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Math Expert V
Joined: 02 Sep 2009
Posts: 58117
If x and y are positive integers and n = 5^x + 7^(y + 15), what is the  [#permalink]

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Difficulty:   55% (hard)

Question Stats: 68% (02:11) correct 32% (02:32) wrong based on 765 sessions

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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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Math Expert V
Joined: 02 Sep 2009
Posts: 58117
Re: If x and y are positive integers and n = 5^x + 7^(y + 15), what is the  [#permalink]

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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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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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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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Math Expert V
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Posts: 7831
Re: If x and y are positive integers and n = 5^x + 7^(y + 15), what is the  [#permalink]

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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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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.
Math Expert V
Joined: 02 Sep 2009
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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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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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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).
Intern  B
Joined: 30 Aug 2017
Posts: 5
Re: If x and y are positive integers and n = 5^x + 7^(y + 15), what is the  [#permalink]

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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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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

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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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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) $$y^2 – 6y + 5 = 0$$

(y-5)(y-1) = 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. Re: If x and y are positive integers and n = 5^x + 7^(y + 15), what is the   [#permalink] 05 Oct 2018, 09:38
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