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A three-digit number is such that when its reverse is subtra [#permalink]
10 Feb 2010, 05:42
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A three-digit number is such that when its reverse is subtracted from it, the result is 297. Also, thrice the tens digit is equal to the difference between its hundreds and units digits. How many possible values are there for the number?
Plz illustrate how to solve this equation question 1) A three-digit number is such that when its reverse is subtracted from it, the result is 297. Also, thrice the tens digit is equal to the difference between its hundreds and units digits. How many possible values are there for the number?
A) 4 B) 5 C) 7 D) 8
Hi, welcome to the Gmat Club.
Solution to your question is as follows:
Three digit number \(abc\) can be represented as \(100a+10b+c\), its reverse number would be \(cba\) or \(100c+10b+a\).
Given: \(100a+10b+c-(100c+10b+a)=297\) --> \(a-c=3\), this gives us 7 values for \(a\) and \(c\): {9,6}{8,5}{7,4}{6,3}{5,2}{4,1}{3,0}.
Also given: \(3b=a-c\), from above we know \(a-c=3\), hence \(3b=3\) --> \(b=1\), only one value for \(b\).
So total of 7 such numbers are possible: {916}{815}{714}{613}{512}{411}{310}.
Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).
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Re: A three-digit number is such that when its reverse is subtra [#permalink]
19 Apr 2014, 07:39
abc - cba = 297
100a+10b+c - 100c - 10b - a = 297
99a-99c = 297
a-c = 3
also, 3b = a-c
b=1
Numbers abc
a1c
ac can take (9,6),(8,5),(7,4),(6,3),(5,2),(4,1),(3,0)
Hence 7 _________________
Rgds, TGC! _____________________________________________________________________ I Assisted You => KUDOS Please _____________________________________________________________________________
Re: A three-digit number is such that when its reverse is subtra [#permalink]
25 May 2015, 07:26
Hello from the GMAT Club BumpBot!
Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).
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Re: A three-digit number is such that when its reverse is subtra [#permalink]
25 May 2015, 22:06
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Expert's post
cleetus wrote:
A three-digit number is such that when its reverse is subtracted from it, the result is 297. Also, thrice the tens digit is equal to the difference between its hundreds and units digits. How many possible values are there for the number?
(A) 4 (B) 5 (C) 7 (D) 8
You can also use reasoning to solve it.
"thrice the tens digit is equal to the difference between its hundreds and units digits."
The difference between any two digits cannot be more than 9-0 = 9. So the tens digit can be 3 at most. But if the difference between the other two digits is 9, their subtraction will give us something around 900. We need something around 300 so the tens digit must be 1 and the difference between the other two digits must be 3.
So the first such number you can have is 310. If you subtract 013 out of it, you get 297 - Correct. Next you can have is 411. If you subtract 114 out of it, you will get 297 - Correct. Next you can have is 512. If you subtract 215 out of it, you will get 297 - Correct. and so on goes the pattern till you have 916.
Re: A three-digit number is such that when its reverse is subtra [#permalink]
16 Aug 2015, 23:54
Bunuel wrote:
cleetus wrote:
Plz illustrate how to solve this equation question 1) A three-digit number is such that when its reverse is subtracted from it, the result is 297. Also, thrice the tens digit is equal to the difference between its hundreds and units digits. How many possible values are there for the number?
A) 4 B) 5 C) 7 D) 8
Hi, welcome to the Gmat Club.
Solution to your question is as follows:
Three digit number \(abc\) can be represented as \(100a+10b+c\), its reverse number would be \(cba\) or \(100c+10b+a\).
Given: \(100a+10b+c-(100c+10b+a)=297\) --> \(a-c=3\), this gives us 7 values for \(a\) and \(c\): {9,6}{8,5}{7,4}{6,3}{5,2}{4,1}{3,0}.
Also given: \(3b=a-c\), from above we know \(a-c=3\), hence \(3b=3\) --> \(b=1\), only one value for \(b\).
So total of 7 such numbers are possible: {916}{815}{714}{613}{512}{411}{310}.
Answer: C.
this gives us 7 values for a and c: {9,6}{8,5}{7,4}{6,3}{5,2}{4,1}{3,0}.
Can you please write how we get 7 possible values? I understood your solution but only 7 possible probability part is unclear.
Re: A three-digit number is such that when its reverse is subtra [#permalink]
17 Aug 2015, 01:19
Expert's post
sashibagra wrote:
Bunuel wrote:
cleetus wrote:
Plz illustrate how to solve this equation question 1) A three-digit number is such that when its reverse is subtracted from it, the result is 297. Also, thrice the tens digit is equal to the difference between its hundreds and units digits. How many possible values are there for the number?
A) 4 B) 5 C) 7 D) 8
Hi, welcome to the Gmat Club.
Solution to your question is as follows:
Three digit number \(abc\) can be represented as \(100a+10b+c\), its reverse number would be \(cba\) or \(100c+10b+a\).
Given: \(100a+10b+c-(100c+10b+a)=297\) --> \(a-c=3\), this gives us 7 values for \(a\) and \(c\): {9,6}{8,5}{7,4}{6,3}{5,2}{4,1}{3,0}.
Also given: \(3b=a-c\), from above we know \(a-c=3\), hence \(3b=3\) --> \(b=1\), only one value for \(b\).
So total of 7 such numbers are possible: {916}{815}{714}{613}{512}{411}{310}.
Answer: C.
this gives us 7 values for a and c: {9,6}{8,5}{7,4}{6,3}{5,2}{4,1}{3,0}.
Can you please write how we get 7 possible values? I understood your solution but only 7 possible probability part is unclear.
We got that a - c = 3: the positive difference between the hundreds and units digits of the number is 3. If a = 9 (max possible value of a), then c is 6, if a = 8, then c = 5, ..., if a = 3, then c = 0 (min possible value of c).
A three-digit number is such that when its reverse is subtra [#permalink]
17 Aug 2015, 21:57
Quote:
Hi, welcome to the Gmat Club.
Solution to your question is as follows:
Three digit number \(abc\) can be represented as \(100a+10b+c\), its reverse number would be \(cba\) or \(100c+10b+a\).
Given: \(100a+10b+c-(100c+10b+a)=297\) --> \(a-c=3\), this gives us 7 values for \(a\) and \(c\): {9,6}{8,5}{7,4}{6,3}{5,2}{4,1}{3,0}.
Also given: \(3b=a-c\), from above we know \(a-c=3\), hence \(3b=3\) --> \(b=1\), only one value for \(b\).
So total of 7 such numbers are possible: {916}{815}{714}{613}{512}{411}{310}.
Answer: C.
Quote:
this gives us 7 values for a and c: {9,6}{8,5}{7,4}{6,3}{5,2}{4,1}{3,0}.
Can you please write how we get 7 possible values? I understood your solution but only 7 possible probability part is unclear.
Quote:
We got that a - c = 3: the positive difference between the hundreds and units digits of the number is 3. If a = 9 (max possible value of a), then c is 6, if a = 8, then c = 5, ..., if a = 3, then c = 0 (min possible value of c).
Hope it's clear.
I am sorry, it was really a silly question and I shouldn't asked it. It was very easy topic but I was just thinking it complicatedly so it was unclear to me. now it's clear , thanks
gmatclubot
A three-digit number is such that when its reverse is subtra
[#permalink]
17 Aug 2015, 21:57
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