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

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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Source : Manhattan Advanced Quant Question No. 4

OFFICIAL SOLUTION

Note that many of the answer choices have four digits, which would require “carrying” from the hundredths digit. This observation gives us a good starting place.

If \(a + c < 10\), no carrying would be required from the ones place to the tens place. The resulting sum would have an even tens digit (\(b + b = 2b\)) and could have either three or four digits. A three digit sum will be a palindrome, as both the ones digit and hundreds digit will be the single-digit \(a + c\). A four digit sum occurs for \(a + c < 10\) only when \(a + c = 9\) and \(2b ≥ 10\) (i.e., carry a 1 from the tens place to the hundreds place), and must have the following digits: 1 0 even 9 .

If \(a + c ≥ 10\), we must “carry” a 1 from the ones place to the tens place. The resulting sum would have an odd tens digit (\(1 + b + b = 2b + 1\)) and must have four digits. Both the ones digit and hundreds digit are formed by summing a + c, so these digits will either be the same (if we don’t carry from the tens digit) or differ by exactly 1 (that is, if a 1 is carried from the tens digit).

We now inspect the answer choices:
(A) OK. Even tens digit in a three digit number.
(B) OK. Odd tens digit in a four digit number with hundreds digit = ones digit + 1.
(C) OK. Odd tens digit in a four digit number with hundreds digit = ones digit.
(D) NO. Even tens digit in a four digit number, yet the first two digits are not 1 and 0.
(E) OK. Odd tens digit in a four digit number with hundreds digit = ones digit + 1.

Alternatively, we could try to produce each sum:
(A) 929 = 316 + 613
(B) 1,110 = 258 + 852
(C) 1,111 = 209 + 902
(D) CANNOT
(E) 1,776 = 789 + 987

The correct answer is D.
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(100a +10b+ c) + (100c + 10b + a) = 101a+20b+101c
101(a+c) + 20b

I got stuck here so I then looked at the answer choices.

I immediately tried c by setting (a+c) = 11. I was able to get 1111
I next tried answer d by setting (a+c) = 13. I noticed 20b is a multiple of 20 so I got 1333. I ran out of time and chose D since I couldn't see how I could get 1322.

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