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

Question

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

Bunuel · Accepted Answer

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.

PareshGmat · Answer

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

chetan2u · Answer

(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

bluesquare · Answer

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.

pacifist85 · Answer

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.

mvictor · Answer

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.

dave13 · Answer

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

adkikani · Answer

dave13 Just keep it simple. You need to take 10^r as common factor from term (y+1) 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.

dave13 · Answer

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)

pushpitkc · Answer

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!

Bunuel · Answer

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.

Kinshook · Answer

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

mysterymanrog · Answer

(x*10^q)-(y*10^r)=10^r 
 10^r(x*10^{q-r}-y)=10^r 
 (x*10^{q-r}-y)=1 
Because we know that q>r,  x*10^{q-r}  will be some multiple of 10, and that whole number will end with digit 0. 
If we subtract a number ending in digit 0 and end with one, it must be subtracted by 9 - take the simplest case of (10-y)=1
y=9 in that case. 
So y will be some number ending with 9.

E.

pearlmotwani · Answer

(x* 10^ q) - (y*10^r) = 10^r
if q > r then I can take 10^r as common

10^r   = 10^r 
cancelling 10^r, 
   = 1
 (x * 10^q-r) - 1 = y 
some power of 10 that mean end would be 0, subtracting 1 from it means 9 will be the units digit

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

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