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

D

KarunMendiratta

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

04 Aug 2021, 23:48

1

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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

B

godxyz

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

07 Aug 2021, 09:15

2

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!

D

KarunMendiratta

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

07 Aug 2021, 09:21

1

godxyz wrote:

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!

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

2

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

D

KarunMendiratta

Retired Moderator

Joined: 16 Apr 2020

Status:Founder & Quant Trainer

Affiliations: Prepster Education

Posts: 1546

Own Kudos [?]: 3224 [2]

Given Kudos: 172

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WE:Education (Education)

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

09 Aug 2021, 20:57

2

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

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

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

B

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

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

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.

D

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