The digits of a three-digit number A are written in reverse order to form another three-digit number B.If B>A

Question

The digits of a three-digit number A are written in reverse order to form another three-digit number B.If B>A and B-A is perfectly divisible by 7,then which of the following is necessarily true
(a) 100 (b) 106 (c) 112 (d) 118

Answer 1

(a)

Answer 2

(b) is the proper choice The smallest solution is A=108
B = 801 and the largest is A = 299 B =992. Therefore (b) is the only possibility. See below analysis.

let A = xyz a 3 digit number.
Then B = zyx

We can express A as:
A = 100x + 10y +z; and likewise:
B = 100z +10y +x; therefore:
B-A = 100(z-x) + x-z =
100(z-x) - (z--x) = 99(z-x)

Therefore, since 7 divides B-A, 7 must divide z-x, since 7 does not divide 99.
Since B>A, we cannot have z = x. (Also x cannot be zero) therefore x=1, or 2, and z = 8 or 9.

As far as solutions go:

We know B-A = 99(z-x) and z-x = 7
Therefore B-A = 99*7 = 693
The last digit of xyz must be 1 or 2 so the smallest solution could be: 108+693 = 801 This is a solution and so A> 106. This solution automatically generates other solutions as follows:
A =108 B=801
A =118 B = 811
A =128 B= 821
A = 138 B= 831
A= 148 B=841
A = 158 B = 851
A= 168 B = 861
A = 178 B 871
A =188 B =881
A = 198 B =981
A = 209 B= 902
A = 219 B =912
A = 229 B = 922
A = 239 B = 932
A= 249 B = 942
A = 259 B = 952
A = 269 B = 962
A = 279 B = 972
A = 289 B = 982
A = 299 B = 992

The largest soultion is A = 299 B= 992 so (a) is ruled out since (a) says A< 299. So the correct answer is (b)
106

Answer 3

1

Answer 4

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Answer 6

2

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Answer 8

(b)

number be of form 100x+10y+z,

so B-A is 99(z-x) and is divisible by 7, so z-x is divisible by 7,
so z,x can be (8,1), (9,2)

108, 118, ...299 fits in this group

Answer 9

It depends on many factors

Answer 10

(c) 112