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If (x × 10^q) – (y × 10^r) = 10^r, where q, r, x, and y are positive

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If (x × 10^q) – (y × 10^r) = 10^r, where q, r, x, and y are positive  [#permalink]

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New post 09 Jun 2015, 02:50
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Re: If (x × 10^q) – (y × 10^r) = 10^r, where q, r, x, and y are positive  [#permalink]

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New post 15 Jun 2015, 01:24
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Bunuel wrote:
If (x × 10^q) – (y × 10^r) = 10^r, where q, r, x, and y are positive integers and q > r, then what is the units digit of y?

(A) 0
(B) 1
(C) 5
(D) 7
(E) 9

Kudos for a correct solution.


MANHATTAN GMAT OFFICIAL SOLUTION:

Group the 10^r terms on one side of the equal sign.
(x × 10^q) – (y × 10^r) = 10^r
x × 10^q = (y × 10^r) + (1 × 10^r)
x × 10^q = (y + 1) × 10^r

Now, solve for y.
\(x * 10^q = (y + 1) * 10^r\)

\(x * \frac{10^q}{10^r} = y + 1\)

\(x * 10^{q-r} = y + 1\)

\(x * 10^{q-r} - 1= y\)

Since q > r, the exponent on 10^(q-r) is positive. Since x is a positive integer, x*10^(q-r) is a multiple of 10 and therefore ends 0. Any multiple of 10 minus 1 yields an integer with a units digit of 9.

The correct answer is E.
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If (x × 10^q) – (y × 10^r) = 10^r, where q, r, x, and y are positive  [#permalink]

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New post 10 Jun 2015, 02:27
5
Answer = E = 9

\(x * 10^q - y * 10^r = 10^r\)

Dividing the equation by 10^r

\(x * 10^{q-r} - y = 1\)

\(x * 10^{q-r}\) >> This will always have a zero at the units place

As the end result is 1, y has to be 9
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Re: If (x × 10^q) – (y × 10^r) = 10^r, where q, r, x, and y are positive  [#permalink]

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New post 09 Jun 2015, 03:01
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Bunuel wrote:
If (x × 10^q) – (y × 10^r) = 10^r, where q, r, x, and y are positive integers and q > r, then what is the units digit of y?

(A) 0
(B) 1
(C) 5
(D) 7
(E) 9

Kudos for a correct solution.


(x × 10^q) – (y × 10^r) = 10^r..
\(\frac{(x × 10^q) – (y × 10^r)}{10^r}\)=1...
or x*10^(q-r)-y=1..
we know q>r, so x*10^(q-r) will have units digit as 0..
therefore units digit of y is 9.. as ...0 - ..9=1
ans E
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If (x × 10^q) – (y × 10^r) = 10^r, where q, r, x, and y are positive  [#permalink]

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New post 10 Jun 2015, 01:46
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\(x*10^q-y*10^r=10^r\)-->
\(x*10^q=10^r+y*10^r\)-->
\(x*10^q=10^r*(1+y)\) -->
\(x*\frac{10^q}{10^r}=1+y\) which has a 0 units digit since \(q>r\) so y must be 9 or E.
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If (x × 10^q) – (y × 10^r) = 10^r, where q, r, x, and y are positive  [#permalink]

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New post 17 Jun 2015, 03:29
I picked numbers to solve it. Is this also a good approach, or knowing that q>r is not enough?

So, I picked q=2 and r=1:

x*10^2 - y*10 = 10
100x - 10y = 10
100x - 10 = 10y
10x - 1 = y

So now, if we choose numbers for x, because that number will be multiplied by 10, the units digit when you subtract 1 will always be 9.
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Re: If (x × 10^q) – (y × 10^r) = 10^r, where q, r, x, and y are positive  [#permalink]

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New post 28 Feb 2016, 11:21
x * 10^q = 10^r + y*10^r
x * 10^q = 10^r(1+y)
(x * 10^q)/10^r = 1+y
x* 10^(q-r) = 1+y
since q>r -> then we definitely have 10 to a positive number. and regardless of x, the last digit of x* 10^(q-r) is always 0.
since we are given that this is equal to 1+y -> it must be true that the last digit of 1+y is a multiple of 10, and the last digit of y must be 9.
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Re: If (x × 10^q) – (y × 10^r) = 10^r, where q, r, x, and y are positive  [#permalink]

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New post 07 Mar 2016, 03:35
Bunuel wrote:
Bunuel wrote:
If (x × 10^q) – (y × 10^r) = 10^r, where q, r, x, and y are positive integers and q > r, then what is the units digit of y?

(A) 0
(B) 1
(C) 5
(D) 7
(E) 9

Kudos for a correct solution.


MANHATTAN GMAT OFFICIAL SOLUTION:

Group the 10^r terms on one side of the equal sign.
(x × 10^q) – (y × 10^r) = 10^r
x × 10^q = (y × 10^r) + (1 × 10^r)
x × 10^q = (y + 1) × 10^r

Now, solve for y.
\(x * 10^q = (y + 1) * 10^r\)

\(x * \frac{10^q}{10^r} = y + 1\)

\(x * 10^{q-r} = y + 1\)

\(x * 10^{q-r} - 1= y\)

Since q > r, the exponent on 10^(q-r) is positive. Since x is a positive integer, x*10^(q-r) is a multiple of 10 and therefore ends 0. Any multiple of 10 minus 1 yields an integer with a units digit of 9.

The correct answer is E.

Excellent Solution Man ..
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If (x × 10^q) – (y × 10^r) = 10^r, where q, r, x, and y are positive  [#permalink]

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New post 27 Jul 2018, 05:29
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Bunuel wrote:
Bunuel wrote:
If (x × 10^q) – (y × 10^r) = 10^r, where q, r, x, and y are positive integers and q > r, then what is the units digit of y?

(A) 0
(B) 1
(C) 5
(D) 7
(E) 9

Kudos for a correct solution.


MANHATTAN GMAT OFFICIAL SOLUTION:

Group the 10^r terms on one side of the equal sign.
(x × 10^q) – (y × 10^r) = 10^r
x × 10^q = (y × 10^r) + (1 × 10^r)
x × 10^q = (y + 1) × 10^r

Now, solve for y.
\(x * 10^q = (y + 1) * 10^r\)

\(x * \frac{10^q}{10^r} = y + 1\)

\(x * 10^{q-r} = y + 1\)

\(x * 10^{q-r} - 1= y\)

Since q > r, the exponent on 10^(q-r) is positive. Since x is a positive integer, x*10^(q-r) is a multiple of 10 and therefore ends 0. Any multiple of 10 minus 1 yields an integer with a units digit of 9.

The correct answer is E.



hi pushpitkc, future world traveler :) Where from did we get 1 here? :? x × 10^q = (y × 10^r) + (1 × 10^r)

(1 × 10^r) modifies the initial question because \(1^1 × 10^r = 10^{1+r}\)

have a great weekend :-)
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Re: If (x × 10^q) – (y × 10^r) = 10^r, where q, r, x, and y are positive  [#permalink]

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New post 28 Jul 2018, 01:47
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dave13


Quote:
Group the 10^r terms on one side of the equal sign.
(x × 10^q) – (y × 10^r) = 10^rx × 10^q = (y × 10^r) + (1 × 10^r)
x × 10^q = (y + 1) × 10^r


Quote:
Where from did we get 1 here? :? x × 10^q = (y × 10^r) + (1 × 10^r)


Just keep it simple. You need to take 10^r as common factor from term (y+1)


Quote:
(1 × 10^r) modifies the initial question because \(1^1 × 10^r = 10^{1+r}\)

Oops !! \(a^m * a^n = a^{m+n}\) There is no thing much to simplify in \(a^m +a^n\).
May be you would like to review exponent rules again. :-)
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If (x × 10^q) – (y × 10^r) = 10^r, where q, r, x, and y are positive  [#permalink]

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New post 28 Jul 2018, 07:21
bluesquare wrote:
\(x*10^q-y*10^r=10^r\)-->
\(x*10^q=10^r+y*10^r\)-->
\(x*10^q=10^r*(1+y)\) -->
\(x*\frac{10^q}{10^r}=1+y\) which has a 0 units digit since \(q>r\) so y must be 9 or E.



pushpitkc :-) can you please remind a formula for factoring in this case, from here we use a factoring formula RHS \(x*10^q=10^r+y*10^r\)--> and get this
---- > \(x*10^q=10^r*(1+y)\)
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Re: If (x × 10^q) – (y × 10^r) = 10^r, where q, r, x, and y are positive  [#permalink]

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New post 29 Jul 2018, 22:37
1
dave13 wrote:
bluesquare wrote:
\(x*10^q-y*10^r=10^r\)-->
\(x*10^q=10^r+y*10^r\)-->
\(x*10^q=10^r*(1+y)\) -->
\(x*\frac{10^q}{10^r}=1+y\) which has a 0 units digit since \(q>r\) so y must be 9 or E.



pushpitkc :-) can you please remind a formula for factoring in this case, from here we use a factoring formula RHS \(x*10^q=10^r+y*10^r\)--> and get this
---- > \(x*10^q=10^r*(1+y)\)


Hey dave13

All you are doing is taking 10^r in common from the right-hand side.
After that, all you need to do is take bring the common terms to each
side

\(x*10^q=10^r*(1+y)\) -> \(x*\frac{10^q}{10^r}=1+y\) -> \(x*10^{q-r}=(1+y)\)

Hope this help you!
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Re: If (x × 10^q) – (y × 10^r) = 10^r, where q, r, x, and y are positive  [#permalink]

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New post 29 Jul 2018, 23:11
1
dave13 wrote:
bluesquare wrote:
\(x*10^q-y*10^r=10^r\)-->
\(x*10^q=10^r+y*10^r\)-->
\(x*10^q=10^r*(1+y)\) -->
\(x*\frac{10^q}{10^r}=1+y\) which has a 0 units digit since \(q>r\) so y must be 9 or E.



pushpitkc :-) can you please remind a formula for factoring in this case, from here we use a factoring formula RHS \(x*10^q=10^r+y*10^r\)--> and get this
---- > \(x*10^q=10^r*(1+y)\)


It the same as in the following example:

b + ab = b(1 + a). This is done by factoring out common b out of b + ab.
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Re: If (x × 10^q) – (y × 10^r) = 10^r, where q, r, x, and y are positive  [#permalink]

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New post 08 Sep 2019, 06:06
Bunuel wrote:
If (x × 10^q) – (y × 10^r) = 10^r, where q, r, x, and y are positive integers and q > r, then what is the units digit of y?

(A) 0
(B) 1
(C) 5
(D) 7
(E) 9

Kudos for a correct solution.


Given: (x × 10^q) – (y × 10^r) = 10^r, where q, r, x, and y are positive integers and q > r,

Asked: What is the units digit of y?

Let q-r =z where z>0 is a positive integer

10^z * x - y = 1
Since unit digit of 10^z * x = 0 since q-r=z>0
y must be 9

IMO E
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Re: If (x × 10^q) – (y × 10^r) = 10^r, where q, r, x, and y are positive   [#permalink] 08 Sep 2019, 06:06
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