The digits of a three-digit number A are written in the reverse order

Question

The digits of a three-digit number A are written in the 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 < A < 299

(B) 106 < A < 305

(C) 112 < A < 311

(D) 118 < A < 317

(E) 122 < A < 337

karthik1488 · Accepted Answer

Let's say A = 100k + 10l + m where k,l,m are single digit positive numbers B = 100m + 10l + k B > A (Given) => B-A = 100(m-k) + (k-m) = 99(m-k) For (B-A) to be divisible by 7, we need (m-k) to be divisible by 7 as 99 is not divisible by 7. Since both m and k are single digit positive numbers, this means that (m-k) = 7 or m = k + 7 So we can have A = 1x8 or 2x9 where 'x' is any number from 0 to 9 Min A = 108 Max A = 299 Option B : 107 < A < 305 is the one that fits.

GMATGuruNY · Answer

Let integer B = 100H + 10T + U, where H, T and U represent the 3 digits.
Since the digits are reversed in A, we get:
A = 100U + 10T + U
Thus:
B-A = (100H+10T+U) - (100U+10T+H) = 99H - 99U = 99(H-U)

Since B-A must be divisible by 7, H-U=7.
Thus, there are two cases:
H=9 and U=2, with the result that B=9_2 and A=2_9
H=8 and U=1, with the result that B=8_1 and A=1_8
Implication:
The smallest option for A = 108
The greatest option for A = 299
Only the range in B includes both 108 and 299.

B

Kinshook · Answer

Given: The digits of a three-digit number A are written in the reverse order to form another three-digit number B. Asked: If B>A and B-A is perfectly divisible by 7, then which of the following is necessarily true? Let 3 -digit number A be xyz where x = 100^{th} digit; y = 10^{th} digit; z = unit digit Then 3 -digit number B after reversing digits be zyx where z = 100^{th} digit; y = 10^{th} digit; x = unit digit A = 100 x + 10 y + z B = 100z + 10y + x B>A => 100z + 10y + x>100 x + 10 y + z => 99z >99x => z>x. (1) B-A = 99 (z-x) is divisible by 7 => z-x=7 (2) since 99 = 3^2*11 does not contain 7 z-x =7 If x = 1 ; z =8 If x = 2; z =9 A = 1y8 or 2y9 min (A) = 108 max (A) = 299 106 < A < 305 Includes both 108 and 299 IMO B

gracie · Answer

let 100x and y=B's 100s digit and 1s digit respectively
let 100y and x=A's 100s digit and 1s digit respectively
B-A=99(x-y)
x-y=7 (because 99 is not a multiple of 7)
two options are:
x=9; y=2 and x=8; y=1
least option for A=108
greatest option for A=292
106 < A < 305
B

amarsinha · Answer

A = xyz ; x - 1st digit, y - 2nd digit, z - 3rd digit. 
So B = zyx

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

So, B-A = 100 (z-x) + x -z
 = 99 (z-x)

If B-A has to be multiple of 7, z-x has to be multiple of 7 (as 99 is not multiple of 7)
For z-x to be multiple of 7, z-x=0, z-x = 7 z-x cannot be 14 or more than that as z and x are single digit number.

z-x cannot be 0, as if z-x =0 z=x so A will be equal to B. But in the question it is given B>A. so z cannot be x.

So z-x = 7
In this case z can be 8,9 and x can be 1,2
So, A = 1y8, 2y9

the minimum and maximum value of A will be when y=0 and y =9 respectively
which will make A = 108 and 299 for minimum and maximum

So, 107 < A < 300
 The only option which fits and takes both 108 and 299 within the range is B.

bumpbot · Answer

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The digits of a three-digit number A are written in the reverse order

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