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in that case divide 27 by 4. The remainder is 3. Now the last digit of 3^27 is same as that of 3^3 and that happens to be 7.

So 7 is the right answer.

why" Now the last digit of 3^27 is same as that of 3^3 and that happens to be 7."

where do you get this from

\(3^1\) =3, \(3^2\) =9, \(3^3\) =27, \(3^4\) =81, \(3^5\) =243, \(3^6\) =729, and so on .....

Now it is not humanly possible to remember all the numbers till \(3^27\)

If you have noticed in the above series the last digit repeats after every 4 terms

the last digit is same for \(3^5\) and \(3^1\) the last digit is same for \(3^6\) and \(3^2\)

If 3 is the unit digit of a number then the unit digit repeats every fourth consecutive term.For our convenience here, lets call it cyclicity. So 3 has a cyclicity of 4.

To find the unit digit of a number having 3 as its last digit and raised to a positive power, divide the power by 4 and find the remainder.

If the remainder is 1 then the unit digit is same as of the unit digit of \(3^1\) If the remainder is 2 then the unit digit is same as of the unit digit of \(3^2\) and so on .....

Note that if the remainder is "0" then the unit digit is same as \(3^4\) since the cyclicity is 4.

Also remember that the numbers 2,3,7 and 8 have cyclicity of 4

in our problem above we have 3^27

3 has a cyclicity of 4 so divide the number 27 by 4. We get a remainder of 3. Now as per cyclicity the last digit of 3^27 is same as that of 3^3. 3^3 is 27 so the last digit of 3^27 is 7.
_________________

"You have to find it. No one else can find it for you." - Bjorn Borg

Last edited by amitdgr on 24 Sep 2008, 05:58, edited 1 time in total.

Now the last digit of 122 is 2. We require only this number to determine the last digit of 122 raised to a positive power.

so the problem is essentially reduced to find the last digit of 2^94.

Now we know 2 has a cyclicity of 4. So we divide 94 by 4. The remainder for 94/4 is 2.

so last digit of 2^94 is same as that of 2^2 which is 4.

so last digit of 122^94 is 4

Remember: 1) Numbers 2,3,7 and 8 have a cyclicity of 4 2) Numbers 0,1,5 and 6 have a cyclicity of 1 (ie) all the powers will have the same unit digit. eg. 5^245 will have "5" as unit digit, 5^2000 will aslo have "5" as unit digit. Same holds for 0,1 and 6 3) If 4 is the number in the unit place of the base number then the unit digit will be "4" if the power is odd and it will be "6" if the power is even. eg. 4^123 will have unit digit of 4 since 123 is odd. 4) Similarly, for 9 the unit digit will be "9" for odd powers and "1" for even powers. eg 9^234 has unit digit as "1" since 234 is even.
_________________

"You have to find it. No one else can find it for you." - Bjorn Borg

Now the last digit of 122 is 2. We require only this number to determine the last digit of 122 raised to a positive power.

so the problem is essentially reduced to find the last digit of 2^94.

Now we know 2 has a cyclicity of 4. So we divide 94 by 4. The remainder for 94/4 is 2.

so last digit of 2^94 is same as that of 2^2 which is 4.

so last digit of 122^94 is 4

Remember: 1) Numbers 2,3,7 and 8 have a cyclicity of 4 2) Numbers 0,1,5 and 6 have a cyclicity of 1 (ie) all the powers will have the same unit digit. eg. 5^245 will have "5" as unit digit, 5^2000 will aslo have "5" as unit digit. Same holds for 0,1 and 6 3) If 4 is the number in the unit place of the base number then the unit digit will be "4" if the power is odd and it will be "6" if the power is even. eg. 4^123 will have unit digit of 4 since 123 is odd. 4) Similarly, for 9 the unit digit will be "9" for odd powers and "1" for even powers. eg 9^234 has unit digit as "1" since 234 is even.

Wow !! Awesome I tried out this thing with a few numbers and matched the results with my scientific calculator. This method gives perfect answers.

You deserve at least a dozen KUDOS for typing out all this patiently and sharing this knowledge with all of us.

Now the last digit of 122 is 2. We require only this number to determine the last digit of 122 raised to a positive power.

so the problem is essentially reduced to find the last digit of 2^94.

Now we know 2 has a cyclicity of 4. So we divide 94 by 4. The remainder for 94/4 is 2.

so last digit of 2^94 is same as that of 2^2 which is 4.

so last digit of 122^94 is 4

Remember: 1) Numbers 2,3,7 and 8 have a cyclicity of 4 2) Numbers 0,1,5 and 6 have a cyclicity of 1 (ie) all the powers will have the same unit digit. eg. 5^245 will have "5" as unit digit, 5^2000 will aslo have "5" as unit digit. Same holds for 0,1 and 6 3) If 4 is the number in the unit place of the base number then the unit digit will be "4" if the power is odd and it will be "6" if the power is even. eg. 4^123 will have unit digit of 4 since 123 is odd. 4) Similarly, for 9 the unit digit will be "9" for odd powers and "1" for even powers. eg 9^234 has unit digit as "1" since 234 is even.

Wow !! Awesome I tried out this thing with a few numbers and matched the results with my scientific calculator. This method gives perfect answers.

You deserve at least a dozen KUDOS for typing out all this patiently and sharing this knowledge with all of us.

+1 from me. Guys pour in Kudos for this

Chayanika

i agree with you. he deserves many kudos.

thanks a million.

the OA is 7 indeed. what a wonderful explanation. where did you read the theory about this one???

in that case divide 27 by 4. The remainder is 3. Now the last digit of 3^27 is same as that of 3^3 and that happens to be 7.

So 7 is the right answer.

why" Now the last digit of 3^27 is same as that of 3^3 and that happens to be 7."

where do you get this from

\(3^1\) =3, \(3^2\) =9, \(3^3\) =27, \(3^4\) =81, \(3^5\) =243, \(3^6\) =729, and so on .....

Now it is not humanly possible to remember all the numbers till \(3^27\)

If you have noticed in the above series the last digit repeats after every 4 terms

the last digit is same for \(3^5\) and \(3^1\) the last digit is same for \(3^6\) and \(3^2\)

If 3 is the unit digit of a number then the unit digit repeats every fourth consecutive term.For our convenience here, lets call it cyclicity. So 3 has a cyclicity of 4.

To find the unit digit of a number having 3 as its last digit and raised to a positive power, divide the power by 4 and find the remainder.

If the remainder is 1 then the unit digit is same as of the unit digit of \(3^1\) If the remainder is 2 then the unit digit is same as of the unit digit of \(3^2\) and so on .....

Note that if the remainder is "0" then the unit digit is same as \(3^4\) since the cyclicity is 4.

Also remember that the numbers 2,3,7 and 8 have cyclicity of 4

in our problem above we have 3^27

3 has a cyclicity of 4 so divide the number 27 by 4. We get a remainder of 3. Now as per cyclicity the last digit of 3^27 is same as that of 3^3. 3^3 is 27 so the last digit of 3^27 is 7.[/quote]

by the way, i have never had such a GMAT rush. it is like a sugar rush. EVERY ONE GIVE MORE KUDOS HERE

Remember: 1) Numbers 2,3,7 and 8 have a cyclicity of 4 2) Numbers 0,1,5 and 6 have a cyclicity of 1 (ie) all the powers will have the same unit digit. eg. 5^245 will have "5" as unit digit, 5^2000 will aslo have "5" as unit digit. Same holds for 0,1 and 6 3) If 4 is the number in the unit place of the base number then the unit digit will be "4" if the power is odd and it will be "6" if the power is even. eg. 4^123 will have unit digit of 4 since 123 is odd. 4) Similarly, for 9 the unit digit will be "9" for odd powers and "1" for even powers. eg 9^234 has unit digit as "1" since 234 is even.

So glad I come across this thread! Great tip! Thanks a bunch +1

Here the given number is \((xyz)^n\) z is the last digit of the base. n is the index

To find out the last digit in \((xyz)^n\), the following steps are to be followed. Divide the index (n) by 4, then

Case I If remainder = 0 then check if z is odd (except 5), then last digit = 1 and if z is even then last digit = 6

Case II If remainder = 1, then required last digit = last digit of the base (i.e. z) If remainder = 2, then required last digit = last digit of the base \((z)^2\) If remainder = 3, then required last digit = last digit of the base \((z)^3\)

Note : If z = 5, then the last digit in the product = 5

Example: Find the last digit in (295073)^130

Solution: Dividing 130 by 4, the remainder = 2 Refering to Case II, the required last digit is the last digit of \((z)^2\), ie \((3)^2\) = 9 , (because z = 3)