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Two different two-digit numbers are written beside each other such tha

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Two different two-digit numbers are written beside each other such tha  [#permalink]

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New post 16 Aug 2019, 09:39
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Two different two-digit numbers are written beside each other such that the larger number is written on the left. When the absolute difference of the two numbers is subtracted from the four-digit number so formed, the number obtained is 5481. What is the sum of the two two-digit numbers?

(A) 68

(B) 70

(C) 73

(D) 118

(E) 187
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Two different two-digit numbers are written beside each other such tha  [#permalink]

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New post Updated on: 22 Aug 2019, 03:39
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let 2 numbers are 'ab' and 'cd', such that ab>cd.
abcd-(ab-cd)=5481
100*ab+cd-ab+cd=5481
99*ab+2*cd=5481

99*ab and 5481 are multiples of 9, hence cd must be a multiple of 9.

At cd=18, we get ab=55

Sum of the two 2-digit numbers= 55+18=73


Hovkial wrote:
Two different two-digit numbers are written beside each other such that the larger number is written on the left. When the absolute difference of the two numbers is subtracted from the four-digit number so formed, the number obtained is 5481. What is the sum of the two two-digit numbers?

(A) 68

(B) 70

(C) 73

(D) 118

(E) 187

Originally posted by nick1816 on 16 Aug 2019, 12:48.
Last edited by nick1816 on 22 Aug 2019, 03:39, edited 1 time in total.
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Two different two-digit numbers are written beside each other such tha  [#permalink]

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New post 16 Aug 2019, 13:15
let ab>cd

To subtract the 4-digit minus the absolute difference of the 2 2-digits:
100ab + cd - (ab-cd) = 5481
99ab + 2cd = 5481

There are two possibilities for ab:
if (ab-cd) > cd ---> in this case, ab = 55 ---> 2cd = 5481 - (55*99) = 36 ---> cd = 18 (possible)
if (ab-cd) < cd ---> in this case, ab = 54 ---> 2cd = 5481 - (54*99) = 135 ---> cd = 67.5 (impossible)

then ab + cd = 55 + 18 = 73
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Re: Two different two-digit numbers are written beside each other such tha  [#permalink]

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New post 22 Aug 2019, 02:27
Hovkial
What is the question source?

nick1816
Why stop at cd=18 and not all other multiples of 9?

Mahmoudfawzy83
What is your reasoning for 'two possibilities for ab'?
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Two different two-digit numbers are written beside each other such tha  [#permalink]

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New post 22 Aug 2019, 02:47
99ab+2cd=5481
\(cd= \frac{-99}{2}ab+\frac{5481}{2}\) is an equation of a straight line, having slope -99/2
two closest integral solutions to (55, 18) on above mentioned line are-

1. cd= 18+99= 117, and ab= 55-2=53
2. cd= 18-99= -81, and ab=55+2=57

As cd is a 2-digit positive integer, hence there is no other integral solution possible in which both ab and cd are positive 2-digit number.


philipssonicare wrote:
Hovkial
What is the question source?

nick1816
Why stop at cd=18 and not all other multiples of 9?

Mahmoudfawzy83
What is your reasoning for 'two possibilities for ab'?
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Re: Two different two-digit numbers are written beside each other such tha  [#permalink]

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New post 22 Aug 2019, 03:03
Hi philipssonicare

(1) it is given that ab>cd, so (ab - cd) is a positive number , let's call this positive number x.
x can be as low as 1 (for example if ab = 99 and cd = 98), and can be as high as 89 (for example if ab = 99 and cd = 10)

(2) it is given that abcd - x = 5481
lets focus on the hundreds and thousands digits while subtracting:
if cd> x, then ab will stay as they are after subtraction (because there will be no need to borrow from the hundreds digit) . i.e ab = 54

if cd< x, then ab will decrease by 1 after subtraction(because we will need to borrow from the hundreds digit). i.e ab = 55 before subtraction

there is no other options
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Two different two-digit numbers are written beside each other such tha  [#permalink]

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New post 22 Aug 2019, 03:30
nick1816 wrote:
let 2 numbers are 'ab' and 'cd', such that ab>cd.
abcd-(ab-cd)=5481
100ab+cd-ab+cd=5481
99ab+2cd=5481

99ab and 5481 are multiples of 9, hence cd must be a multiple of 9.

At cd=18, we get ab=55

Sum of the two 2-digit numbers= 55+18=73



nick1816
how did u decide that 99ab is multiple of 9
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Re: Two different two-digit numbers are written beside each other such tha  [#permalink]

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New post 22 Aug 2019, 03:38
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99ab is not a 4 digit-number; it's 99*ab, that is obviously a multiple of 9.
I'll put \(*\) in my solution , in order to remove ambiguity.
vanam52923 wrote:
nick1816 wrote:
let 2 numbers are 'ab' and 'cd', such that ab>cd.
abcd-(ab-cd)=5481
100ab+cd-ab+cd=5481
99ab+2cd=5481

99ab and 5481 are multiples of 9, hence cd must be a multiple of 9.

At cd=18, we get ab=55

Sum of the two 2-digit numbers= 55+18=73



nick1816
how did u decide that 99ab is multiple of 9
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Re: Two different two-digit numbers are written beside each other such tha  [#permalink]

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New post 22 Aug 2019, 04:13
nick1816 wrote:
99ab is not a 4 digit-number; it's 99*ab, that is obviously a multiple of 9.
I'll put \(*\) in my solution , in order to remove ambiguity.
vanam52923 wrote:
nick1816 wrote:
let 2 numbers are 'ab' and 'cd', such that ab>cd.
abcd-(ab-cd)=5481
100ab+cd-ab+cd=5481
99ab+2cd=5481

99ab and 5481 are multiples of 9, hence cd must be a multiple of 9.

At cd=18, we get ab=55

Sum of the two 2-digit numbers= 55+18=73



nick1816
how did u decide that 99ab is multiple of 9


Ty so much
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Re: Two different two-digit numbers are written beside each other such tha   [#permalink] 22 Aug 2019, 04:13
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