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If x and y are positive integers and n = 5^x + 7^(y + 15), what is the
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23 Oct 2014, 00:02
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Re: If x and y are positive integers and n = 5^x + 7^(y + 15), what is the
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23 Oct 2014, 00:03



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Re: If x and y are positive integers and n = 5^x + 7^(y + 15), what is the
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23 Oct 2014, 23:42
Bunuel wrote: Tough and Tricky questions: Remainders. If x and y are positive integers and n = 5^x + 7^(y + 15), what is the units digit of n? (1) y = 2x – 15 (2) y^2 – 6y + 5 = 0 Sol:n = 5^x + 7^(y + 15) Note that 5^ (any positive Integer) gives unit digit of 5...So we need to know only y to be able solve the problem.. Cylcity of 7 is 4 7^1=7 7^2=4 97^3=34 37^4=XXX1 7^5=XXXX7 St 1 says y=2x15 or y+15=2x..thus we know power of 7 in the expression is even so unit digit for 7^(y+15) will be =9 or 1.. We have 2 answers possible for unit digit of n i.e 5+9=1 4(unit digit) or 5+ 1=6. Not sufficient St 2 says gives value of y=1,5 so we have either 7^16 or 7^20...for each expression unit digit is 1.. Sufficient Ans B
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Re: If x and y are positive integers and n = 5^x + 7^(y + 15), what is the
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24 Oct 2014, 04:36
Bunuel wrote: Tough and Tricky questions: Remainders. If x and y are positive integers and n = 5^x + 7^(y + 15), what is the units digit of n? (1) y = 2x – 15 (2) y^2 – 6y + 5 = 0 B. n = 5^x + 7^(y+15) 5^x always ends in 5. so n = 5 + 7^(y+15) 1) y = 2x15 => y+15 = 2x so (y+15) is always even but 7^2x can end in 9 or 1 so insufficient. 2) y^2  6y + 5 = 0 y = 1 or 5 => y+15 = 16, 20 (which always ends in 1) so sufficient.
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Re: If x and y are positive integers and n = 5^x + 7^(y + 15), what is the
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09 Aug 2017, 14:17
Hi.
I did this question and I don't understand why the answers ignore the units digit of 5^x.
if x is even, the units of 5^x is 0, but if x is odd, it is 5. The question is about the units of n, not 7^(y+15), and only with statement (2) you don't have information about x.



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Re: If x and y are positive integers and n = 5^x + 7^(y + 15), what is the
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07 Nov 2017, 10:10
Edivar wrote: Hi.
I did this question and I don't understand why the answers ignore the units digit of 5^x.
if x is even, the units of 5^x is 0, but if x is odd, it is 5. The question is about the units of n, not 7^(y+15), and only with statement (2) you don't have information about x. If x is a positive integer, the units digit of \(5^x\) is always 5: 5^1 = 5, 5^2 = 25, 5^3 = 125, 5^4 = 625, ... I think you are mixing \(5^x\) (5 to the power of x) with \(5x\) (5 multiplied by x).
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If x and y are positive integers and n = 5^x + 7^(y + 15), what is the
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18 Dec 2017, 00:51
Bunuel wrote: Tough and Tricky questions: Remainders. If x and y are positive integers and \(n = 5^x + 7^{(y + 15)}\), what is the units digit of n? (1) \(y = 2x – 15\) (2) \(y^2 – 6y + 5 = 0\) Can you please tell me, what would be the answer if roots of quadratic equation were (2,1)? I believe, answer should be B. Even in this new case we can derive value for (n), however, unit digit for 7 will not match (as in the present case when roots are 5,1).



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Re: If x and y are positive integers and n = 5^x + 7^(y + 15), what is the
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18 Dec 2017, 01:33
anubhavece wrote: Bunuel wrote: Tough and Tricky questions: Remainders. If x and y are positive integers and \(n = 5^x + 7^{(y + 15)}\), what is the units digit of n? (1) \(y = 2x – 15\) (2) \(y^2 – 6y + 5 = 0\) Can you please tell me, what would be the answer if roots of quadratic equation were (2,1)? I believe, answer should be B. Even in this new case we can derive value for (n), however, unit digit for 7 will not match (as in the present case when roots are 5,1). hi.. In this case the statement II would read \(y^23y+2=0\) so when y = 1 or 2, the units digit will change.. in \(n = 5^x + 7^{(y + 15)}\).. 5^x will always give you 5 \(7^{(y+15)}\) will become 7^(16) or 7^(17) 7^16 = 7^(4*4+0) so will give you same units digit as 7^4 or 1 7^17 will give 7^(4*4+1) so will give same units digit as 7^1 or 7 so ans will be 5+1=6 OR 5+7=12 or 2 thus insuff
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Re: If x and y are positive integers and n = 5^x + 7^(y + 15), what is the
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18 Dec 2017, 01:37
chetan2u wrote: anubhavece wrote: Bunuel wrote: Tough and Tricky questions: Remainders. If x and y are positive integers and \(n = 5^x + 7^{(y + 15)}\), what is the units digit of n? (1) \(y = 2x – 15\) (2) \(y^2 – 6y + 5 = 0\) Can you please tell me, what would be the answer if roots of quadratic equation were (2,1)? I believe, answer should be B. Even in this new case we can derive value for (n), however, unit digit for 7 will not match (as in the present case when roots are 5,1). hi.. In this case the statement II would read \(y^23y+2=0\) so when y = 1 or 2, the units digit will change.. in \(n = 5^x + 7^{(y + 15)}\).. 5^x will always give you 5 \(7^{(y+15)}\) will become 7^(16) or 7^(17) 7^16 = 7^(4*4+0) so will give you same units digit as 7^4 or 1 7^17 will give 7^(4*4+1) so will give same units digit as 7^1 or 7 so ans will be 5+1=6 OR 5+7=12 or 2 thus insuff Ok Thanks a lot My Bad ..... I forgot that the purpose in DS is not just to find a answer but also remove ambiguity.



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If x and y are positive integers and n = 5^x + 7^(y + 15), what is the
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18 Dec 2017, 05:23
Bunuel wrote: Tough and Tricky questions: Remainders. If x and y are positive integers and \(n = 5^x + 7^{(y + 15)}\), what is the units digit of n? (1) \(y = 2x – 15\) (2) \(y^2 – 6y + 5 = 0\) Question : Unit Digit of n = ?Given: \(n = 5^x + 7^{(y + 15)}\) Point to Note: \(5^x\) will always have unit digit 5 for any positive integer value of x (as given) Hence, Calculating value of x is completely immaterial for us
All we need is the unit digit of \(7^{(y + 15)}\) to find the unit digit of nStatement 1: \(y = 2x – 15\) \(7^{(y + 15)}\) becomes \(7^{(2x)}\) for x = 1, \(7^{(2x)}\) will have unit digit = 9 for x = 2, \(7^{(2x)}\) will have unit digit = 1 NOT SUFFICIENT Statement 2: \(y^2 – 6y + 5 = 0\) i.e. y = 1 or 5 for y = 1, \(7^{(y+15)}\) will have unit digit = 1 for y = 5, \(7^{(y+15)}\) will have unit digit = 1 SUFFICIENT Answer: option B
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Re: If x and y are positive integers and n = 5^x + 7^(y + 15), what is the
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05 Oct 2018, 08:38
Bunuel wrote: Tough and Tricky questions: Remainders. If x and y are positive integers and \(n = 5^x + 7^{(y + 15)}\), what is the units digit of n? (1) \(y = 2x – 15\) (2) \(y^2 – 6y + 5 = 0\) given that x and y are positive integers they are >0 \(5^1\) has a unit digit of 5 \(5^2\) has a unit digit of 5. This means that regardless of the exponential power of 5 it will always have a digit of 5, as long as the power is a positive integer. So we are interested in knowing the y value. Statement 1) does not provide us any information. Insufficient Statement 2) \(y^2 – 6y + 5 = 0\) (y5)(y1) = 0 y = 1 or 5 The unit digit of powers of 7 are below. \(7^1\)=7 \(7^2 = 49\) \(7^3 = 343\) \(7^4 = 2401\) This cycle repeats in multiples of 4, so 8,12,16,20 would have a unit digit of 1. if y =1 then the power is 16, meaning it will have a units digit of 1 if y = 5 then the power is 20, meaning again it will have a digit of 1. sufficient. B is the answer.




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