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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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If x and y are positive integers and n = 5^x + 7^(y + 15), what is the [#permalink]

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New post 23 Oct 2014, 01:02
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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) \(y^2 – 6y + 5 = 0\)

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Re: If x and y are positive integers and n = 5^x + 7^(y + 15), what is the [#permalink]

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New post 23 Oct 2014, 01:03
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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


Check Units digits, exponents, remainders problems directory in our Special Questions Directory.
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Re: If x and y are positive integers and n = 5^x + 7^(y + 15), what is the [#permalink]

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New post 24 Oct 2014, 00:42
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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=49
7^3=343
7^4=XXX1

7^5=XXXX7

St 1 says y=2x-15 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=14(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 [#permalink]

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New post 24 Oct 2014, 05:36
1
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 = 2x-15
=> 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 [#permalink]

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New post 09 Aug 2017, 15:17
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 [#permalink]

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New post 07 Nov 2017, 11:10
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 [#permalink]

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New post 18 Dec 2017, 01:51
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 [#permalink]

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New post 18 Dec 2017, 02:33
1
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 \(y^2-3y+2=0\)
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 [#permalink]

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New post 18 Dec 2017, 02:37
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 \(y^2-3y+2=0\)
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 [#permalink]

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New post 18 Dec 2017, 06:23
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 n


Statement 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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