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

Question

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

(1) The ten's digit of 11^n is 4
(2) The hundred's digit of 5^n is 6

vaibhav1221 · Answer

Should be E

Statement 1: The ten's digit of  11^n  is 4
 in 11^4 ten's digit is 4. BUT in 11^14 ten's digit is also 4.
( 11^1 =  1 1; 11^2 = 1 2 1; 11^3 = 13 3 1 ... forms a pattern and will repeat)
  Insufficient

Statement 2: The hundred's digit of 5^n is 6
in 5^4 hundred's digit is 6. BUT in 5^6 hundred's digit is also 6.
( 5^4 =  6 25; 5^6 = 15 6 25; 5^8 = 390 6 25 ... forms a pattern and will repeat)
  Insufficient

Statement 1 and 2:
N could be 4, 14, 24... No definite answer.
  Insufficient

Therefore, E.

Hope this helps.

Regards,
V

PKN · Answer

St1:-
The ten's digit of  11^n  is 4.
So n=4,14,24 etc
So insufficient.

St2:-
The hundred's digit of  5^n  is 6.
So, n=4,6,8,10,12,14,16,18,20,24 etc
So, insufficient.

Combining, we have n=4,14 etc.

Hence, insufficient.

Ans (E)

asthagupta · Answer

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

GyMrAT · Answer

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

CounterSniper · Answer

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

kys123 · Answer

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.

amanvermagmat · Answer

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 .

shuvodip04 · Answer

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.

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

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

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