Question

The difference between 4 digit number and the number obtained by reversing its digits is always divisible by

Answer 1

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$\underline{\underline{\huge{\textbf{Solution:}}}}$

$\underline{\large{\textsf{Step-1:}}}$
Assume the digits of 4-digit Number

Let Unit's place digit be d
Ten's place Digit be c
Hundred's Place digit be b
and Thousand's Place digit be a

$\underline{\large{\textsf{Step-2:}}}$
Express the 4-digit Number in terms of assumed digits.

A Number is equal to sum of product of weight of the digit at each place and Face value at that place.

$4-digit\: No. = 1000a+100b+10c+1d$------(1)

$\underline{\large{\textsf{Step-3:}}}$
Express the 4-digit Number when the digits are reversed.

When a 4-digit No. is reversed,
The digit at Thousand's place and Unit's place
are Interchanged.
Also, the digit at Hundred's place and Ten's place are interchanged.

$\therefore$
For Reversed 4-digit No.,

Unit's place digit = d
Ten's place Digit = c
Hundred's Place digit = b
and Thousand's Place digit = a

Thus,
$Reversed\:4-digit\: No. = 1000d+100c+10b+1a$-----(2)

$\underline{\large{\textsf{Step-4:}}}$.
Find the difference between the 4-digit No. and Reversed 4-digit No.

Do (1) - (2),

$(1000a+100b+10c+d)-(a+10b+100c+1000d)$

$\implies (1000-1)a+(100-10)b+(10-100)c+(1-1000)d$

$\implies (999)a+(90)b+(-90)c+(-999)d$

$\implies 999(a-d)+90(b-c)$

$\underline{\large{\textsf{Step-5:}}}$
Take a Number common from the difference obtained.

Find HCF(999, 90), which gives the common factor out of the two terms in the difference.


$999 = 3^{3} \times 37^{1}$

$90 = 2 \times 3^{2}\times 5^{1}$

$\therefore HCF = 3^{2} = 9$

$\underline{\large{\textsf{Step-5:}}}$
Express the Difference in terms of the common factor in the two terms of difference.

$\implies 9[111(a-d)+10(b-c)]$

Clearly,
The difference between 4-digit number and the number obtained by reversing its digits is always divisible by $\underline{\large{\mathcal{9}}}$

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