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A three digit number A with hundreds digit a, tens digit b and units

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failatmath
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A three digit number A with hundreds digit a, tens digit b and units [#permalink] New post   27 May 2018, 05:31
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A three digit number A with hundreds digit a, tens digit b and units digit c is written in reverse order to form another three digit number B. If B > A , is B + A > 1200?

(1) B - A is divisible by 8
(2) A is a multiple of 5
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chetan2u
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Re: A three digit number A with hundreds digit a, tens digit b and units [#permalink] New post   06 Jun 2018, 22:13
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GMATSkilled wrote:
A three digit number A with hundreds digit a, tens digit b and units digit c is written in the reverse order to form another three digit number B. If B > A, is B + A > 1200?

A. B - A is divisible by 8

B. A is a multiple of 5



A is 100a+10b+c
B is 100c+10b+a

Let's see the statements

1) B-A is div by 8
So (100c+10b+a)-(100a+10b+c)=99(c-a) is div by 8...
Since C and a are digits, ways the equation is satisfied
a) c is 9 and a is 1 , number becomes 9b1 and 1b9
Even you take b as max possible 9, sum becomes 991+199=1190, <1200
b) c is 8 and a is 0, but then A becomes two digit number so not possible..

So number can be only 9b1 and 1b9, and answer will be NO irrespective of value of B

Sufficient

2) A is a multiple of 5..
So C can be either 5 or 0 but 0 is not possible as the other number B becomes two digit number.
Hence ab5 and 5ba where 5ba>ab5 but the larger number has to be > 1200/2 or 600.. hence Ans is NO
Sufficient

D
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Re: A three digit number A with hundreds digit a, tens digit b and units [#permalink] New post   27 May 2018, 06:50
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Statement I alone is not sufficient to answer the question, since it gives us multiple possibilities.
Let's take statement II:
Since A is a multiple of 5, it should end with 5 or 0, since the reverse order number should also be a three digit number, the number cannot end with 0. Hence, A is of the form ab5. Then B should be 5ba, then B+A=5ba + ab5. Since B>A, a can take values (5,4,3,2,1), in any case A+B will always be <1200. Hence statement II alone is sufficient to answer.
IMO, option B.


Thanks and regards,
Sharan Salem
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xalemsharan
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Re: A three digit number A with hundreds digit a, tens digit b and units [#permalink] New post   27 May 2018, 07:11
Sorry mate, i missed some information.
From statement I, we get:
B-A = (100c+10b+a) - (100a+10b+c) = 99c-99a. This should be a multiple of 8, this is possible only when c is 9 and a is 1, then in such case A+B is always <1200.
Hence, both the statements individually are sufficient to answer.
Imo, option D.

Thanks and regards,
Sharan Salem
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Shobhit7
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A three digit number A with hundreds digit a, tens digit b and units [#permalink] New post   10 Jun 2018, 12:50
Hi


Is it possible for digit c to have different values in different Statements.

As per St.1, c=9
As per St.2, c=5
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Re: A three digit number A with hundreds digit a, tens digit b and units [#permalink] New post   13 Mar 2021, 05:53
chetan2u wrote:
GMATSkilled wrote:
A three digit number A with hundreds digit a, tens digit b and units digit c is written in the reverse order to form another three digit number B. If B > A, is B + A > 1200?

A. B - A is divisible by 8

B. A is a multiple of 5



A is 100a+10b+c
B is 100c+10b+a

Let's see the statements

1) B-A is div by 8
So (100c+10b+a)-(100a+10b+c)=99(c-a) is div by 8...
Since C and a are digits, ways the equation is satisfied
a) c is 9 and a is 1 , number becomes 9b1 and 1b9
Even you take b as max possible 9, sum becomes 991+199=1190, <1200
b) c is 8 and a is 0, but then A becomes two digit number so not possible..

So number can be only 9b1 and 1b9, and answer will be NO irrespective of value of B

Sufficient

2) A is a multiple of 5..
So C can be either 5 or 0 but 0 is not possible as the other number B becomes two digit number.
Hence ab5 and 5ba where 5ba>ab5 but the larger number has to be > 1200/2 or 600.. hence Ans is NO
Sufficient

D


Hi, why c is either 5 or 0?
Why c cannot have value larger than 5?
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Re: A three digit number A with hundreds digit a, tens digit b and units [#permalink] New post   16 Mar 2021, 21:46
I don't understand why the only option for statement 1 is 9 and 1

It could also be:-

1,9
8,0
0,8

All of them are divisible by 8. Statement 1 should not be sufficient.
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Re: A three digit number A with hundreds digit a, tens digit b and units [#permalink] New post   31 Mar 2021, 08:59
Bunuel can you please help with this Q.
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Re: A three digit number A with hundreds digit a, tens digit b and units [#permalink] New post   12 Feb 2022, 22:12
failatmath wrote:
A three digit number A with hundreds digit a, tens digit b and units digit c is written in reverse order to form another three digit number B. If B > A , is B + A > 1200?

(1) B - A is divisible by 8
(2) A is a multiple of 5


Let me help..
Let the hundreds digit be a ,10's digit be b and unit's digit be c.
So A becomes 100*a + 10*b +c .
B is in the reverse order. SO B= c b a = 100*c +10*b + a

B- A = 100*(c-a) + (a-c ) = 99*(c-a)

Statement 1 :-
B-A is divisible by 8.So c and a can be 9 , 1 ( 9 -1 = 8 ) or 8 , 0 (8-0=8)

But 8 and 0 are not possible as then the B will not remain a three digit number.

hence 9 ,1 are possible.
Hence A = 9b1
To maximise A , lets take b = 9 as well..So A= 991
B is in reverse order. B = 199.
So the max. (B+A) = 991 +199 = 1190
So B + A cant be greater than 1200.

Statement 2 : Since A is a multiple of 5 so A = ab5 . A cant be A cant be ab0 as then the B will not remain a two digit number.
SO A = ab5
B = 5ba

Now a has to be lesser than c as B >A.
So the maximum value a can take is 4.
So A = 4b5 = 495
B=5b4 = 594

So the max. B+A = 495+594 = 1089
So B + A cant be greater than 1200.

Hence Each statement is sufficient . Answer is option D.
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Re: A three digit number A with hundreds digit a, tens digit b and units [#permalink] New post   04 Nov 2023, 03:39
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Re: A three digit number A with hundreds digit a, tens digit b and units [#permalink]
04 Nov 2023, 03:39
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