# Quarter Wit, Quarter Wisdom: Cyclicity of Units Digits on the GMAT (Part 2)

As discussed last week, all units digits have a cyclicity of 1 or 2 or 4. Digits 2, 3, 7 and 8 have a cyclicity of 4, i.e. the units digit repeats itself every 4 digit:

Cyclicity of 2: 2, 4, 8, 6

Cyclicity of 3: 3, 9, 7, 1

Cyclicity of 7: 7, 9, 3, 1

Cyclicity of 8: 8, 4, 2, 6

Digits 4 and 9 have a cyclicity of 2, i.e. the units digit repeats itself every 2 digits:

Cyclicity of 4: 4, 6

Cyclicity of 9: 9, 1

Digits 0, 1, 5 and 6 have a cyclicity of 1, i.e. the units digit is 0, 1, 5, or 6 respectively.

Now let’s take a look at how to apply these fundamentals:

What is the units digit of 813^(27)?

To get the desired units digit here, all we need to worry about is the units digit of the base, which is 3.

Remember, our cyclicity of 3 is 3, 9, 7, 1 (four numbers total).

We need the units digit of 3^(27). How many full cycles of 4 will be there in 27? There will be 6 full cycles because 27 divided by 4 gives 6 as quotient and 3 will be the remainder. So after 6 full cycles of 4 are complete, a new cycle will start:

3, 9, 7, 1

3, 9, 7, 1

… (6 full cycles)

3, 9, 7 (new cycle for remainder of 3)

7 will be the units digit of 3^(27), so 7 will be the units digit of 813^(27).

Let’s try another question:

What is the units digit of 24^(1098)?

To get the desired units digit here, all we need to worry about is the units digit of the base, which is 4.

Remember, our cyclicity of 4 is 4 and 6 (this time, only 2 numbers).

We need the units digit of 24^(1098) – every odd power of 24 will end in 4 and every even power of 24 will end in 6.

Since 1098 is even, the units digit of 24^(1098) is 6.

Not too bad; let’s try something a little harder:

What is the units digit of 75^(25)^5

Note here that you have 75 raised to power 25 which is further raised to the power of 5.

25^5 is not the same as 25*5 – it is 25*25*25*25*25 which is far more complicated. However, the simplifying element of this question is that the last digit of the base 75 is 5, so it doesn’t matter what the positive integer exponent is, the last digit of the expression will always be 5.

Now let’s take a look at a Data Sufficiency question:

Given that x and y are positive integers, what is the units digit of (5*x*y)^(289)?

Statement 1: x is odd.

Statement 2: y is even.

Here there is a new complication – we don’t know what the base is exactly because the base depends on the value of x and y. As such, the real question should be can we figure out the units digit of the base? That is all we need to find the units digit of this expression.

When 5 is multiplied by an even integer, the product ends in 0.

When 5 is multiplied by an odd integer, the product ends in 5.

These are the only two possible cases: The units digit must be either 0 or 5.

With Statement 1, we do not know whether y is odd or even, we only know that x is odd. If y is odd, x*y will be odd. If y is even, x*y will be even. Since we don’t know whether x*y is odd or even, we don’t know whether 5*x*y will end in 5 or 0, so this statement alone is not sufficient.

With Statement 2, if y is even, x*y will certainly be even because an even * any integer will equal an even integer. Therefore, it doesn’t matter whether x is odd or even – regardless, 5*x*y will be even, hence, it will certainly end in 0.

As we know from our patterns of cyclicity, 0 has a cyclicity of 1, i.e. no matter what the positive integer exponent, the units digit will be 0. Therefore, this statement alone is sufficient and the answer is B (Statement 2 alone is sufficient but Statement 1 alone is not sufficient).

Finally, let’s take a question from our own book:

If n and a are positive integers, what is the units digit of n^(4a+2) – n^(8a)?

Statement 1: n = 3

Statement 2: a is odd.

We know that the cyclicity of every digit is either 1, 2 or 4. So to know the units digit of n^{4a+2} – n^{8a}, we need to know the units digit of n. This will tell us what the cyclicity of n is and what the units digit of each expression will be individually.

Statement 1: n = 3

As we know from our patterns of cyclicity, the cyclicity of 3 is 3, 9, 7, 1

Plugging 3 into “n”, n^{4a+2} = 3^{4a+2}

In the exponent, 4a accounts for “a” full cycles of 4, and then a new cycle begins to account for 2.

3, 9, 7, 1

3, 9, 7, 1

3, 9

The units digit here will be 9.

Again, plugging 3 into “n”, n^{8a} = 3^{8a}

8a is a multiple of 4, so there will be full cycles of 4 only. This means the units digit of 3^{8a} will be 1.

3, 9, 7, 1

3, 9, 7, 1

3, 9, 7, 1

3, 9, 7, 1

Plugging these answers back into our equation: n^{4a+2} – n^{8a} = 9 – 1

The units digit of the combined expression will be 9 – 1 = 8.

Therefore, this statement alone is sufficient.

In Statement 2, we are given what the exponents are but not what the value of n, the base, is. Therefore, this statement alone is not sufficient, and our answer is A (Statement 1 alone is sufficient but Statement 2 alone is not sufficient).