Question

A three-digit number has digits in strictly descending order and is 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?

• A. 5
• B. 6
• C. 7
• D. 8
• E. None of the above

Hint:

When both numbers must be divisible by 10, force the last digit to be $0$ in both. Then only the other two digits matter. Always reduce differences into modulus conditions to count valid pairs.

Answer 1

The Correct Option is B

Step 1: Structure of the number

Since the number is a 3-digit number divisible by 10, it must end with \(0\). Let the number be: \[ N = 100a + 10b + 0, \quad \text{with } a > b > 0. \] So the digits are in strictly descending order: \(a > b > 0\).

Step 2: Possible rearrangement

The rearranged number must also be divisible by 10 \(\Rightarrow\) it must also end in \(0\). Hence the new number is either: \[ N' = 100b + 10a \quad \text{or} \quad N' = 100a + 10\cdot 0. \] But the second choice equals the original, so not valid. Therefore, the valid rearrangement is: \[ N' = 100b + 10a. \]

Step 3: Difference condition

We need: \[ N - N' \equiv 0 \pmod{40}. \] That is: \[ (100a + 10b) - (100b + 10a) = 90a - 90b = 90(a-b). \] So, \[ 90(a-b) \equiv 0 \pmod{40}. \]

Step 4: Simplify divisibility

\[ 90(a-b) \equiv 0 \pmod{40} \quad \Rightarrow \quad (a-b) \equiv 0 \pmod{4}. \]

Step 5: Count solutions

Since \(a > b > 0\), digits \(a,b\) must be from 9 down to 1. Possible pairs \((a,b)\) with \(a-b = 4, 8\):

• \(a-b=4:\ (5,1),(6,2),(7,3),(8,4),(9,5)\) → 5 cases.
• \(a-b=8:\ (9,1)\) → 1 case.

Total = 6 cases.

Final Answer

\[ \boxed{6} \]