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abc and cba are 3-digit positive integers. D = abc - cba

KarunMendiratta

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Re: abc and cba are 3-digit positive integers. D = abc - cba [#permalink]

04 Aug 2021, 23:48

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Explanation:

Any 3-digit number \(xyz\) can be written as \(100x + 10y + z\)

\(abc = 100a + 10b + c\)
\(cba = 100c + 10b + a\)

So, \(D = 99a - 99c = 99(a - c)\)

If \(D\) is a multiple of \(7\) then \((a - c)\) must be \(7\)

Let us look for cases for which \((a - c) = 7\);

Case I: a = 9 and c = 2
Case II: a = 8 and c = 1
Case III: a = 7 and c = 0

Now, we can have two feasible cases and \(b\) can take 10 values
So, Number of +ve values of \(abc\) which are divisible by \(7 = (2)(10) = 20\)

Col. A: 20
Col. B: 15

Hence, option A

NOTE: Case III is invalid as c = 0, \(cba\) becomes a 2-digit integer

godxyz

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Re: abc and cba are 3-digit positive integers. D = abc - cba [#permalink]

07 Aug 2021, 09:15

KarunMendiratta,

Hi sir,

I'm getting the column A value as 40. Can we not consider 982 and 289 as two separate cases since the sign of D is not mentioned? Please let me know if I'm missing something here.

Thanks in advance!

KarunMendiratta

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Re: abc and cba are 3-digit positive integers. D = abc - cba [#permalink]

07 Aug 2021, 09:21

godxyz wrote:

I think you have got a valid point.
Col. A must read as Number of +ve values of D
Kudos to you Sir

nakedsnake2614

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Re: abc and cba are 3-digit positive integers. D = abc - cba [#permalink]

09 Aug 2021, 20:20

KarunMendiratta wrote:

Explanation:

Any 3-digit number \(xyz\) can be written as \(100x + 10y + z\)

\(abc = 100a + 10b + c\)
\(cba = 100c + 10b + a\)

So, \(D = 99a - 99c = 99(a - c)\)

If \(D\) is a multiple of \(7\) then \((a - c)\) must be \(7\)

Let us look for cases for which \((a - c) = 7\);

Case I: a = 9 and c = 2
Case II: a = 8 and c = 1
Case III: a = 7 and c = 0

Now, we can have two feasible cases and \(b\) can take 10 values
So, Number of values of D which are divisible by \(7 = (2)(10) = 20\)

Col. A: 20
Col. B: 15

Hence, option A

NOTE: Case III is invalid as c = 0, \(cba\) becomes a 2-digit integer

The question asks about the different values of D. No matter what value of b you consider, the value of D remains the same. Why would you then consider them as 10 different values?

Posted from my mobile device

KarunMendiratta

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abc and cba are 3-digit positive integers. D = abc - cba [#permalink]

09 Aug 2021, 20:57

nakedsnake2614 wrote:

KarunMendiratta wrote:

Explanation:

Any 3-digit number \(xyz\) can be written as \(100x + 10y + z\)

\(abc = 100a + 10b + c\)
\(cba = 100c + 10b + a\)

So, \(D = 99a - 99c = 99(a - c)\)

If \(D\) is a multiple of \(7\) then \((a - c)\) must be \(7\)

Let us look for cases for which \((a - c) = 7\);

Case I: a = 9 and c = 2
Case II: a = 8 and c = 1
Case III: a = 7 and c = 0

Now, we can have two feasible cases and \(b\) can take 10 values
So, Number of values of D which are divisible by \(7 = (2)(10) = 20\)

Col. A: 20
Col. B: 15

Hence, option A

NOTE: Case III is invalid as c = 0, \(cba\) becomes a 2-digit integer

nakedsnake2614
Rightly said, I've made the changes

The question should read "Number of values of \(abc\) which are divisible by 7"

Apologies everyone for the inconvenience

test01

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Re: abc and cba are 3-digit positive integers. D = abc - cba [#permalink]

15 Oct 2023, 06:36

KarunMendiratta wrote:

nakedsnake2614 wrote:

KarunMendiratta wrote:

Explanation:

Any 3-digit number \(xyz\) can be written as \(100x + 10y + z\)

\(abc = 100a + 10b + c\)
\(cba = 100c + 10b + a\)

So, \(D = 99a - 99c = 99(a - c)\)

If \(D\) is a multiple of \(7\) then \((a - c)\) must be \(7\)

Let us look for cases for which \((a - c) = 7\);

Case I: a = 9 and c = 2
Case II: a = 8 and c = 1
Case III: a = 7 and c = 0

Now, we can have two feasible cases and \(b\) can take 10 values
So, Number of values of D which are divisible by \(7 = (2)(10) = 20\)

Col. A: 20
Col. B: 15

Hence, option A

NOTE: Case III is invalid as c = 0, \(cba\) becomes a 2-digit integer

nakedsnake2614
Rightly said, I've made the changes

The question should read "Number of values of \(abc\) which are divisible by 7"

Apologies everyone for the inconvenience

Sir, I think the changes made have accidentally removed the question. Kindly point me to where the question is.

Carcass

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Re: abc and cba are 3-digit positive integers. D = abc - cba [#permalink]

15 Oct 2023, 06:46

Expert Reply

I hardly believe that we could recover the question itself because

1) it was a question created by karun and not taken from a book
2) Once deleted, the message is unrecoverable , neither via google cache

I archive the discussion