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

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.

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).

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.

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

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.

If x and y are positive integers and \(n = 5^x + 7^{(y + 15)}\), what is the units digit of n?

Quick notes:
5 to any power = units digit of 5.
Cyclicity of 7 = 7, 9, 3, 1

To determine the units digit of n, we need to determine the value of y+15. Lets look at the choices:


(1) \(y = 2x – 15\)

We have no idea what x is -- Insufficient.

(2) \(y^2 – 6y + 5 = 0\)

y = 5, 1

y + 15 = 16 or 20.

Since y has a cyclicity of 4, y raised to any multiple of 4 will give us a units digit of 1.

Statement 2 is sufficient.

Answer is B.

Trick here is not to fall into C trap
5^anything = 5
Hence finding x is not important

y^2 – 6y + 5 = 0
We get solutions as 5 and 1
for both, the unit digit ie 7^20 and 7^16 is the same ie 9
Hence B

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