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rthothad
MA can you be more brief - how did you arrive at 7




rthoad,

i have edited my answer.

to get the answer conseider only the unit digits of these numbers. they both have 3 in unit digit.

in first, the unit digit of the 33^33 is 1 where as the unit digit of 43^43 is 7. the sum of these two digits is 0.

moreover,
1. 3 = 3
2. 3X3=9
3. 3X3X3=27
4. 3X3X3X3=81
5.3X3X3X3X3=243

from the above, we can generalize that the unit digit of the multiple of four 3s is 1.

in thirty-three 3s, we have 32 (4X8) 3s, which has 1 as unit digit of the resulting value if we multiply these 3s, and one extra 3, whose unit digit is obviously 3.

similarly in fourty-three 3s, we have 40 (4X10) 3s, which has 1 as unit digit, and remaining three 3s has 7 as unit digit of the value.

then the sum of these two unit digits has 0 u nit digit value.
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MA, now I get it, your first reply made me unsure of my way of approaching this problem.
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FN
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MA can you please explain again, I am having a hard time following your explanation....thanks!

MA
rthothad
MA can you be more brief - how did you arrive at 7



rthoad,

i have edited my answer.

to get the answer conseider only the unit digits of these numbers. they both have 3 in unit digit.

in first, the unit digit of the 33^33 is 1 where as the unit digit of 43^43 is 7. the sum of these two digits is 0.

moreover,
1. 3 = 3
2. 3X3=9
3. 3X3X3=27
4. 3X3X3X3=81
5.3X3X3X3X3=243

from the above, we can generalize that the unit digit of the multiple of four 3s is 1.

in thirty-three 3s, we have 32 (4X8) 3s, which has 1 as unit digit of the resulting value if we multiply these 3s, and one extra 3, whose unit digit is obviously 3.

similarly in fourty-three 3s, we have 40 (4X10) 3s, which has 1 as unit digit, and remaining three 3s has 7 as unit digit of the value.

then the sum of these two unit digits has 0 u nit digit value.
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33^33 + 43^43
All we care for is the units digit because the tenths digit is obviously divisible by 5 and will leave no remainder.

Let's only focus on the units digit of 33^33:
Let's look at the pattern:
3^1 = 3 -- 3^5 = 3
3^2 = 9 -- 3^6 = 9
3^3 = 7 -- 3^7 = 7
3^4 = 1 -- 3^8 = 1

All we care for is the units digit. Forget about the rest. As you can see, units digit pattern repeats after every power of 4. Hence, divide the power value by 4 and you will get the units digit. 33/3 is 8 remainder 1.
When you get remainder 1, units value will be 3
When you get remainder 2, units value will be 9
When you get remainder 3, units value will be 7
When you get remainder 0, units value will be 1
Since the remainder is 1, you will have a units value of 3. You do the same thing to 43/3 and you will get 10 remainder 3. Since remainder is 3, units value will be 7. Answer is 3+7 = 10. Since you are interested only in units value, 0 is the answer.
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OA is 0 - real smooth guys...real smooth. :-D
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thanks paul...nice way to put it....
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fresinha12
thanks paul...nice way to put....


Paul is our man...Two Thumbs up!
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we are interested only in the unit digit, so we can write:

43^43 + 33^33 => 3^43+3^33=3^33*(3^10+1)

3^10=3^2*3^(4*2)=9*81^2 => unit digit 9 => 3^10+1 unit digit 0

(without relevance, but 3^33=3*3^32=3*3^(4*8)=3*81^8=> unit digit=3)

So the unit digit is 0.



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