If X/Y=63^33×36^195, what is the remainder of X/(10Y)?

Question

If  \frac{X}{Y}=63^{33}×36^{195} , what is the remainder of  \frac{X}{(10Y)} ?

A. 3 
B. 5 
C. 6 
D. 8
E. 9

arhumsid · Accepted Answer

Its basically asking the remainder of  \frac{63^{33}×36^{195}}{10} 
And it all depends on the unit's digit of  63^{33}×36^{195}

Units digit of  63^{33}  depends on cyclicity of 3. 
 63^{1}  = 3
 63^{2}  = 9
 63^{3}  = 7
 63^{4}  = 1
 63^{5}  = 3
.
.
.
 63^{32}  = 1
 63^{33}  = 3

Hence the units digit would be:  3

On the other hand, the units digit of  36^{195} , will depend on cyclicity of 6, which is 1 and the units digit it always 6.

Hence the unit's digit of the expression  63^{33}×36^{195}  is  3x6=18   =>8
 
Therefore, the remainder when we divide the above expression by 10 would be 8

Option  D

jfranciscocuencag · Answer

Hello Bunuel !

I am just wondering...

Is it the same approach if in the statement would have been written without the "Y"?

\frac{X}{(10)}

Thank you so much in advance!

Anukriti04 · Answer

Keep in mind - Any number when divided by 10 has remainder same as the last digit of that number. Given that, we can break it down by considering the last digits for 3^33 and 6^195. The cycle for 3 repeats after 4th power which will give that the last digit for 3^33 will be the same as the last digit for 3^1 i.e. 3, and we know for 6 raised to any power it will be 6. 
Now we have 3x6 = 18. Again look at the last digit = 8 Hence Answer is D.

Kudos

kiranbhasker · Answer

Hi,

I have a doubt as to how you arrived at cyclicity of 3^3 = 3.

i had learnt about method wherein we divide the power (33) by the unit digit (3) which gives us remainder as 0, therefore 3^0 =1 hence i considered 1 as remainder.

Was it by dividing the power (33) by the entire number (63) and not just the units digit as that is when you would arrive at the remainder 3,

Thanks in advance.

EgmatQuantExpert · Answer

Given

•  \frac{X}{Y } =  63^{33}  ×  36^{195}

To Find

• Remainder when X/Y is divided by 10.

Approach and Working Out

• We basically need to find the unit digit of X/Y.

•  63^{33}  ×  36^{195} 
 o  3^{33}  × 6
  As 6 has a cyclicity of 1.  
 o  3^1  × 6 
   As 3 has a cyclicity of 4. 
o 18

Correct Answer: Option D

enayat · Answer

Thanks for the solution, btw you already had divided 63^33 and 36^195 to 10 to find the unit's digits, would you please explain why did you again divide 18/10 in the and to find the remainder?

Anukriti04 · Answer

Hi enayat pardon me as i don't know whom you asked this question to, taking a stab at answering your question.
Its not about dividing 18 again by 10, its about getting the last digit of the end result(which in our case is 18, and hence 8 is the last digit and the answer). Hope it makes sense now? You can check out my solution, might help you clear the doubt. Have a good day!

MathRevolution · Answer

\frac{X}{Y}  =  63^{33}  *  36^{195}

=> \frac{X}{Y}  =  3^{33}  *  6^{195}

=> Unit digit of  3^{33}  =  3^{4 * 8}  * 3 =  3^4  * 3 =  3^5  = unit digit is 3.

=> Unit digit of  6^{195}  = 6

=> 3 * 6 = 18

=>  \frac{18 }{ 10}  = Remainder will be 8

Answer D

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If X/Y=63^33×36^195, what is the remainder of X/(10Y)?

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If X/Y=63^33×36^195, what is the remainder of X/(10Y)?

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