There is a 4 digit card number,
4th digit is double of 1st digit,
2nd n 3rd digits are same,
And last digits together are double of first two digits
Can you guess the car number?
Check your answer:-
Click here to see Explanation
lets assume first digit is x, then the 4th digit will be 2x.
also assume 2nd and 3rd digits are y.
then from => last two digits are double of first two digits, we can say
y(2x) = 2*(xy)
also assume 2nd and 3rd digits are y.
then from => last two digits are double of first two digits, we can say
y(2x) = 2*(xy)
We can rewrite is as 10y + 2x = 20x + 2 y 8y = 18x 4y = 9x; => y/x = 9/4 as x and y are integers y => 9, x => 4
thus the car number is 4998
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Can 0000 be also an answer ? I think it holds good all conditions 🙂
how can you write last two digits with “y(2x)”?
y is 3rd letter and 2x is 4th one… and as given in puzzle last 2 digit means y(2x)
how the 10c+d=2(10a+b) derrvied
The puzzle asks for the “car” number and proceeds to lay out clues to come up with a 4-digit “card” number. No clues about a car, but a card number result could be 1002. 1002 works as well as 4998, so does 2004, 3006, 4008, and so on. Puzzle’s got too many typos.
I think when they say (last two digits together, they mean concatenation, not sum. i.e. 49*2 = 98, not 2*(2+0) = (0+4)
Thanks.
Let t number be ABCD
Now d= 2a
C=b
10c + d = 2(10a + b)