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Re: Sum of the Digits [#permalink]
18 Nov 2012, 16:40

4

This post received KUDOS

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

Re: If a, b, and c are different nonnegative digits, which of [#permalink]
12 Feb 2014, 04:20

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Re: Sum of the Digits [#permalink]
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 _________________

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

1

This post received KUDOS

Expert's post

1

This post was BOOKMARKED

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.

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