If r, s, and t are all positive integers, what is the remainder of 2^p

Sol: The questions becomes what is the remainder when 2^(rst)/10 ?

St 1 says: s is even that means r*s*t=even

Now 2^(Even number) can end is 4 or 6..So we have to remainders possible. A and D ruled out

St 2 says p=4*t or p is a multiple of 4 so we will have expression of the form 2^4 or 2^8 or 2^12...all end up in 6..

So remainder will be 6..

Ans is B

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Bunuel how do we know whether or not the question reads 2^(p/10) pr (2^p) /10?

2^p/10 can mean nothing but (2^p)/10. If it were 2^(p/10), it would have been written that way.
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Hi,

I think I have missed some concept . can you help me with how 2^4 or 2^8 or 2^12...all end up in 6..?

p=rst

Statement2: p=4t
—> \(\frac{2^p}{10}\)=\(\frac{2^{4t}}{10}\)= \(\frac{16^t}{10}\)= \(\frac{(10+6)^t}{10}\)

—> \(\frac{(10^t+10^{t-1}*6^1+...+6^t)}{10}\)

Each term is divided by 10, except the last one, \(6^t.\)

If you divide \(6^t\) by 10,
—> no matter what kind of positive integer t is, the last digit of it will be 6.
The remainder will be always 6.
Sufficient

The answer choice is B.

I also observed that the unit digit of the numerator (cyclicity of numerator) is deciding the Remainder when we divide the numerator by 10.
FOR EG 2^even no. would give us 4/10 or 16/10 or 64/10 so REMAINDER will be 4,6,4 and so on . i.e unit digits .

In option B 2^4t /10 = 16^t/10 and 6 being the unit digit has only 6 in cycle and hence the remainder will be 6 always .

Concept Tested

Cyclicity

2 has a cyclicity of 4: 2^1 => 2, 2^2 => 4, 2^3 => 8, 2^4 => 6 (Focus on the unit digit)

1) Remainder could be 4 or 6 => Not sufficient
2) Remainder would always be 6. => Sufficient => B

Any number when divided by 10, remainder will be unit's digit. For example, 51 divided by 10 has a
remainder of 1. This question asks for the remainder when an
integer power of 2 is divided by 10. Powers of 2
(2, 4, 8, 16, 32, 64…), we see that the units digit
alternates in a consecutive pattern of 2, 4, 8, 6. Thus, we need to know which of the above mentioned possible units digits we
have with 2p.

(1) INSUFFICIENT: If s is even, we know that the product rst is
even and so is p. Then units digit will be either 4 or 6 (22 = 4, 24 = 16....).

(2) SUFFICIENT: If p = 4t and t is an integer, p must be a multiple
of 4. Since every fourth power of 2 ends with 6 (4 = 16, 8
= 256...), we know that the remainder when 2p is divided by 10
is 6.
The correct answer is B.

Kudos if you like the explanation.

If r, s, and t are all positive integers, what is the remainder of \(\frac{2^{p}}{10}, if p = rst\)?

(1) s is even

This statement only tells us that rst = even. We can get multiple values of rst, resulting in multiple values of 2^{rst}. Insufficient.

(2) p = 4t

If p = 4t, we know that p = multiple of 4.

The cyclicity of 2 = 2, 4, 8, 6.

Knowing that p = a multiple of four means that the 2^p ALWAYS has a units digit of 6.

6 / 10 = remainder 6. Statement 2 is sufficient.

\(Answer is B.\)

VeritasKarishma plz explain

Remainder of 2^p/10 will be decided by the units digit of 2^p.
Units digit of 2^p will be decided by where p appears in the cycle of multiples of 4.

(1) s is even

If s is even, p is even. But do we know whether it is of the form 4a or 4a+2? No. 2^{4a} has units digit of 6 and 2^{4a+2} has units digit of 4. Not sufficient.

(2) p = 4t

p is a multiple of 4 which means that it is of the form 4a. So 2^p will have units digit of 6. So remainder when 2^p is divided by 10 will be 6.
Sufficient

Answer (B)
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