If r, s, and t are all positive integers, what is the

If r, s, and t are all positive integers, what is the remainder when 2^(rst) is divided by 10?

First of all, when a positive integer is divided by 10, the remainder is the units digit of that integer. For example, 30 divided by 10 yields the remainder of 0, 31 divided by 10 yields the remainder of 1, 32 divided by 10 yields the remainder of 2, ...

Next, the units digit of 2 in positive integer power repeats in blocks of 4: {2, 4, 8, 6}

The units digit of 2^1 is 2;
The units digit of 2^2 is 4;
The units digit of 2^3 is 8;
The units digit of 2^4 is 6;
The units digit of 2^5 is 2, AGAIN;
...

(1) s is even --> rst is even, hence the units digit of 2^(rst) is either 4 or 6. Not sufficient.

(2) rs = 4 --> rst is a multiple of 4, hence the units digit of 2^(rst) is the same as the units digit of 2^4 so 6, which means that the remainder upon division of 2^(rst) by 10 is 6. Sufficient.

Answer: B.

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r,s,t are +ve

REM(2^rst/10) ?

(1).

s is even also even * even = even and even*odd=even

But REM(2^2/10) and REM(2^4/10) are different hence insufficient .

(2).

rs=4

REM(2^4t/10)

REM(2^4/10) ....REM(2^8/10).......REM(2^12/10) .... All are same

Hence sufficient

(B). it is !

a) if s is even, i.e. rst = even -> 2^even/10 -> can't determine
b) rs = 4, i.e. rst = 4t -> 2^4t/10 -> 2^4t will always have 6 in unit's place(always the multiplication for unit place will be 6*6), so remainder will be 6 -> determined.

Hence, B is the answer.

since we are dividing by \(10\) means remainder will depend on only the unit digit of \(2^{rst}\)

moreover we know unit digit of\(2^{4n} = 6\)
unit digit of \(2^{4n+1} = 2\)
unit digit of \(2^{4n+2} = 4\)
unit digit of \(2^{4n+3} = 8\)

so determining the unit digit we should be able to make \(2^{rst}\) in any of the above form.

(1) s is even
not clear it can be of form \(2^{4n} or 2^{4n+2}\)
hence not sufficient.

2)\(rs = 4\)
clearly we will get \(2^{4s}\)==>hence unit digit will be\(6\)
hence remainder will be \(6\).
hence sufficient.

hence B

Any integer that does not end in 0 will have a positive remainder when divided by 10. Specifically, the remainder will be equal to the ones column. No power of 2 ends in 0. We need the units digit of 2^(rst).

2^1 = 2
2^2 = 4
2^3 = 8
2^4 = 16
2^5 = 32
2^6 = 64

Every fourth power of 2 repeats.

Statement 1 tells us the units digit will be 4 or 6. Not sufficient.

Statement 2 is sufficient. rst will be a multiple of 4, with units digit 6. Sufficient.

The answer is B.

If r, s, and t are all positive integers, what is the remainder of 2^p/10, if p = rst?

(1) s is even

(2) p = 4t

Hi everyone, i have a doubt with statement B. since p=4t so when divided by 10 we can cancel a 2 from both numerator and denominator so we have 2^3t/5 which is 8^t/5 so in this case we have different remainders each time.

Please advice.

\(\frac{2^{4t}}{10}=\frac{2^{4t-1}}{5}\) not 2^3t/5. Also, when we are asked to find the remainder of a/b it's not correct to reduce the fraction and find the remainder of the resulting fraction. For example, the remainder when 15 is divided by 6 is 3, but if you reduce that by 3 and find the remainder of 5 by 2 you'd get the remainder of 1.

Hope its clear.
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Thanks Bunuel for your reply. I understood that one should not cancel out however cant understand 2^4t/10 is simplified into 2^4t-1/5 and not 2^3t/5. Don't we cancel the powers. for Example 2^3/2= 2^2. then why cant it be in the previous one.Please help.Thanks

\(\frac{a^n}{a^m}=a^{n-m}\). Hence, \(\frac{2^3}{2^2}=2^{3-2}=2\) the same way: \(\frac{2^{4t}}{2}=2^{4t-1}\).



Hope this helps.
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Remainder when divided by 10 = last digit

Last digit of 2 ^4 = 2^8 = 2^16 and so on.....

Statement I in insufficient:

If rst = 2 then 2 ^2 = 4 and rst =4 then 2 ^4 = last digit is 6

Statement II is sufficient:

If rs = 4 then rst is a multiple of 4 which means rst = 4k hence 2^4, 2^8, 2^12 will give the same last digit.

Hence the answer is B.

This is a question from MGMAT. I think there is a little bit confusion by saying rst since it can be a number rst or r*s*t. Nevertherless it comes after a question of 1n5, find the value of n.

1) The remainders when exponentials of 2 are divided by 10 are: 2, 4, 8, 6, 2, 4, 8....so it follows cycle of 4. when s is even, rst will also be even, so the remainder can be either 4 or 6. not sufficient.

2) if rst = 4, the remainder is 6. When rst =8, again the remainder is 6. All the values of rst will result in multiples of 4. So the remainder will always be 6. Sufficient.

B is the answer.

If r, s, and t are all positive integers, what is the remainder when 2^{rst} is divided by 10?

(1) s is even

Right away we can tell this answer is insufficient since we don't know the values of r or t.

(2) rs = 4

Cyclicity of 2: 2, 4, 8, 6

\(2^{4t}\) = 2 is raised to a multiple of 4.

Therefore, the units digit will be 6.

6 / 10 = remainder 6.

Statement 2 is sufficient. Answer is B.
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If r, s, and t are all positive integers, what is the remainder when 2^(rst) is divided by 10?

(1) s is even
No idea about then answer

(2) rs = 4
t=1 -->2^rst = 2^4 = 32
32/10 remainder 2
t = 2--> 2^rst = 2^8 = 512
512/10 remainder 2

-> B is the correct answer

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