can some one elaborate?

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can some one elaborate?

by swapna » Mon Feb 01, 2010 11:33 am
If r, s, and t are all positive integers, what is the remainder when 2rst is divided by 10?

(1) s is even

(2) rs = 4

When a number is divided by 10, the remainder is simply the units digit of that number. For example, 256 divided by 10 has a remainder of 6. This question asks for the remainder when an integer power of 2 is divided by 10. If we examine the powers of 2 (2, 4, 8, 16, 32, 64, 128, and 256...), we see that the units digit alternates in a consecutive pattern of 2, 4, 8, 6. To answer this question, we need to know which of the four possible units digits we have with 2rst.

(1) INSUFFICIENT: If s is even, we know that the product rst is even. Knowing that rst is even tells us that 2rst will have a units digit of either 4 or 6 (22 = 4, 24 = 16, and the pattern continues).

(2) SUFFICIENT: If rs = 4, then rst = 4t. Because t is an integer, rst must be a multiple of 4. Since every fourth integer power of 2 ends in a 6 (24 = 16, 28 = 256, etc.), we know that the remainder when 2rst is divided by 10 is 6.

The correct answer is B.

Can some one elaborate on this explanation....??
Source: — Data Sufficiency |

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by ajith » Mon Feb 01, 2010 11:39 am
swapna wrote:If r, s, and t are all positive integers, what is the remainder when 2rst is divided by 10?

(1) s is even

(2) rs = 4

When a number is divided by 10, the remainder is simply the units digit of that number. For example, 256 divided by 10 has a remainder of 6. This question asks for the remainder when an integer power of 2 is divided by 10. If we examine the powers of 2 (2, 4, 8, 16, 32, 64, 128, and 256...), we see that the units digit alternates in a consecutive pattern of 2, 4, 8, 6. To answer this question, we need to know which of the four possible units digits we have with 2rst.

(1) INSUFFICIENT: If s is even, we know that the product rst is even. Knowing that rst is even tells us that 2rst will have a units digit of either 4 or 6 (22 = 4, 24 = 16, and the pattern continues).

(2) SUFFICIENT: If rs = 4, then rst = 4t. Because t is an integer, rst must be a multiple of 4. Since every fourth integer power of 2 ends in a 6 (24 = 16, 28 = 256, etc.), we know that the remainder when 2rst is divided by 10 is 6.

The correct answer is B.

Can some one elaborate on this explanation....??
Honestly it does't make sense, are you sure that you did not miss anything on the question

If not, I will provide you an example to prove the explanation is nonsense

say r=2; s=2, t =10 (r, s, t all +ve integers)

satisfies 1

satisfies 2

and the remainder is not 6

it is zero
Last edited by ajith on Mon Feb 01, 2010 11:44 am, edited 1 time in total.
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by swapna » Mon Feb 01, 2010 11:43 am
yeah...i hvnt missed out on anythin....ths is from manhattan free CAT..Hope some one else is able to figure out

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by ajith » Mon Feb 01, 2010 11:46 am
swapna wrote:yeah...i hvnt missed out on anythin....ths is from manhattan free CAT..Hope some one else is able to figure out
If the question is the same, I am sure there is something wrong with the solution.
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by ajith » Mon Feb 01, 2010 11:52 am
swapna wrote:If r, s, and t are all positive integers, what is the remainder when 2rst is divided by 10?

(1) s is even

(2) rs = 4

When a number is divided by 10, the remainder is simply the units digit of that number. For example, 256 divided by 10 has a remainder of 6. This question asks for the remainder when an integer power of 2 is divided by 10. If we examine the powers of 2 (2, 4, 8, 16, 32, 64, 128, and 256...), we see that the units digit alternates in a consecutive pattern of 2, 4, 8, 6. To answer this question, we need to know which of the four possible units digits we have with 2rst.

(1) INSUFFICIENT: If s is even, we know that the product rst is even. Knowing that rst is even tells us that 2rst will have a units digit of either 4 or 6 (22 = 4, 24 = 16, and the pattern continues).

(2) SUFFICIENT: If rs = 4, then rst = 4t. Because t is an integer, rst must be a multiple of 4. Since every fourth integer power of 2 ends in a 6 (24 = 16, 28 = 256, etc.), we know that the remainder when 2rst is divided by 10 is 6.

The correct answer is B.

Can some one elaborate on this explanation....??
The explanation makes sense only if there is a power function somewhere it could be rs^t or something like that

otherwise I see no reason why 24=16(see the OA) it must be 2^4 =16 and the power sign has gone missing
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by sars72 » Mon Feb 01, 2010 12:00 pm
This question asks for the remainder when an integer power of 2 is divided by 10. I
it appears that the question is what is the reminder when 2^(rst) is divided by 10.

We know that the possible units digits of powers of 2 are : 2,4,8,6
(1) s is even
for even powers of 2, we have units digits 4 and 6, hence we cannot determine the reminder. Insufficient!
(2) rs = 4
for powers of 2 which are multiples of 4, we have
2^(4*1) = 16
2^(4*2) = 256
... so all 2 raised to powers of multiples of 4 will have the units digits as 6, hence the reminder when divided by 10 will be 6. Therefore statement 2 alone is sufficient and the answer is B

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by thephoenix » Mon Feb 01, 2010 12:09 pm
swapna wrote:If r, s, and t are all positive integers, what is the remainder when 2rst is divided by 10?

(1) s is even

(2) rs = 4

When a number is divided by 10, the remainder is simply the units digit of that number. For example, 256 divided by 10 has a remainder of 6. This question asks for the remainder when an integer power of 2 is divided by 10. If we examine the powers of 2 (2, 4, 8, 16, 32, 64, 128, and 256...), we see that the units digit alternates in a consecutive pattern of 2, 4, 8, 6. To answer this question, we need to know which of the four possible units digits we have with 2rst.

(1) INSUFFICIENT: If s is even, we know that the product rst is even. Knowing that rst is even tells us that 2rst will have a units digit of either 4 or 6 (22 = 4, 24 = 16, and the pattern continues).

(2) SUFFICIENT: If rs = 4, then rst = 4t. Because t is an integer, rst must be a multiple of 4. Since every fourth integer power of 2 ends in a 6 (24 = 16, 28 = 256, etc.), we know that the remainder when 2rst is divided by 10 is 6.

The correct answer is B.

Can some one elaborate on this explanation....??
i think there is a typo
its 2^rst
s1)insuff
for 2^even=2,4,6,8 unit dig are 4,6,4,6
hence rem will be 4 or 6

s2) r*s=4--->r*s*t=4t
2^4t can be 2^4,2^8,2^12......to the power of multiples of 4

for each no. unit dig is 6
and as the no. is div by10 remanider will be the unit dig which is 6
hence sufff

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by swapna » Mon Feb 01, 2010 12:30 pm
other than manually checkin the last digit in this case, is there a way 2 figure out the last digit widout calculating?

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by sars72 » Mon Feb 01, 2010 12:36 pm
swapna wrote:other than manually checkin the last digit in this case, is there a way 2 figure out the last digit widout calculating?
**sarcasm alert**
sure, they will also give the answer at the end of the question. All you have to do is select the choice that has the answer. GMAC is als working towards having the correct answer choice selected by default, so that all you will have to do is keep clicking Next for all the questions before the 75 minutes are up.

Just playful banter... trying to drill home a point

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by thephoenix » Mon Feb 01, 2010 12:38 pm
swapna wrote:other than manually checkin the last digit in this case, is there a way 2 figure out the last digit widout calculating?
yeah def

2^4t=16^t
any no. with unit dig as 6 will have unit dig as 6 raised to any integer power

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by ajith » Mon Feb 01, 2010 12:39 pm
swapna wrote:other than manually checkin the last digit in this case, is there a way 2 figure out the last digit widout calculating?
Last digits of powers shows a pattern always..... so it is very easy to spot em

3^1 =3
3^2 =9
3^3 =7
3^4 = 1
3^5 = 3 ... Repeats


4^1 = 4
4^2 =6
4^3 =4 ... repeats


5^1 =5
5^2 =5 ..


6^1 =6
6^2 =6....

7^1 = 7
7^2 = 9
7^3 = 3
7^4 = 1
7^5 = 7...


8^1 =8
8^2 =4
8^3 = 2
8^4 =6
8^5 =8 ...

9^1 =9
9^2 =1
9^3 =9 ...
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by indiheats » Tue Jan 01, 2013 8:00 pm
Why can T not be zero - isnt T a positive integer?