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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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09 Jun 2015, 01:50
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Difficulty:

55% (hard)

Question Stats:

64% (01:56) correct 36% (01:58) wrong based on 384 sessions

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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.

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Math Expert
Joined: 02 Sep 2009
Posts: 52431
Re: If (x × 10^q) – (y × 10^r) = 10^r, where q, r, x, and y are positive  [#permalink]

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15 Jun 2015, 00:24
4
5
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.

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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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10 Jun 2015, 01:27
5

$$x * 10^q - y * 10^r = 10^r$$

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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##### General Discussion
Math Expert
Joined: 02 Aug 2009
Posts: 7213
Re: If (x × 10^q) – (y × 10^r) = 10^r, where q, r, x, and y are positive  [#permalink]

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09 Jun 2015, 02:01
3
1
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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Joined: 29 Mar 2015
Posts: 44
Location: United States
If (x × 10^q) – (y × 10^r) = 10^r, where q, r, x, and y are positive  [#permalink]

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10 Jun 2015, 00:46
2
$$x*10^q-y*10^r=10^r$$-->
$$x*10^q=10^r+y*10^r$$-->
$$x*10^q=10^r*(1+y)$$ -->
$$x*\frac{10^q}{10^r}=1+y$$ 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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17 Jun 2015, 02:29
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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28 Feb 2016, 10:21
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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07 Mar 2016, 02:35
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.

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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27 Jul 2018, 04:29
1
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.

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
Study Buddy Forum Moderator
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Posts: 1299
Location: India
WE: Engineering (Other)
Re: If (x × 10^q) – (y × 10^r) = 10^r, where q, r, x, and y are positive  [#permalink]

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28 Jul 2018, 00:47
1
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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VP
Joined: 09 Mar 2016
Posts: 1287
If (x × 10^q) – (y × 10^r) = 10^r, where q, r, x, and y are positive  [#permalink]

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28 Jul 2018, 06:21
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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29 Jul 2018, 21:37
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

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

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Math Expert
Joined: 02 Sep 2009
Posts: 52431
Re: If (x × 10^q) – (y × 10^r) = 10^r, where q, r, x, and y are positive  [#permalink]

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29 Jul 2018, 22:11
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 &nbs [#permalink] 29 Jul 2018, 22:11
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