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ABC is a three-digit number such that ABC = 5(AB + BC + CA), where AB

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ABC is a three-digit number such that ABC = 5(AB + BC + CA), where AB  [#permalink]

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New post 25 Nov 2019, 01:50
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  95% (hard)

Question Stats:

36% (03:52) correct 64% (03:04) wrong based on 22 sessions

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ABC is a three-digit number such that ABC = 5(AB + BC + CA), where AB, BC and CA are all two-digit numbers. Find the total number of possible values for the number ABC.

(A) 3
(B) 4
(C) 12
(D) 13
(E) 18
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ABC is a three-digit number such that ABC = 5(AB + BC + CA), where AB  [#permalink]

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New post 26 Nov 2019, 06:09
CaptainLevi wrote:
ABC is a three-digit number such that ABC = 5(AB + BC + CA), where AB, BC and CA are all two-digit numbers. Find the total number of possible values for the number ABC.

(A) 3
(B) 4
(C) 12
(D) 13
(E) 18


ABC = 5(AB + BC + CA) tells us that ABC is a multiple of 5, so C can be 0 or 5.
But CA will be 2-digit number only when C=5.

Now, AB can be written as 10A+B..
\(5(AB + BC + CA)=5(10A+B+10B+C+10C+A)=5(11(A+B+C))=55(A+B+C)=ABC\)
Thus, ABC is a multiple of 11, and the units digit of A+B should be even, as then only 55(A+B+5) will give AB5..

For AB5 to be multiple of 11, 5+A-B should be multiple of 11...
So
=>5+A-B=0......A-B=-5 ..Discard as A+B is EVEN, which means A-B is also EVEN
or 5+A-B=11......A-B=6..This means least value is when A=7 and B=1 as B cannot be 0
Thus, check for values of A+B=8 onwards

(I) A+B=8
Let us solve
A-B=6 and A+B=8...Add both...2A=14...A=7 and B=1...Number is 715

(II) A+B=10
Let us solve
A-B=6 and A+B=10...Add both...2A=16...A=8 and B=2...Number is 825

(III) A+B=12
Let us solve
A-B=6 and A+B=12...Add both...2A=18...A=9 and B=3...Number is 935

ALL values above 12 can be discarded as it will give A as 2-digit.
for example
A+B=16
Let us solve
A-B=-5 and A+B=16...A and B will come out as fraction..Discard
A-B=6 and A+B=16...Add both...2A=22...A=11 and B=5..Discard

So 3 cases

A
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Re: ABC is a three-digit number such that ABC = 5(AB + BC + CA), where AB  [#permalink]

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New post 26 Nov 2019, 05:33
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ABC is a 3-digit no.
So it could be written as
ABC= (100A+10B+C)
Similarly 2-digit numbers
AB= (10A+B)
BC= (10B+C)
CA= (10C+A)
Given that:
100A+10B+C= 5*(10A+B+10B+C+10C+A)
=5*(11A+11B+11C)
=55A+55B+55C
Or
45A-45B=54C
A-B=6C/5
Now since A,B,C are unit, tens or hundreds digit and each of them happen to be first digit(which cannot be zero) of a two digit number ie AB, BC, CA.
So clearly A, B, C lie between 1 & 9
Now again:
A-B=6C/5 ( A-B will also be an integer, hence C needs to be a multiple of 5, Only possible value of C=5)
Now A-B=6
If
A=9 B=3
A=8 B=2
A=7 B=1
A=6 B=0 Not possible.
Therefore only 3 possible values of ABC are: ( 935, 825 & 715)

Answers: A

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Re: ABC is a three-digit number such that ABC = 5(AB + BC + CA), where AB   [#permalink] 26 Nov 2019, 05:33
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