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How many such positive integers exist so that if the unit digit of the

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How many such positive integers exist so that if the unit digit of the [#permalink]

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New post 23 Jun 2017, 08:34
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How many such positive integers exist so that if the unit digit of the original integer is removed, the ratio of the new number to the original one becomes \(\frac{1}{14}\) ?

(A) 0
(B) 1
(C) 2
(D) 3
(E) 4

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Re: How many such positive integers exist so that if the unit digit of the [#permalink]

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New post 23 Jun 2017, 09:37
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You should start by translating the problem to symbols:

old= 10r + u (i.e. r=1 u = 3 so 10*1 +3 = 13)
new= r

\(\frac{new}{old} = \frac{1}{14}\) so:

\(\frac{10r + u}{r} = \frac{1}{14}\)

which simplifies to: \(4r = u\)

now we can list posible solutions starting:

\(r=1 => u = 4\)
\(r=2 => u = 8\)
\(r=3 => u = 12\) wrong because u is unit number so must be smaller or equal to 9

hence we have 2 solutions
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How many such positive integers exist so that if the unit digit of the [#permalink]

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New post 23 Jun 2017, 11:57
Vardan95 wrote:
How many such positive integers exist so that if the unit digit of the original integer is removed, the ratio of the new number to the original one becomes \(\frac{1}{14}\) ?

(A) 0
(B) 1
(C) 2
(D) 3
(E) 4


let x=tens digit
y=units digit
(10x+y)/x=14
4x=y
y must be a one digit multiple of 4, either 4 or 8
x must be either 1 or 2
(10*1+4)/1=14
(20*1+8)/2=14
14 and 28 are 2 positive integers
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Re: How many such positive integers exist so that if the unit digit of the [#permalink]

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New post 23 Jun 2017, 12:16
The original number must be a multiple of 14..

14, 28, 42, 56, ....

let's start converting
= \(\frac{1}{14}\)( dropping 4 from 14) = \(\frac{1}{14}\) = solution
= \(\frac{2}{28}\)( dropping 2 from 28) = \(\frac{1}{14}\) = solution
= \(\frac{4}{42}\)( dropping 4 from 42) = \(\frac{2}{21}\) = Not a solution
= \(\frac{5}{56}\)( dropping 5 from 56) = \(\frac{5}{56}\) = Not a solution

so the denominator must be a multiple of numerator which is not possible as we proceed further : hence only 2 solution
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Re: How many such positive integers exist so that if the unit digit of the [#permalink]

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New post 10 Aug 2017, 14:17
lbowl wrote:
You should start by translating the problem to symbols:

old= 10r + u (i.e. r=1 u = 3 so 10*1 +3 = 13)
new= r

\(\frac{new}{old} = \frac{1}{14}\) so:

\(\frac{10r + u}{r} = \frac{1}{14}\)

which simplifies to: \(4r = u\)

now we can list posible solutions starting:

\(r=1 => u = 4\)
\(r=2 => u = 8\)
\(r=3 => u = 12\) wrong because u is unit number so must be smaller or equal to 9

hence we have 2 solutions


Haven't you reversed the old/new when applying the ratio?
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Re: How many such positive integers exist so that if the unit digit of the [#permalink]

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New post 10 Aug 2017, 23:45
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Vardan95 wrote:
How many such positive integers exist so that if the unit digit of the original integer is removed, the ratio of the new number to the original one becomes \(\frac{1}{14}\) ?

(A) 0
(B) 1
(C) 2
(D) 3
(E) 4


The given ratio is 1/14. This is a solution. If the units digit of 14 is removed, we get 1. Let's look at the multiples of this ratio.
2/28 - Solution
3/42
4/56
5/70
6/84
7/98
8/112
...

Note the pattern. In the denominator, if you remove the units digit, the gap between what you get and the numerator is increasing. So you will not have another solution.

Answer (C)
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Re: How many such positive integers exist so that if the unit digit of the   [#permalink] 10 Aug 2017, 23:45
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