If n > 2 where n is an integer, what is the value of n?

isnt there any other way, other than finding all the initial values for both statement, as it took a lot of time?
looking forward for OE

Given n is an integer & n > 2

Question: n = ?


Statement 1: Tens digit of 11^n is 4

The tens digits of 11^ n will be 4 if units digit of n is 4.

But we don't know whether n is a single digit number or 2 digits or more.

Statement 1 is Not Sufficient.

Additional Tip: In general last two digits of any power of a number ending in 1 are found by multiplying the units digit of the power by the tens digit of the number for the tens place & units place is always 1.

e.g. 11^12 has last two digits as 21

41^23 has last two digits as 21

241^2234 has last two digits as 61


Statement 2: The hundreds digit of 5^n is 6

We have 5^3 = 125
5^4 = 625, hence n could be 4
5^5 = 3125
5^6 = 15625, hence n could be 6

we can see the pattern here that hundreds digit is either 1 or 6, for subsequent powers.

Hence n = 4, 6, 8,...

Statement 2 is Not Sufficient.

In general, for n>=3, hundreds digit for 5^n will be 1 if n is odd & it will be 6 if n is even.

Combining 1 & 2,

We don't get any new information. Since n can still be a single digit or a two digit or more number.

Combining is not sufficient.


Answer E.



Thanks,
GyM

1.Tens digit of 11^n is 4
since n > 2
let n be 3 , then 11^3=1331
let n be 4, then 11^4,last two digits =41 - Tens digit in this case is 4
let n be 5 , then 11^5,last two digits = 51
There seems to be a pattern such that the value of n and the tens digit is same
based on above n can be 4 or 44 or probably 444 who knows.
insufficient.

2.hundred's digit of 5^n is 6
will start off with n=4 , 5^4 is 625
625*25 last three digits =625
so n =4 or n=6 or n=8
insufficient

combining both n=4 or n=44 insufficient

Therefore E

It's E. Wow Bunuel your still here. I am retaking my GMAT since it has been 5 years. Hopefully I can score 750ish this time. 50 points increase should be okay, I hope.

I think there is another approach possible to this question. For every base and exponent of the form a^n, where a and n are positive integers, there always exists a certain cyclicity in powers for units digit, tens digit, hundreds digit etc.

Eg, consider 2^n. For various values of 'n', 2^n will take various values of last digit (units digit) but at a certain point it will start getting repeated. 2^1, 2^2, 2^3, 2^4 end in 2, 4, 8, 6 respectively. After that, it starts getting repeated - so 2^5, 2^6, 2^7, 2^8 also end in 2, 4, 6, 8 respectively.

Similarly 11^n will take various values of tens digit depending on the value of n. So 11^1, 11^2, 11^3, 11^4, 11^5, 11^6, 11^7, 11^8, 11^9, 11^10 respectively take the last two digits as 11, 21, 31, 41, 51, 61, 71, 81, 91, 01. After that it starts getting repeated. So 11^11, 11^12, 11^13.... also end in 11, 21, 31.. respectively.

Similarly 5^n, once it starts getting into 3 digits, will take its hundreds place digit as 1 and 6 only. 5^3 = 125, 5^4 = 625, 5^5 = 3125... and so on.

So, the approach is as follows:
Knowing the fact that units or tens or hundreds digit for a^n will start repeating after a certain fixed value of power 'n' (depending on the value of base 'a' also), we can never uniquely determine a single value for which the tens place digit will be 4 or for which the hundreds place digit will be 6. We are sure to get various values for which the same tens digit and hundreds digit will occur.

Since there is no unique value, answer has to be straight E.

From statement 1 - It can be observed that whenever we multiply a number with 11 the ten's place digit gets added to the unit's place digit to form the ten's place digit of the resulting product. From these observation last 2 digits of the following can be written as:
11^3 =121*11= ..31,
11^4= ...31*11 = ..41
11^5= ..41*11 = ..51
11^6 =...51*11 = ..61
11^7=...61*11=..71
11^8=...71*11 = ..81
11^9= ...81*11= ..91
11^10= ..91*11= ..01
11^11= ..01*11= ..11
11^12 = ..11*11= ...21
11^13= ...21*11= ..31
So the ten's digit number starts to repeat.So a particular value of n for which ten's place digit will be 6 cannot be determined. Hence 1) Not sufficient.

From statement 2- Similary 5^3 = 25*5=125, 5^4=625 5^5 = 3025

Last three digits are 025 if we keep on multiplying the hundred's digit will keep giving value of 6 for n= 4,7,10,13..so on.Hence statement 2) insufficient.