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If n > 2 where n is an integer, what is the value of n?
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06 Jul 2018, 02:29
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Re: If n > 2 where n is an integer, what is the value of n?
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06 Jul 2018, 02:59
Bunuel wrote: 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 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. (\(11^1 = 11; 11^2 = 1 21; 11^3 = 13 31\)... forms a pattern and will repeat) InsufficientStatement 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. (\(5^4 = 625; 5^6 = 15 625; 5^8 = 390 625\)... forms a pattern and will repeat) InsufficientStatement 1 and 2: N could be 4, 14, 24... No definite answer. InsufficientTherefore, E.Hope this helps. Regards, V
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Re: If n > 2 where n is an integer, what is the value of n?
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06 Jul 2018, 02:59
Bunuel wrote: 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 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)
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Re: If n > 2 where n is an integer, what is the value of n?
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10 Jul 2018, 00:13
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



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Re: If n > 2 where n is an integer, what is the value of n?
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10 Jul 2018, 00:43
Bunuel wrote: 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 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



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Re: If n > 2 where n is an integer, what is the value of n?
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10 Jul 2018, 02:01
Bunuel wrote: 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 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



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Re: If n > 2 where n is an integer, what is the value of n?
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10 Jul 2018, 11:15
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.



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Re: If n > 2 where n is an integer, what is the value of n?
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10 Jul 2018, 21:35
asthagupta wrote: 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 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.




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