Question

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?

Accepted Answer

B. Units digit must be $0$ for both numbers. Let original number be $ab0$ ($a > b > 0$). New number is $ba0$. Difference $= (100a + 10b) - (100b + 10a) = 90(a - b)$. For $90(a - b)$ to be divisible by $40$, $a - b$ must be a multiple of $4$. Possible $(a, b)$ pairs: - $a - b = 4$: $(5,1), (6,2), (7,3), (8,4), (9,5)$ - $a - b = 8$: $(9,1)$ Total valid numbers $= 5 + 1 = 6$ numbers.

3-Digit Number Divisibility Conditions

Explanation

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