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If a, b, and c are different nonnegative digits, which of

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If a, b, and c are different nonnegative digits, which of [#permalink] New post 18 Nov 2012, 12:22
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If a, b, and c are different nonnegative digits, which of the following CANNOT be a solution to the addition problem below?

abc +
cba
------------

A) 929
B) 1,110
C) 1,111
D) 1,322
E) 1,776
[Reveal] Spoiler: OA

Last edited by Bunuel on 19 Nov 2012, 02:38, edited 1 time in total.
Renamed the topic and edited the question.
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Re: Sum of the Digits [#permalink] New post 18 Nov 2012, 16:40
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monsoon1 wrote:
If a, b, and c are different nonnegative digits, which of the following CANNOT be a solution to the addition problem below?

abc +
cba
------------

A) 929
B) 1,110
C) 1,111
D) 1,322
E) 1,776


abc can be written as = 100a +10b+c
similarly cba = 100c+10b+a

Hence sum = 101a+101c+20b = 101(a+c) +20b
Sum has to be of the form of 101 *Number1 + 20 *Number2

Choices,
A) 929 = 101 *9 + 20*1 => possible
B) 1,110 = 101*10+20*5 =>possible
C) 1,111= 101*11+20*0 => possible
D) 1,322 = 101*13+9 or 101*12+110 => not possible
E) 1,776 = 101*17 + 59 or 101*16+20*8 = possible

Ans D it is!
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Re: If a, b, and c are different nonnegative digits, which of [#permalink] New post 12 Feb 2014, 04:20
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Re: Sum of the Digits [#permalink] New post 21 Mar 2014, 00:11
Vips0000 wrote:
monsoon1 wrote:
If a, b, and c are different nonnegative digits, which of the following CANNOT be a solution to the addition problem below?

abc +
cba
------------

A) 929
B) 1,110
C) 1,111
D) 1,322
E) 1,776


abc can be written as = 100a +10b+c
similarly cba = 100c+10b+a

Hence sum = 101a+101c+20b = 101(a+c) +20b
Sum has to be of the form of 101 *Number1 + 20 *Number2

Choices,
A) 929 = 101 *9 + 20*1 => possible
B) 1,110 = 101*10+20*5 =>possible
C) 1,111= 101*11+20*0 => possible
D) 1,322 = 101*13+9 or 101*12+110 => not possible
E) 1,776 = 101*17 + 59 or 101*16+20*8 = possible

Ans D it is!


I randomly took 415 & 514 to give 929 ; so ruled out option A

Reached upto this point:

abc can be written as = 100a +10b+c
similarly cba = 100c+10b+a
Hence sum = 101a+101c+20b = 101(a+c) +20b

Cant understand how the numbers were picked / tested to check the results??

Can someone please explain? Thanks
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Re: Sum of the Digits [#permalink] New post 21 Mar 2014, 00:41
1
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Expert's post
PareshGmat wrote:
Vips0000 wrote:
monsoon1 wrote:
If a, b, and c are different nonnegative digits, which of the following CANNOT be a solution to the addition problem below?

abc +
cba
------------

A) 929
B) 1,110
C) 1,111
D) 1,322
E) 1,776


abc can be written as = 100a +10b+c
similarly cba = 100c+10b+a

Hence sum = 101a+101c+20b = 101(a+c) +20b
Sum has to be of the form of 101 *Number1 + 20 *Number2

Choices,
A) 929 = 101 *9 + 20*1 => possible
B) 1,110 = 101*10+20*5 =>possible
C) 1,111= 101*11+20*0 => possible
D) 1,322 = 101*13+9 or 101*12+110 => not possible
E) 1,776 = 101*17 + 59 or 101*16+20*8 = possible

Ans D it is!


I randomly took 415 & 514 to give 929 ; so ruled out option A

Reached upto this point:

abc can be written as = 100a +10b+c
similarly cba = 100c+10b+a
Hence sum = 101a+101c+20b = 101(a+c) +20b

Cant understand how the numbers were picked / tested to check the results??

Can someone please explain? Thanks


abc can be written as 100a + 10b + c.
cba can be written as 100c + 10b + a.

The sum = (100a + 10b + c) + (100c + 10b + a) = 20(5a + 5c + b) + (a + c) = {a multiple of 20} + (a + c).

A. 929 --> 920 + 9 = {a multiple of 20} + (a + c) --> a + c can be 9;
...
D. 1,322 --> 1,320 + 2 = 1,300 + 22 --> a + c can be neither 2 (because a and c are different, non negative digits) nor 22.

Answer: D.

Similar questions to practice: given-that-a-b-c-and-d-are-different-nonzero-digits-and-126865.html

Hope it helps.
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Re: Sum of the Digits   [#permalink] 21 Mar 2014, 00:41
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