A three-digit number has digits in strictly descending order and divisible by 10. By changing the places of the digits a new three-digit number is constructed in such a way that the new number is divisible by 10. The difference between the original number and the new number is divisible by 40. How many numbers will satisfy all these conditions?
Explanation:
Units digit of the number must be 0.
Let 100x + 10y is the number such that x > y.
New number obtained by changing the digits is also divisible by 10.
So, only x and y are to be interchanged
∴ New number is of the form = 100y + 10x
Difference = 90x – 90y = 90(x – y)
For the difference to be divisible by 4,
(x – y) has to be divisible by 4.
(x – y) = 4 or 8
So, y = 1 to 5
For y = 1 to 5, x = (1+4) to (5+4) i.e., 5 to 9
One more possibility for y = 1 is x = 9.
Thus, in all 6 numbers satisfy the given conditions.
Hence, option (b).
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