Question: What is the units digit of 2222^333 ∗ 3333^222?

1. 0
2. 2
3. 4
4. 6
5. 8

“What is the units digit of 2222^333 ∗ 3333^222 ?''- is a topic of the GMAT Quantitative reasoning section of the GMAT exam. GMAT quant section analyzes the quantitative knowledge and rational skills of the candidates. The candidates must select the appropriate option by undergoing proper calculations with mathematical knowledge. The candidates must know the basic numerical concepts in order to solve GMAT Problem Solving questions. The GMAT Quant topic in the problem-solving part notes mathematical problems that must be solved with better mathematical skills.

Solution and Explanation:

Approach Solution 1:

The problem statement asks to find out the units digit of 2222^333 ∗ 3333^222.

Here we can say that the units digit of 2222^333 is similar to that of 2^333;
Further, we can say that the units digit of 3333^222 is similar to that of 3^222;
Hence, it can be inferred that the units digit of 2222^333 ∗333^222 is the same as that of 2^333 ∗ 3^222;

The units digit of both the numbers 2 and 3 in positive integer power gets repeated in 4 patterns.
In the case of 2, it is {2, 4, 8, 6} and in the case of 3, it is {3, 9, 7, 1}.
Since the unit digit of 2^333 will be similar to that of 2^1, so the unit digit is 2. This is because if 333 is divided by a cyclicity of 4, it gives a remainder of 1. This denotes that the unit digit of 2^333 is the first number from the pattern.

Since the unit digit of 3^222 will be similar to that of 3^2, so the unit digit is 9.
This is because if 222 is divided by a cyclicity of 4, it gives a remainder of 2. This denotes that the units digit 3^222 is the second number from the pattern.

Therefore, we get 2*9=18
Thus, the units digit of 2222^333 ∗ 3333^222 is 8.

Correct Answer: (E)

Approach Solution 2:

The problem statement asks to find out the units digit of 2222^333 ∗ 3333^222.

The question states the equation [(2222)^333][(3333)^222]

We can merge some elements and derive this product as:
([(2222)(3333)]^222) [(2222)^111]

Therefore, we can say (2222)(3333) is a big number whose last digit will be 6
Thus taking a number whose last digit is 6 and presenting it to a power makes a nice pattern:

6^1 = 6
6^2 = 36
6^3 = 216 so on.
Thus, we can infer that ([(2222)(3333)]^222) will be a gigantic number whose last digit will be 6.

Now, 2^111 gives us to deduce 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, for every "powers" of 4, the units digits pattern gets repeated (2, 4, 8, 6.....2, 4, 8, 6....) and so on.
Therefore 111 can be derived as 27 sets of 4 with a remainder of 3.
This implies that 2^111 is a big number whose last digit is 8

So it is required to multiply a big number whose last digit is 6 and a big number whose last digit is 8.
=> (6)(8) = 48

Therefore, the final product will be a gigantic number whose last digit is 8.
Thus the units digit of 2222^333 ∗ 3333^222 is 8.

Correct Answer: (E)

Approach Solution 3:

The problem statement asks to find out the units digit of 2222^333 ∗ 3333^222.

We can say that the unit digit of 2222^333 is similar to that of the unit digit of 2^333.
We know that the unit digit of powers of 2 pursues a pattern: 2, 4, 8, 6

Now, we can write that 4*83 = 332 i.e. 2^332 has 6 as its unit digit.
Hence, 2^333 will hold a unit digit as 2.
Moreover, we can say that the unit digit of 3333^222 is similar to that of the unit digit of 3^222.
We know that the unit digit of powers of 3 pursues a pattern: 3, 9, 7, 1
Now, we can write that 4*55 = 220 i.e. 3^220 has 1 as its unit digit.
Hence, 3^221 will hold a unit digit as 3.
And, 3^222 will hold a unit digit as 9.

Now we can derive that 2 * 9 = 18
Therefore, the final unit digit of 2222^(333)*3333^(222) = 8.

Correct Answer: (E)

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