FOR GENIUS PPL,answer this???

Question

If A is a 3 digit no and if 3 digits of are written in reverse we get B
IF B>A and B-A is perfecly divisible by 7,then which of foll. is true

1)100
2))106
3))112
4))118

Louise and volterwd bothof u r correct , ur method is absolutely correct but the correct answer is 2 , louise u got it but explain it lil more and volter u got the answer still y did u chose 1 the answer is 2 ,by the way explanation is very good

Answer 1

Let a = 100 x + 10 y + z where x y and z are digits
then b = 100 z + 10 y + x

since b > a then z > x

b - a = 100 (z - x) + (x - z)
= 100 (z - x) - (z-x)
= 99 ( z - x )
= 3 * 3 * 11 * (z - x)

But b-a is divisible by 7 which is prime thus z - x is divisible by 7 so z - x = 7

So x = 0, 1, 2 which gives z = 7, 8 , 9

Which means that a is either

a = 0y7
a= 1y8
a = 2y9

Now if you dont allow a leading 0 in a then the correct answer would be

1)) 100 < A < 299

EDIT I chose 1 cause i cant read haha... i make stupid mistakes like this all the time. Yes... clearly 299 is a solution and thus it couldnt be 1

ALSO

Doug you forgot the case with 0.

The restriction to not including that isnt explicitly in the question though which it clearly shoud be based on the question.

Note 7y0 - 0y7 = 700 - 7 = 99 * 7 which satisties our conditions

Answer 2

Multiple answers.

Let A=a2*100 +a1*10+a0 then B=a0*100 + a1*10 + a2

Since B>A a0 must be > a1

If 7 | (B-A) ==> 7 | 99*(a0 - a2) ==> (a0-a2) = 7

therefore, (a0 = 8, a2 = 1) or (a0 = 9, a2 = 2) will satisfy the requirements. (Note that a1 has no special requirements placed on it. It can be any integer 0 - 9)

If we are looking for the smallest such A, then choose
A = 108 check B = 801 801-108 = 693 = 7*99

But note

A = 118 ==> B = 811 ==> (811-118) = 693 = 7*9

also for 128, 138, etc. and if

A = 209 ==> B = 902 ==> (902-209) = 693 = 7*9

that works, as does A = 219, 229, etc.

So answer (2) is correct

Doug

Note to louise: (1) is *not* correct since it says A < 299 which excludes 299.


Doug

Answer 3

A = 100x + 10y + z
B = 100z + 10y + x

B - A = 99z - 99x = 99(z - x)

(B - A) can only be divisible by 7 if (z - x) = 0 or 7.
At most, B - A = 99(7) = 693.

B = 693 + A.
Since B is a 3-digit number, the greatest of which is 999, then
693 + A < 999
A < 306.

I like answer (2), then... 106 < A < 305.

[Edited to add] Now that I read the question again, it says which of the following are true?
Both (1) and (2) are true. "A" can take on the following values:
{108, 118, 128, ... 198, 209, 219, 229, ... ,299}

Answer 4

r u sure that's the complete question and that you haven't missed out some information in it?
If 3 digits of ??? are written in reverse, do u get B??
If you complete the question, perhaps I can try to answer it.

Answer 5

oops... i meant 1...