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Let abc and dcb represent three-digit positive integers.

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Let abc and dcb represent three-digit positive integers.  [#permalink]

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New post 03 Jan 2014, 09:07
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Let abc and dcb represent three-digit positive integers. If abc+dcb=598, then which of the following must be equivalent to a?

A. d-1
B. d
C. 3-d
D. 4-d
E. 5-d

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Re: Let abc and dcb represent three-digit positive integers.  [#permalink]

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New post 03 Jan 2014, 09:25
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gmat6nplus1 wrote:
Let abc and dcb represent three-digit positive integers. If abc+dcb=598, then which of the following must be equivalent to a?

A. d-1
B. d
C. 3-d
D. 4-d
E. 5-d


\(abc\)
+
\(dcb\)
_____
\(598\)

Notice that for the units and tens digit we have the same sum: \(c+b\). But the result is 8 for the units digit and 9 for tens digit. This implies that there is a carry over 1 from units to tens, thus there is a carry over 1 from units to hundreds. Therefore \(a+d=4\) --> \(a=4-d\).

Answer: D.
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Re: Let abc and dcb represent three-digit positive integers.  [#permalink]

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New post 03 Jan 2014, 10:06
Bunuel wrote:
gmat6nplus1 wrote:
Let abc and dcb represent three-digit positive integers. If abc+dcb=598, then which of the following must be equivalent to a?

A. d-1
B. d
C. 3-d
D. 4-d
E. 5-d


\(abc\)
+
\(dcb\)
_____
\(598\)



Notice that for the units and tens digit we have the same sum: \(c+b\). But the result is 8 for the units digit and 9 for tens digit. This implies that there is a carry over 1 from units to tens, thus there is a carry over 1 from units to hundreds. Therefore \(a+d=4\) --> \(a=4-d\).

Answer: D.



Hi Bunuel,

I have followed the following method to solve the problem

100(a+d)+11(b+C) = 598

a+d can be anything like 5 or 4 or 3 etc not sure what it is..
if we leave out the hunderds digit and concentrate on the rest of the number... it should be a multiple of 11...
98 is not a multiple of 11... multiples of 11 are 11, 22, 33, 44... so on 198

598 - 198 = 400

so the hundreds digit is 4

a+d=4
a=4-d

let me know if this approach is good....
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Re: Let abc and dcb represent three-digit positive integers.  [#permalink]

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New post 21 Apr 2015, 00:17
1
gmat6nplus1 wrote:
Let abc and dcb represent three-digit positive integers. If abc+dcb=598, then which of the following must be equivalent to a?

A. d-1
B. d
C. 3-d
D. 4-d
E. 5-d


Answer :D .
c+b gives a unit digit of 8
b+c gives a unit digit of 9

clearly c,b=9 i.e. a+d=4
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Re: Let abc and dcb represent three-digit positive integers.  [#permalink]

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New post 13 Oct 2015, 19:29
Bunuel wrote:
gmat6nplus1 wrote:
Let abc and dcb represent three-digit positive integers. If abc+dcb=598, then which of the following must be equivalent to a?

A. d-1
B. d
C. 3-d
D. 4-d
E. 5-d


\(abc\)
+
\(dcb\)
_____
\(598\)

Notice that for the units and tens digit we have the same sum: \(c+b\). But the result is 8 for the units digit and 9 for tens digit. This implies that there is a carry over 1 from units to tens, thus there is a carry over 1 from units to hundreds. Therefore \(a+d=4\) --> \(a=4-d\).

Answer: D.

how do you conclude that a carry over 1 from units to hundreds since a and d is different , it could be 3+2 or 1+4
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Re: Let abc and dcb represent three-digit positive integers.  [#permalink]

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New post 13 Oct 2015, 23:11
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Hi anik19890,

To start, you have to look at the UNITS digits:

We know that C+B = a number that ends in 8...
So C+B = 8 OR
C+B = 18

Now look at the TENS digits:
Here, we ALSO have B+C....but the total ends in 9. How can THAT happen (when the total ended in 8 with the UNITs digit)?

IF, in the UNITS digit, C+B = 8, then in the TENS digit, B+C should also equal 8. Since that does NOT happen, C+B CANNOT = 8 (it must equal 18).

So, there must be a 'carryover' of 1 from the UNITS to the TENS.....and there must ALSO be a 'carryover' of 1 from the TENS to the HUNDREDS.

Thus, A+D+1 = 5

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Re: Let abc and dcb represent three-digit positive integers.  [#permalink]

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New post 26 Dec 2015, 11:24
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ok, so we have
abc
dcb
----
598.

we see that b+c must be 18, otherwise we have twice b+c with different results. since the max value for digits could be 9, and since 18 is the maximum value of 2 digits, it must be the case that b and c are both 9.
we have
a99
d99
----
598.

ok, so 5 = a+d+1. since we have to carry 1 from 9+9 in the tens digit.
subtract 1:
4=a+d.
a=4-d.

we have only one answer choice that satisfies this condition.
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Re: Let abc and dcb represent three-digit positive integers.  [#permalink]

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New post 18 Mar 2017, 09:41
abc
dcb
598

notie that though b, c is both in units and tens, and that the sum is different=>their sum is over 10=>+1 for the hundreds=>4-d is a
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Re: Let abc and dcb represent three-digit positive integers.  [#permalink]

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Re: Let abc and dcb represent three-digit positive integers.   [#permalink] 23 Apr 2018, 12:25
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