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

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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

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 [(2222)^333][(3333)^222]

We can 'combine' some of the pieces and rewrite this product as....
([(2222)(3333)]^222) [(2222)^111]

(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 ([(2222)(3333)]^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:

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

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.

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

Refer photo attached below:

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

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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.

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.

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

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

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 = [Unit digit of 2222^333]*[Unit digit of 3333^222] = 2*9 = 18 = 8

Hence E

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