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Bunuel
Take a four-digit number and reverse its digits. The difference between the original number and the new number must be divisible by which of the following?

A) 2
B) 4
C) 5
D) 6
E) 9

­

For any number with even number of digits, the difference between the number and its reverse is divisible by 9.
Example:
52-25=27, divisible by 9
4321-1234=3087, divisible by 9

For any number with odd (3 or above) number of digits, the difference between the number and its reverse is divisible by 99
Example:
321-123=198, divisible by 99

Reason: if the 4 digit number is 1000a+100b+10c+d
Reverse: 1000d+100c+10b+a
Subtract: 999a+90b-9c-999d, divisible by 9

Ans E
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Bunuel
Take a four-digit number and reverse its digits. The difference between the original number and the new number must be divisible by which of the following?

A) 2
B) 4
C) 5
D) 6
E) 9

­
Usual method to solve problems with reverse digits is:

TWO DIGIT NUMBERS

ab = 10a+ b
ba = 10b +a

subtracting it, we get

ab - ba = (10a+ b ) - (10b +a ) = 9a -9b = 9 (a-b).

Adding it, we get

ab + ba = (10a+ b ) + (10b +a ) = 11a + 11b = 11 (a+b).


Similarly, for three digit number it’s

abc = 100 a + 10 b + c
cba = 100 c + 10 b+ a

subtracting it, we get

abc - cba = (100 a + 10 b + c) - (100 c + 10 b + a) = 99a -99c = 99 (a-c)

Adding it, we get

abc + cba = (100 a + 10 b + c) + (100 c + 10 b + a) = 101 (a+c) + 20 b.

Now let’s come to the QUESTION , FOUR DIGIT NUMBER

abcd = 1000a+100b+10c+d

dcba = 1000d+100c+10b+a

subtracting it we get,

abcd - dcba = (1000a+100b+10c+d) - ( 1000d+100c+10b+a)

= 999 a + 90b - 90 c - 999 d

= 999 (a-d) + 90 (b-c)

Both, 999 and 90 are divisible by 9. Hence option E

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Let, original number = 1000a+100b+10c+d
Reverse number = 1000d+100c+10b+a

Original number - Reverse number = (1000a+100b+10c+d) - (1000d+100c+10b+a)
= 1000a -1a +100b -10b +10c -100c +1d -1000 d
= 999a + 90b - 90c - 999d
= 9(111a+10b-10c-111d)
The difference is always divisible by 9.

Answer: E
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