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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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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
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15 Jun 2015, 01:24
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. \(x * 10^q = (y + 1) * 10^r\) \(x * \frac{10^q}{10^r} = y + 1\) \(x * 10^{qr} = y + 1\) \(x * 10^{qr}  1= y\) Since q > r, the exponent on 10^(qr) is positive. Since x is a positive integer, x*10^(qr) 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
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10 Jun 2015, 02:27
Answer = E = 9 \(x * 10^q  y * 10^r = 10^r\) Dividing the equation by 10^r \(x * 10^{qr}  y = 1\) \(x * 10^{qr}\) >> 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
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09 Jun 2015, 03:01
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^(qr)y=1.. we know q>r, so x*10^(qr) 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
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10 Jun 2015, 01:46
\(x*10^qy*10^r=10^r\)> \(x*10^q=10^r+y*10^r\)> \(x*10^q=10^r*(1+y)\) > \(x*\frac{10^q}{10^r}=1+y\) 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
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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
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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^(qr) = 1+y since q>r > then we definitely have 10 to a positive number. and regardless of x, the last digit of x* 10^(qr) 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
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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. \(x * 10^q = (y + 1) * 10^r\) \(x * \frac{10^q}{10^r} = y + 1\) \(x * 10^{qr} = y + 1\) \(x * 10^{qr}  1= y\) Since q > r, the exponent on 10^(qr) is positive. Since x is a positive integer, x*10^(qr) 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
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27 Jul 2018, 05:29
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. \(x * 10^q = (y + 1) * 10^r\) \(x * \frac{10^q}{10^r} = y + 1\) \(x * 10^{qr} = y + 1\) \(x * 10^{qr}  1= y\) Since q > r, the exponent on 10^(qr) is positive. Since x is a positive integer, x*10^(qr) 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
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28 Jul 2018, 01:47
dave13Quote: 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
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28 Jul 2018, 07:21
bluesquare wrote: \(x*10^qy*10^r=10^r\)> \(x*10^q=10^r+y*10^r\)> \(x*10^q=10^r*(1+y)\) > \(x*\frac{10^q}{10^r}=1+y\) 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
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29 Jul 2018, 22:37
dave13 wrote: bluesquare wrote: \(x*10^qy*10^r=10^r\)> \(x*10^q=10^r+y*10^r\)> \(x*10^q=10^r*(1+y)\) > \(x*\frac{10^q}{10^r}=1+y\) 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 dave13All you are doing is taking 10^r in common from the righthand side. After that, all you need to do is take bring the common terms to each side \(x*10^q=10^r*(1+y)\) > \(x*\frac{10^q}{10^r}=1+y\) > \(x*10^{qr}=(1+y)\) 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
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29 Jul 2018, 23:11




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