Question

Let X be a four digit number with exactly three consecutive digits being the same, and X is a multiple of 9. How many such X's are possible?

• A. 12
• B. 16
• C. 19
• D. None of the above.

Hint:

Split the count into the two shapes AAAB and ABBB, and use the fact that a number is a multiple of 9 only when its digit sum is a multiple of 9.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
A 4-digit number has a digit sum between $1$ (for $1000$) and $36$ (for $9999$), so the only sums that are multiples of $9$ worth checking are $9$, $18$ and $27$. The sum $36$ needs all four digits to be $9$, which has all four digits equal, not exactly three, so it is not counted.

Step 2: Key Formula or Approach:
For the pattern $aaab$ (repeated digit $a$ from $1$-$9$ in the first three places, different digit $b$ from $0$-$9$ last), the digit sum is $3a+b$. For the pattern $cbbb$ (different digit $c$ from $1$-$9$ first, repeated digit $a$ from $0$-$9$ in the last three places), the digit sum is $c+3a$. In both cases we look for the value that has to be added to $3a$ to reach the nearest multiple of $9$.

Step 3: Detailed Explanation:
For $aaab$: taking $3a$ for $a=1$ to $9$ gives $3,6,9,12,15,18,21,24,27$. The amount needed to reach the next multiple of $9$ (this is the required $b$) is $6,3,0,6,3,0,6,3,0$ in order. When this gap is $0$, both $b=0$ and $b=9$ work, unless one of them equals $a$; this removes one option only at $a=9$, since $b=9$ would then repeat $a$. Adding up the valid $b$ values for each $a$ gives $1+1+2+1+1+2+1+1+1=11$.
For $cbbb$: taking $3a$ for $a=0$ to $9$ gives $0,3,6,9,12,15,18,21,24,27$, and the same gap logic gives the required $c$, except $c$ must be from $1$ to $9$ and never $0$. This removes the case $a=9$, where the only fitting $c$ would be $9$ itself, equal to $a$, and so not allowed. This leaves $9$ working numbers.

Step 4: Final Answer:
Total $= 11+9 = 20$, which is not one of $12$, $16$, $19$ or $21$, so the answer is option E, "none of the above".