What is the units digit of 2222^(333)*3333^(222) ?

Question

What is the units digit of  2222^{333}*3333^{222}  ?

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

Bunuel · Accepted Answer

What is the units digit of 2222^{333}*3333^{222} ? A. 0 B. 2 C. 4 D. 6 E. 8 The units digit of 2222^{333} is the same as that of 2^{333} ; The units digit of 3333^{222} is the same as that of 3^{222} ; Hence, the units digit of 2222^{333}*333^{222} is the same as that of 2^{333}*3^{222} ; Now, the units digits of both 2 and 3 in positive integer power repeat in patterns of 4. For 2 it's {2, 4, 8, 6} and for 3 it's {3, 9, 7, 1}. The units digit of 2^{333} will be the same as that of 2^1 , so 2 (as 333 divided by cyclicity of 4 yields remainder of 1, which means that the units digit is first # from pattern); The units digit of 3^{222} will be the same as that of 3^2 , so 9 (as 222 divided by cyclicity of 4 yields remainder of 2, which means that the units digit is second # from pattern); Finally, 2*9=18 --> the units digit is 8. Answer: E. For more on this check Number Theory chapter of Math Book: https://gmatclub.com/forum/math-number-theory-88376.html Hope it helps.

LalaB · Answer

I used the method described by Bunuel.
but not to be repetitive, I will post another solution -

(2222^333)*(3333^222)=2222^111*(2222^222*3333^222)

here please pay attention to the fact that the unit digit of multiplication of 2222 and 3333 is 6 (2222^222*3333^222). since 6 powered in any number more than 0 results in 6 as a units digit, as a result we have-
6*2222^111

2 has a cycle of 4 . 111=27*4+3 . 2^3=8 
6*8=48
so the units digit is 8, and the answer is E

Bunuel · Answer

For more on this kind of questions check  Units digits, exponents, remainders problems collection .

adityadon · Answer

Use cyclicity rule here , 
2^1=2, 
2^2=4, 
2^3=8
2^4=6 
2^5=2
We can see here unit digit repeated after 4 powers so cyclicity of 2 is 4 . devide power by 4 and check above .
Ans E

EMPOWERgmatRichC · Answer

Hi All,

Each of the other explanations to this question has properly explained that you need to break down the calculation into pieces and figure out the repeating "pattern" of the units digits.

Here's another way to organize the information.

We're given

We can 'combine' some of the pieces and rewrite this product as....
( ^222)

(2222)(3333) = a big number that ends in a 6

Taking a number that ends in a 6 and raising it to a power creates a nice pattern:
6^1 = 6
6^2 = 36
6^3 = 216
Etc.
Thus, we know that ( ^222) will be a gigantic number that ends in a 6.

2^111 requires us to figure out the "cycle" of the units digit...

2^1 = 2
2^2 = 4
2^3 = 8
2^4 = 16

2^5 = 32
2^6 = 64
2^7 = 128
2^8 = 256

So, every 4 "powers", the pattern of the units digits repeats (2, 4, 8, 6.....2, 4, 8, 6....).

111 = 27 sets of 4 with a remainder of 3....

This means that 2^111 = a big number that ends in an 8

So we have to multiply a big number that ends in a 6 and a big number that ends in an 8.

(6)(8) = 48, so the final product will be a gigantic number that ends in an 8.

Final Answer:  E

GMAT assassins aren't born, they're made,
Rich

OptimusPrepJanielle · Answer

Unit digit of 2222^333 is same as unit digit of 2^333. Unit digit of powers of 2 follows a pattern: 2, 4, 8, 6 Now, 4*83 = 332 i.e. 2^332 uses 6 as its unit digit. Hence, 2^333 will have unit digit as 2. Unit digit of 3333^222 is same as unit digit of 3^222. Unit digit of powers of 3 follows a pattern: 3, 9, 7, 1 Now, 4*55 = 220 i.e. 3^220 uses 1 as its unit digit. Hence, 3^221 will have unit digit as 3. And, 3^222 will have unit digit as 9. Now we have 2 * 9 = 18 So the final unit digit of 2222^(333)*3333^(222) = 8. Hence option E. -- Optimus Prep's GMAT On Demand course for only $299 covers all verbal and quant. concepts in detail. Visit the following link to get your 7 days free trial account: https://www.optimus-prep.com/gmat-on-demand-course

shahsuhailp · Answer

This type of sums can also be solved using Fermet's Theorem approach.

Refer photo attached below:

Hadiagh · Answer

Answered in 47 sec to be exact. 
Approach: 
Since only the unit digits matter here, 
 —2^333 * —3^222 = ?
Now, 
We know cycle of 2 follows: 2,4,8,6 pattern
And the exponent we want is 333, we also know that any multiple of 4 will give us a 6 in unit digit. 333 is not a multiple of 4 but 332 is, so 333 will give us the unit digit of 2

And if you repeating this exact same technique for —3^222 you’ll get the unit digit of 9

Now, we can multiply the the unit digits to find the unit digit of the product of those numbers.
Here: —2*—9=18
Therefore correct answer is 8

Posted from my mobile device

Basshead · Answer

2 exponent pattern = (2, 4, 6, 8)
3 exponent pattern = (3, 9, 7, 1)

333/4 gives us remainder 1. Thus 2^333 = units digit 2.
222/4 gives us remainder 2. Thus 3^222 = units digit 9

2 * 9 = units digit 8.

Answer E.

Basshead · Answer

Cyclicity of 2: 2, 4, 8, 6
Cyclicity of 3: 3, 9, 7, 1

333 / 4 = remainder 1. 
Units digit of  2222^{333} = 2

222 / 4 = remainder 2.
Units digit of  3333^{222} = 9

2 * 9 = units digit of 8.

Answer is E .

CrackverbalGMAT · Answer

Cyclicity of 2 is 4 (2,4,8,6) & Cyclicity of 3 is also 4 (3,9,7,1)

2222 is raised to power 333.

333 when divided by Cyclicity of 2 gives us a remainder of 2 and hence the unit digit for 2222^333 is 2.

3333 is raised to power 222.

222 when divided by Cyclicity of 3 gives us a remainder of 2 and hence the unit digit for 3333^222 is 3^2 or 9.

Hence the unit digit of the product is 2 * 9 =1(8) or 8.

(option e)
 
  D.S
GMAT SME

Rohstar750 · Answer

Hi Bunnel,

I took time to study your this trick. It works like magic. Gained a lot of confidence in approaching these questions.

Thanks a lot.

Regards,
Rohan

Paras96 · Answer

What is the unit digit of 2222^333∗3333^222?

Unit digit of 2222^333 = Unit digit of 2^333 = Unit digit of (2^332)(2) = Unit digit of (16)(2) = Unit digit of (32) = 2
Unit digit of 3333^222 = Unit digit of 3^222 = Unit digit of (3^220)(3^2) = Unit digit of 81*9 = Unit digit of 729 = 9

The unit digit of 2222^333∗3333^222 =  *  = 2*9 = 18 = 8

Hence E

Abhishek009 · Answer

Units digit cyclicity of 2 & 3 both are 4.

Thus,  2222^{333}*3333^{222}=2222^{83*4+1}*3333^{55*4+2}

2222^{83*4} =  Units digit  6  
 2222^{1} =   Units digit  2

Thus, units digit of  2222^{83*4+1} = 2*6 = 2

3333^{55*4} =  Units digit  1  
 3333^{2} =  Units digit  9

Thus, units digit of  3333^{55*4+2} = 9*1 = 9

Finally  2222^{333}*3333^{222}  = Units digit of  2*9 = 18 , Answer must be (E) 8

ManifestDreamMBA · Answer

2222^{333}*3333^{222} 
 2^{333}*3^{222} 
 2^{332+1}*3^{220+2}

Both have cyclicity of 4

2^{1}*3^{2} 
 2*9 
1 8

What is the units digit of 2222^(333)*3333^(222) ?

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What is the units digit of 2222^(333)*3333^(222) ?

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