Problem: The four-digit number $25 A B$ is divisible by nine, with $A$ the tens digit and $B$ the units digit. How many different such four-digit numbers could $25 A B$ represent?

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∻Solution by Steps∻ ‖ step 1 ⋮ To find the number of 4-digit numbers of the form $25AB$ that are divisible by 9, we need to use the divisibility rule for 9. ‖ ‖ step 2 ⋮ The divisibility rule for 9 states that a number is divisible by 9 if the sum of its digits is divisible by 9. ‖ ‖ step 3 ⋮ The sum of the digits of the number $25AB$ is $2 + 5 + A + B$. ‖ ‖ step 4 ⋮ For $25AB$ to be divisible by 9, the sum $2 + 5 + A + B$ must be divisible by 9. ‖ ‖ step 5 ⋮ Simplifying the sum gives us $7 + A + B$, which must be a multiple of 9. ‖ ‖ step 6 ⋮ The smallest value for $A + B$ that is a multiple of 9 and greater than 7 is 9 (when $A = 0$ and $B = 2$), and the largest value for $A + B$ that is less than 18 (since $A$ and $B$ are digits) is 17 (when $A = 8$ and $B = 9$). ‖ ‖ step 7 ⋮ We need to find all pairs of $(A, B)$ such that $7 + A + B$ is 9, 18, or any multiple of 9 in between. ‖ ‖ step 8 ⋮ The possible sums for $7 + A + B$ are 9, 18, and no more since the next multiple, 27, is not possible with single-digit $A$ and $B$. ‖ ‖ step 9 ⋮ For the sum of 9, there is only 1 possibility: $A = 0$ and $B = 2$. ‖ ‖ step 10 ⋮ For the sum of 18, we have the following possibilities for $(A, B)$: $(1, 0), (2, 9), (3, 8), (4, 7), (5, 6), (6, 5), (7, 4), (8, 3), (9, 2)$. ‖ ‖ step 11 ⋮ Counting all the possibilities from steps 9 and 10, we have a total of 1 + 9 = 10 possible 4-digit numbers. ‖ ∻Answer∻ ⚹ There are 10 different four-digit numbers of the form $25AB$ that are divisible by 9. ⚹ ∻Key Concept∻ ⚹Divisibility by 9⚹ ∻Explanation∻ ⚹A number is divisible by 9 if the sum of its digits is divisible by 9. We applied this rule to find all possible values of $A$ and $B$ such that the number $25AB$ is divisible by 9.⚹◊What is the rule for determining if a number is divisible by nine in terms of the sum of its digits?⍭ Generate me a similar question◊

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