What is the remainder when X^Y is divided by Z , Where X = 32, Y = (32

Question

Weekly Quant Quiz Question -9

What is the remainder when X^Y is divided by Z , Where X = 32, Y = (32)^(32) and Z = 13?
a) 4
b) 5
c) 6
d) 7
e) 8

ONLY TEXT SOLUTIONS ARE ALLOWED.

chetan2u · Accepted Answer

rvgmat12

32^{32^{32}}  can be written as  (26+6)^{32*32^{31}}

As we are looking at only the remainder, so in the expansion all terms except last will be divisible by 13

Last term =  (6)^{2*16*32^{31}}

(36^{16*32^{31}}=(39-3)^{16*32^{31}}

Again the last term =  (-3)^{16*32^{31}}=(3)^{16*32^{31}}=3^{2^4*2^{5*31}}=3^{2^{159}}

Now  3^{2^1}=9  and will leave a remainder of 9 when divided by 13.
 3^{2^2}=3^4=81=13*6+3  and will leave a remainder of 3 when divided by 13.
Next,  3^{2^3}=3^8=(3^4)^2  and as  3^4  leaves a remainder 3,  (3^4)^2  will leave a remainder of  (3)^2  when divided by 13.

Thus the pattern is 9,3,9,3….
 3^{2^{odd}}  will leave 9 and  3^{2^{even}}  will leave 3 as the remainder.

Thus 9 is the answer.

Options are wrong.

Aashutoshkumar · Answer

Should be 6, since 32÷13 gives 6 As remainder, and then x^y would be in the form of 32*32*32.... hence comes out to be 6

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Gladiator59 · Answer

What is the remainder when X^Y is divided by Z , Where X = 32, Y = (32)^(32) and Z = 13?
a) 4
b) 5
c) 6
d) 7
e) 8

Ans:
x^y => 32^(32^32)

Write 32 as 26+6

The exponent is a very large integer call it M.

x^y => (26+6)^M

In the binomial expansion of the above binomial entity every term will have a factor of 26 except the last term. Hence 13 will divide all the terms except 6.

The remainder will be 6.

Hence IMO Option (C) is correct.

Best,
Gladi

honneeey · Answer

What is the remainder when X^Y is divided by Z , Where X = 32, Y = (32)^(32) and Z = 13?
a) 4
b) 5
c) 6
d) 7
e) 8

By calculation the cyclicity of remainders when 13 divides 32 and powers of 32= 6,10,8,4 therefore as 32 is a factor of therefore the remainder is 
4
Answer=A

u1983 · Answer

Ans C-----6 (13*2 +6) =32

mongmat27 · Answer

Ans is C, 6

X* Y = (32) raised to the power of 33

This can be written as (26+6) to the power of 33.

Using binomial expression, we will get (6) power of 33

We now have to check it’s divisibity by 13. Thus remainder is 6

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MattyE · Answer

Hey all, hoping someone can clarify one aspect of this for me as I'm hitting my head against a wall here!

My thought process is: When expanding the binomial (26+6)^M the last term will be 6^M.

6^M/13 doesn't always leave remainder 6. e.g. 6^2/13 = remainder 10... So how can we confidently say it does.

What simple fact am i missing here!

SchruteDwight · Answer

This question is incorrect.

32^{{32}^{32}}  mod  13  is  9  (checked with wolfram alpha)

So how do we get to  R=9 ?

Kem12 · Answer

Hi @gladiator56, thanks for your explanation but I will appreciate if you can explain a bit on this part ""

The exponent is a very large integer call it M.

x^y => (26+6)^M

In the binomial expansion of the above binomial entity every term will have a factor of 26 except the last term. Hence 13 will divide all the terms except 6.""

""From what I know 13 will be divisible by all terms except 6^M and not just 6. Did you just assume or check for cyclicity of remainder of 6^M as well before arriving at OA. I am simply asking the same question  MattyE  is asking ?. Thanks.

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SchruteDwight · Answer

Kem12  Question is incorrect IMO, as are all answers. The remainder is 9 which you can easily check if you plug in the numbers as x mod y into wolfram alpha

PearlRay · Answer

The answer is 9 as per my deduction.  gmatbusters  please confirm?[

ridwanmd · Answer

i cannot understand any of above solutions

rvgmat12 · Answer

chetan2u  could you please solve this?

Wolfram alpha gives 9 as answer

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Ertucinar · Answer

X = 32, Y = 32^{32}, Z = 13

We need to compute:

32 ^ { (32^{32}) } \mod 13

Step 1: Reduce  X \mod Z

First, simplify  32 \mod 13 :

32 \div 13 = 2 	ext{ remainder } 6

Thus,  32 \equiv 6 \mod 13

So, the problem simplifies to:

6 ^ { (32^{32}) } \mod 13

Step 2: Find the Cycle of  6^n \mod 13

To determine the pattern in powers of 6 modulo 13:

6^1 \equiv 6 \mod 13

6^2 \equiv 36 \equiv 10 \mod 13

6^3 \equiv 60 \equiv 8 \mod 13

6^4 \equiv 48 \equiv 9 \mod 13

6^5 \equiv 54 \equiv 2 \mod 13

6^6 \equiv 12 \equiv -1 \mod 13

6^7 \equiv -6 \equiv 7 \mod 13

6^8 \equiv 42 \equiv 3 \mod 13

6^9 \equiv 18 \equiv 5 \mod 13

6^{10} \equiv 30 \equiv 4 \mod 13

6^{11} \equiv 24 \equiv 11 \mod 13

6^{12} \equiv 66 \equiv 1 \mod 13

Since  6^{12} \equiv 1 \mod 13 , the powers of 6 repeat every 12 cycles.

Step 3: Reduce the Exponent Modulo 12

Since we need to compute  32^{32} \mod 12 , we first reduce 32 modulo 12:

32 \equiv 8 \mod 12

Thus:

32^{32} \equiv 8^{32} \mod 12

We analyze the cycle of  8^n \mod 12 :

8^1 \equiv 8 \mod 12

8^2 \equiv 64 \equiv 4 \mod 12

8^3 \equiv 32 \equiv 8 \mod 12

8^4 \equiv 256 \equiv 4 \mod 12

So,  8^n \mod 12  cycles between 8, 4, 8, 4, ...

If  n  is odd:  8^n \equiv 8 \mod 12

If  n  is even:  8^n \equiv 4 \mod 12

Since 32 is even, we conclude:

32^{32} \equiv 4 \mod 12

Step 4: Find  6^4 \mod 13

From Step 2, we found:

6^4 \equiv 9 \mod 13

Conclusion:
Thus, the remainder when  32 ^ { (32^{32}) }   is divided by 13 is:  9

If you think the answer is correct please give a kudo

What is the remainder when X^Y is divided by Z , Where X = 32, Y = (32

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What is the remainder when X^Y is divided by Z , Where X = 32, Y = (32

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