A Number Theory Question: “If abcd x 4 = dcba what are the values of a b c and d?”
Answer
Let us reason out completely and not just “guess” at an answer.
We use base 10.
abcd = a*10^3 + b*10^2 + c*10^1 + d
and
dcba = d*10^3 + c*10^2 + b*10^1 + a
abcd x 4 = dcba implies,
4 x [ a*10^3 + b*10^2 + c*10^1 + d ]= d*10^3 + c*10^2 + b*10^1 + a
Now, 4a < 10. Why? if 4a > 10 then the LHS > 10^4 [ being a 5 digit number ] while the RHS will be <10^4 [ being a 4 digit number ] contradicting the equality above.
The only multiples of 4 that are less than 10 are 4 and 8. Hence 4a = 4,8 giving, a = 1 or 2.
Note also that 4d must ‘end’ in a ( by considering the last digits ). Also 4a must end in d ( by considering the the first digits ) in the above equation.
This means that if a =2, then d =8 [ obtained by equating the coefficient of 10^3 in the equation above ], but 4d = 32 ends in ‘a’ or ‘2’ as well and thus this could be a solution. On the other hand, if a =1 then 4a = 4 and thus d = 4 [ obtained by equating the coefficient of 10^3 in the equation above ]. But
4d = 16 and does NOT end in ‘a’ or ‘1’ as described in the earlier paragraph. Hence a =2 and d =8.
Our equation now becomes,
4 x [ 2*10^3 + b*10^2 + c*10^1 + 8 ]= 8*10^3 + c*10^2 + b*10^1 + 2
which reduces to
4 x [ b*10^2 + c*10^1 + 8 ]= c*10^2 + b*10^1 + 2
400b + 40c + 32 = 100c + 10b +2
390b + 30 = 60c
13b + 1 = 2c …(1)
Consider (1). Remember that b and c are all integers between 0 to 9 because they are digits of an integer.
2c < 2 (10) = 20.
13b + 1 < 20 using (1) and the above.
b < 19/13 = 1.46…
meaning b = 0 or b =1 since b is an integer less than 1.47.
If b = 0, equation (2) gives us that 2c = 1 or c =1/2 ( a contradiction since c is not integer ).
Hence b = 1. equation (2) gives us 14 = 2c implying that c=7.
To recap, we have found the values of a,b,c,d as follows:
(a,b,c,d ) = ( 2, 1, 7, 8 )
Indeed to check :
2178 x 4 = 8712
STAY CONNECTED