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

Question

ANSWER:

0

Y is the ans 0..
i have an explanation to the question but i donot understand it please can some 1 explain

EXPLANATION:

1)Units digit for power of any number with 3 in the units place follows pattern 3,9,7,1. 
2)Units digit for 43^43 = 7 (remainder of 43 divided by 3 is 1. So the units digit will be 3)
3)Units digit for 33^33 = 7 
4)Units digit of (43^43 + 33^33) = Units digit for 43^33 + Units digit for 33^33 = 0 (discard the 1).
5)Units digit when it is divided by 10 is same as the units digit of the number itself = 0.

i did not understand line 4, why will we do 43^33?

Bunuel · Answer

Integer is divisible by 10 if the units digit is zero. Next, units digit of  (xyz)^n  is the same as units digit of  z^n , so for example units digit of  43^{43}  is the same as the units digit of  3^{43} .

There are 2 typos in explanation:
3. Units digit of 33^33 =  3  --> units digit of  33^{33}  is the same as the units digit of  3^{33} . Cyclicity of 3 in power is 4 so  3^{33}  will have the same units digit as  3^1 , so 3.

4. 43^43 + 33^33 --> some number with units digit 7 plus some number with units digit 3 = some numer with units digit 0. Number with units digit zero is divisible by 10.

Hope it helps.

anshumishra · Answer

last digit of 43^43 = last digit of 43^(10*4+3)= last digit of 43^3=7

last digit of 33^33= last digit of 33^(8*4+1) = last digit of 33^1= 3

please note that I have expressed the power in form of 4x+k, because power of 3 repeats after every 4 times.

So, the last digit of the sumis 7+3 = 0
therefore, the raminder will be 0.

Posted from my mobile device

amitjash · Answer

Understood now.. initially i was confused.. thanks

rashi22 · Answer

undersood!! thanks alot! appreciated!

prashantbacchewar · Answer

Thanks for the explaination Bunuel

AtifS · Answer

The question says what is the remainder when divided by  10 ?
I know that the unit digit remainder is  0  as sum of remainders is  7+3=10  but when divided by  10 , shouldn't it be some other digit after dividing the number by  10  (actually  10th  digit remainder in this case)?

Let's suppose that last 2-digits for  43^{43}  are  27  and for  33^{33}  are  33  then the last 2-digits of the sum would be  60  and when divided by  10  it gives  \frac{60}{10}=6  as a remainder (not essentially  6 , it can be any  10th  digit remainder).
Am I right or not?? Any expert??

lauramo · Answer

HI,

Could we do the following?

last digit of 43^43 = last digit of 43^(10*4+3)= last digit of 43^3=7

last digit of 33^33= last digit of 33^(10*3+3) = last digit of 33^3= 7

If not, what are the rules to the break down the power?

Thanks

bumpbot · Answer

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

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

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