What is the remainder when (2^16)(3^16)(7^16) is divided by 10?

What is the remainder when (2^16)(3^16)(7^16) is divided by 10?

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

Remainder when divided by 10 is the units digit and remainders can be multiplied (when adjusted for excess)

Cyclicity of 2 -> 2,4,8,6 -> exponent/cyclicity - remainder 0 -> Units digit of 2 is 6 -> Remainder = 6
Cyclicity of 3 -> 3,9,7,1 -> exponent/cyclicity - remainder 0 -> Units digit of 3 is 1 -> Remainder = 1
Cyclicity of 7 -> 7,9,3,1 -> exponent/cyclicity - remainder 0 -> Units digit of 2 is 1 -> Remainder = 1

Therefore remainder = 6*1*1 = 6 -> Answer - D

What is the remainder when \((2^{16})(3^{16})(7^{16})\) is divided by 10?

A. 0
B. 2
C. 4
D. 6 --> correct
E. 8

Solution:
N = \((2^{16})(3^{16})(7^{16})\)
= \((2*3*7)^{16}\)
= \((42)^{16}\)
= \((40+2)^{16}\)
Reminder of \((40+2)^{4*4}\) divided by 10 = Reminder of \(2^{16}\) divided by 10

2^1 = 2, reminder = 2
2^2=4, reminder = 4
2^3=8, reminder = 8
2^4=16, reminder = 6 --> cyclicity of 4,
2^5 = 32, reminder = 2
Reminder of \(2^{16}\) divided by 10 = 6

Solution:

Let’s look at some quick examples: If we divide 76 by 10, the quotient is 7, remainder 6. If we divide 23 by 10, the quotient is 2, remainder 3. Thus, we see that when we divide a number by 10, the remainder is always equal to the units digit.

Thus, to find the remainder when (2^16)(3^16)(7^16) is divided by 10, we simply have to find the units digit of (2^16)(3^16)(7^16). Now, let’s look at the units digit of each factor.

The pattern of units digits for a base of 2 is 2-4-8-6.

So, 2^16 has a units digit of 6.

The pattern of units digits for a base of 3 is 3-9-7-1.

So, 3^16 has a units digit of 1.

The pattern of units digits for a base of 7 is 7-9-3-1.

So, 7^16 has a units digit of 1.

Thus, the remainder of 2^16 x 3^16 x 7^16 when divided by 10 is 6 x 1 x 1 = 6.

Alternate solution:

Again, to find the remainder when (2^16)(3^16)(7^16) is divided by 10 is to find the units digit of (2^16)(3^16)(7^16). However, we can see that (2^16)(3^16)(7^16) = (2 x 3 x 7)^16 = 42^16 and 42^16 has the same units digit as 2^16. Since the pattern of units digits for a base of 2 is 2-4-8-6,
2^16 has a units digit of 6.

Answer: D

Euler of 10= 4. so 3^16 mod 10=0 and 7^16 mod 10 = 0
Only left part is 2^16 mod 10=> 16^4 mod 10=> 6^4 mod 10=>36*36 mod 10=>6*6 mod 10=> 36 mod 10=6
So answer is 6.

We need to find What is the remainder when \((2^{16})(3^{16})(7^{16})\) is divided by 10

\((2^{16})(3^{16})(7^{16})\) = \((2 * 3 * 7)^{16}\) = \(42^{16}\) = \((40 + 2)^{16}\)

Now, we have split 42 into two numbers, one (40) is a number closer to 42 and a multiple of 10 and other is a small number

Now, if we expand this using Binomial theorem then we will get all terms except the last term as a multiple of 40 => A multiple of 10

=> All terms except the last term will give us a remainder of 0 when divided by 10

=> Remainder of \((2^{16})(3^{16})(7^{16})\) by 10 is same as remainder of the last term = 16C16 * 2^16 * 40^0 = 2^16 by 10

Theory: Remainder of a number by 10 is same as remainder of the unit's digit of that number by 10

Now, Let's find the unit's digit of \(2^{16}\) first.

We can do this by finding the pattern / cycle of unit's digit of power of 2 and then generalizing it.

Unit's digit of \(2^1\) = 2
Unit's digit of \(2^2\) = 4
Unit's digit of \(2^3\) = 8
Unit's digit of \(2^4\) = 6
Unit's digit of \(2^5\) = 2

So, unit's digit of power of 2 repeats after every \(4^{th}\) number.
=> We need to divided 16 by 4 and check what is the remainder
=> 16 divided by 4 gives 0 remainder

=> \(2^{16}\) will have the same unit's digit as \(2^4\) = 6
=> Unit's digits of \(2^{16}\) = 6

But remainder of \(2^{16}\) by 10 = 6

So, Answer will be D
Hope it helps!

Learn How to Find Remainders with 2, 3, 5, 9, 10 and Binomial Theorem

ScottTargetTestPrep BrushMyQuant Bunuel chetan2u yashikaaggarwal

I was able to arrive at the answer without breaking apart the denominator 10 into 2 x 5.

However, when I do break it apart: (2^15 x 3^16 x 7^16) produces units digits of 6, and 1 and 1. But dividing the product by the denominator 5 produces a remainder of 1.

Could anyone help me understand why this method doesn't work?

Hey achloes

While other experts whom you've tagged to the post respond, sharing my two cents if that helps.

Before we figure out the reason why you're seeing a difference in answer let's take a general example to understand the concept -

What's the remainder when 10 is divided by 12 ?

The remainder (\(\frac{10}{12}\)) = 10, we can reason this well enough using logic right.

Now, let's say we divide the numerator and the denominator by 2

Remainder (\(\frac{10}{12}\)) = Remainder (\(\frac{5}{6}\)) = 5

Is the remainder when 10 divided by 12, 5? Of course, not!

What's the difference, the remainder got divided by 2 as well. In the first case, the remainder is 10, and in the second case, the remainder is 5.

The general rule is: Whenever you divide the numerator and denominator by a common factor, the common factor has to be multiplied (accounted for) to obtain the net remainder.

In our example, the common factor is 2.

Remainder (\(\frac{10}{12}\)) = Remainder (\(\frac{5}{6}\)) = 5

The final remainder = 5 * 2 = 10. This is what we're getting in the earlier example.

Now let's extend this concept to the question -

What is the remainder when \((2^{16})(3^{16})(7^{16})\) is divided by 10 ?

If we find the unit's digit and use the concept of cyclicity we will get the unit digit of the product \((2^{16})(3^{16})(7^{16})\) = 6 and that's our reminder. I guess you were able to get the answer without cancellation.

Now, let's see what happens when you cancel 2 from the numerator and denominator

Remainder (\(\frac{2^{16} * 3^{16}*7^{16}}{10}\))

= Remainder (\(\frac{2^{15}*3^{16}*7^{16}}{5}\))

Using the concept of cyclicity, we can find the units' digits

= Remainder (\(\frac{8* 1 * 1}{5}\)) = 3

Just as in the previous example, the remainders are no longer the same. In fact, the new remainder is half of the original remainder and this is expected

So to get the actual remainder, we need to multiply the obtained value by the factor used, i.e. multiply the new remainder by 2 to get the actual remainder.

3 * 2 = 6

Now by both methods, the remainders are the same.

Hope this help.

achloes Let's take an example to understand this.

Let's say we wanted to find reminder of 32 by 10. Remainder is 2.
But if we tried cancelling 2 in both 32 and 10 and tried finding remainder of 16 by 5 then the remainder will be 1, which is not equal to 2.
So, for finding remainders you CANNOT cancel out terms and then find remainders of remaining number.

What you can do though is split 32 into product of two numbers and find remainder of those individual numbers with 10 and then multiply the remainders.
Ex: 32 = 16*2
So, remainder of 32 by 10 = (Remainder of 16 by 10) * (Remainder of 2 by 10) = 6 * 2 = 12
Now, remainder cannot be more than 10, so we again divide 12 by 10 to get the final remainder as 2

Let's apply the same logic here.
Remainder of 2^16 x 3^16 x 7^16 by 10 = (Remainder of 2^16 by 10) * (Remainder of 3^16 by 10) * (Remainder of 7^16 by 10)
= Unit's digit of 2^16 * Units digit of 3^16 * Unit's digit of 7^16 = 6 * 1 * 1 = 6

So, Answer will be D
Hope it helps!

I have also explained this and other concepts of Remainders in below video, hope you find it helpful!

Hey, I am sure you find a detailed explanation in the above two answers.
The way I solved it is simply by putting the cyclicity of unit digits as power increases.

2^n, 3^n, 7^n gives 2, 3, & 7 respectively, after every 4th multiple of power.
So, unit digit of 2^16 = unit digit of 2^4. Similar for all.

unit digit of 2^4 = 6
unit digit of 3^4 = 1
unit digit of 7^4 = 1

So, when you divide 6/10, answer will be 6.

It took me approx 30 seconds or so to mentally calculate.
You, can improve your speed by solving more questions. But whenever you have this type of questions. You can use cyclicity.

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