The “hash” of a three-digit integer with three distinct integers is de

Question

The “hash” of a three-digit integer with three distinct integers is defined as the result of interchanging its units and hundreds digits. The absolute value of the difference between a three-digit integer and its hash must be divisible by

A. 9
B. 7
C. 5
D. 4
E. 2

A. 9
B. 7
C. 5
D. 4
E. 2

Hoozan · Answer

Let the three digit number = ABC

Hence the "hash" will = CBA

Now a three digit number can be written as 100A+10B+C

Hence the difference will be

100A+10B+C - (100C+10B+A)

Upon simplification we can write the above as

100(A-C) + (C-A)

100A-100C+C-A = 99(A-B)

Cleary 9 is divisible

Alternate method (not recommended)

Plug in values for the three digit number.

Eg-
302-203= 99

99/9 gives an integer value

(A)

Posted from my mobile device

KanishkM · Answer

IMO A

Let 3 digit number be 100x + 10y + z

hash = 100z +10y + x

difference between a three-digit integer and its hash = 100x + 10 y + z - 100z - 10y -x
= 99x -99z 
= 99 (x-z)

Divisible by 9

saurabh9gupta · Answer

My soln

let the number be abc

also expressed as = 100a + 10b + c

given the hash (#)
100c + 10b + a

as per q
100c + 10b + a - 100a - 10b - c

99c - 99a

99(c-a)

only A works

Kinshook · Answer

Given:The “hash” of a three-digit integer with three distinct integers is defined as the result of interchanging its units and hundreds digits. 
Asked: The absolute value of the difference between a three-digit integer and its hash must be divisible by

Let the 3 digit number be xyz where x is hundredth digit, y is tenth digit and z is unit digit.
It hash will be zyx.

The absolute value of the difference between a three-digit integer and its hash = |(100x + 10y +z) - (100z +10y+x)| = |99x -99z|
Which is divisible by 99 = 9×11

IMO A

Posted from my mobile device

bumpbot · Answer

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The “hash” of a three-digit integer with three distinct integers is de

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