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![If [x(10^q)]-[y(10^r)]=10^r, where q r x and](/forum/styles/gmatclub_light/imageset/icon_post_target_unread.svg "If [x(10^q)]-[y(10^r)]=10^r, where q r x and"){kind=icon}
01 Jan 2014, 10:47
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E
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
(N/A)
Question Stats:
50% (01:18) correct
50%
(01:42)
wrong
based on 2
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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
A. 0
B. 1
C. 5
D. 7
E. 9
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01 Jan 2014, 10:47
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Since we are told that \(q>r\)
\(10^r[x(10^q^-^r)-y]=10^r\) and this happens when \([x(10^q^-^r)-y]=1\) They told us that q>r this conveys us that \(q-r\) is at least 10. For this reason the units digit of x will always be zero and at the same time the expression \(x(10^q^-^r)\) is greater than y, let's represent x as 10.
10-y=1 -----> y=9
the answer is E.
\(10^r[x(10^q^-^r)-y]=10^r\) and this happens when \([x(10^q^-^r)-y]=1\) They told us that q>r this conveys us that \(q-r\) is at least 10. For this reason the units digit of x will always be zero and at the same time the expression \(x(10^q^-^r)\) is greater than y, let's represent x as 10.
10-y=1 -----> y=9
the answer is E.
01 Jan 2014, 13:56
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gmat6nplus1
Perhaps I am getting use to those tricky question but now only 1:18 to do this type of question! #ThanksGmatClub !
The solution relies on two things:
A) All numbers are positive integers
B) X10 - y = 1 (you find this one by factorizing by 10^r)
Do you find the pattern? What will be the number that will allow you to have 10 - y = 1?
Well 9 of course!
Answer E!
