Etim changed his pin code from 0000 to another code which he had forgotten.Luckily,he remember that if you multiply the pin code by 4 the answer is reverse the pin code.Can you help Etim to find the code
Let the number be abcd.
The original code was changed from 0000, so abcd cannot also be 0000.
If a = 0, then b (i.e. 4a) must also be zero.
But if a = 0 and b = 0, then c = 4b + carry and b = 4c
i.e. b = 16b + 4*carry, carry = 0,1,2,3
which does not have a positive integer as a solution,
except b = 0 and carry = 0, => c = 0.
Hence neither a nor b nor c nor d can be zero.
Since a =/= 0 and 4a < 10, then a = 1 or 2.
The number then may be: {1bcd}
Multiply by 4 gives {4 + carry, 4b + carry – 10n, 4c + carry – 10n, 4d – 10n}, n,carry = 0,1,2,3
Reverse this giving, {4d – 10n, 4c + carry – 10n, 4b + carry – 10n, 4 + carry}
Compare with original. {1,b,c,d} => 4d – 10n = 1, but no solution. (lhs = even, rhs = odd)
Therefore a = 1 not a solution.
Therefore a = 2.
The number then is: {2bcd}
Multiply by 4 gives {8 + carry, 4b + carry – 10n, 4c + carry – 10n, 4d – 10n}, n,carry = 0,1,2,3
Reverse this giving, {4d – 10n, 4c + carry – 10n, 4b + carry – 10n, 8 + carry}
Compare with original. {2,b,c,d} => 4d – 10n = 2
4d – 10n = 2 has solutions (n,d) = (1,3) and (n,d) = (3,8), i.e. d = 3, or d = 8
Let d = 3
The number then may be: {2bc3}
Multiply by 4 gives {8 + carry, 4b + carry – 10n, 4c + 1 – 10n, 2}, n,carry = 0,1,2,3
Reverse this giving, {2, 4c + 1 – 10n, 4b + carry – 10n, 8 + carry}
Compare with original. {2,b,c,3} => 8 + carry = 3, therefore d = 3 not a solution.
Let d = 8
The number then may be: {2bc8}
Multiply by 4 gives {8 + carry, 4b + carry – 10n, 4c + 3 – 10n, 2}, n,carry = 0,1,2,3
Reverse this giving, {2, 4c + 3 – 10n, 4b + carry – 10n, 8 + carry}
Compare with original. {2,b,c,8} => 8 + carry = 8, therefore d = 8 is a solution for carry = 0.
The number then is: {2bc8}
Since the previous carry = 0, (for 8 + carry)
Then 4b + carry < 10
Assume b = 0,1,2
Using b = 4c + 3 – 10n (see 6th and 7th line above)
At b = 0, 4c + 3 = 10n, but no solution since lhs = odd and rhs = even
At b = 2, 4c + 1 = 10n, but no solution since lhs = odd and rhs = even
At b = 1, 4c + 2 = 10n, Solutions are: (n,c) = (1,2), (3,7)
Using (n,c) = (1,2), then b = 1, c = 2, and the number is {2128}
4 * {2128} = {8512}, which is not the reverse, therefore c = 2 not a solution.
Using (n,c) = (3,7), then b = 1, c = 7, and the number is {2178}
4 * {2178} = {8712}, which IS the reverse, therefore c = 7 is a solution.
Etim’s changed Pin Code is: 2178