given above.

You should start with power of 1: 3^1=3 (the units digit is 3) 3^2=9 (the units digit is 9) 3^3=27 (the units digit is 7) 3^4=81 (the units digit is 1) 3^5=243 (the units digit is 3 again!) ... _________________

If n = (33)^43 + (43)^33 what is the units digit of n? [#permalink]

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05 Oct 2014, 05:40

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Based on the Number Theory > Last digit of a power (in the GMATclub forum, I cannot link to the post)

Step 1: Find the last digit of 33^43

find the cyclicity of 33 33^1= _3 33^2= _9 33^3= _7 33^4= _1 33^5= _3 -> 33 has the cyclicity of 4

Divide the power by the cyclicity: 43/4 -> remainder is 3 (refer to the 3rd position in the cyclicity) -> the last digit of 33^43 is the same as that of 33^3 -> 7 is the last digit

Step 2: Similarly, find the last digit of 43^33 43 also has the cyclicity of 4 (ends with 3) Divide 33 by 4 -> remainder is 1 -> refer to the 1st position in the cyclicity -> last digit is 3

If n = (33)^43 + (43)^33 what is the units digit of n? [#permalink]

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06 Feb 2015, 17:38

I am completely lost on this question.

My (incorrect) approach was to drop the ten units digit of each number and of each exponent, so that I ended up with 3^3 + 3^3 - therefore the answer would be 7+7 = 14 or answer choice C. I'm at a complete loss as to why this is incorrect and am having trouble understanding the explanations above.

Is anyone able to kindly provide an explanation for the theory behind this, in layman's terms for this poor layman?

Re: If n = (33)^43 + (43)^33 what is the units digit of n? [#permalink]

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09 Feb 2015, 01:59

Expert's post

Anonamy wrote:

I am completely lost on this question.

My (incorrect) approach was to drop the ten units digit of each number and of each exponent, so that I ended up with 3^3 + 3^3 - therefore the answer would be 7+7 = 14 or answer choice C. I'm at a complete loss as to why this is incorrect and am having trouble understanding the explanations above.

Is anyone able to kindly provide an explanation for the theory behind this, in layman's terms for this poor layman?

Thanks

Please read the solutions above and follow the links provided. _________________

Re: If n = (33)^43 + (43)^33 what is the units digit of n? [#permalink]

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10 May 2015, 01:50

Bunuel wrote:

Thus: The units digit of 3^43 is the same as the units digit of 3^3, so 7 (43 divided by the cyclicity of 4 gives the remainder of 3). The units digit of 3^33 is the same as the units digit of 3^1, so 3 (33 divided by the cyclicity of 4 gives the remainder of 1).

Therefore the units digit of (33)^43 + (43)^33 is 7 + 3 = 0.

Answer: A.

I do not understand this part with remainder and cyclicity, why do i need to divide 43 by 4 to get a remainder of one which is then the units digit??

Re: If n = (33)^43 + (43)^33 what is the units digit of n? [#permalink]

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11 May 2015, 03:03

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noTh1ng wrote:

Bunuel wrote:

Thus: The units digit of 3^43 is the same as the units digit of 3^3, so 7 (43 divided by the cyclicity of 4 gives the remainder of 3). The units digit of 3^33 is the same as the units digit of 3^1, so 3 (33 divided by the cyclicity of 4 gives the remainder of 1).

Therefore the units digit of (33)^43 + (43)^33 is 7 + 3 = 0.

Answer: A.

I do not understand this part with remainder and cyclicity, why do i need to divide 43 by 4 to get a remainder of one which is then the units digit??

You take the remainder obtained by dividing the power by cyclicity. For example, 3^43 --> 43 (power) divided by the cyclicity of 4 gives the remainder of 3 --> the units digit of 3^43 is the same as the units digit of 3^3.

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