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: Set up the two possible patterns.
Call the digits of the four digit number \(d_1 d_2 d_3 d_4\), with \(d_1 \neq 0\). "Exactly three consecutive digits the same" can happen in only two ways: either the first three digits match and the last is different (\(d_1=d_2=d_3=a\), \(d_4=b\), \(b \neq a\)), or the last three digits match and the first is different (\(d_1=c\), \(d_2=d_3=d_4=a\), \(c \neq a\)). A number cannot satisfy both patterns at once unless all four digits are equal, and that case is excluded since we need exactly three matching digits, not four.
A number is a multiple of \(9\) exactly when the sum of its digits is a multiple of \(9\).

Step 2: Count the pattern \(aaab\) (first three digits same).
Here \(a\) runs from \(1\) to \(9\) (it is the leading digit) and \(b\) runs from \(0\) to \(9\) with \(b \neq a\). We need \(3a+b\) to be a multiple of \(9\).
Checking each \(a\) from \(1\) to \(9\): \(a=1,b=6\) (1116); \(a=2,b=3\) (2223); \(a=3,b=0\) and \(b=9\) (3330, 3339); \(a=4,b=6\) (4446); \(a=5,b=3\) (5553); \(a=6,b=0\) and \(b=9\) (6660, 6669); \(a=7,b=6\) (7776); \(a=8,b=3\) (8883); \(a=9,b=0\) only, since \(b=9\) would repeat \(a\) (9990).
That is \(1+1+2+1+1+2+1+1+1 = 11\) numbers.

Step 3: Count the pattern \(cbbb\) (last three digits same).
Here the repeated digit \(a\) runs from \(0\) to \(9\), and the leading digit \(c\) runs from \(1\) to \(9\) with \(c \neq a\). We need \(c+3a\) to be a multiple of \(9\).
Checking each \(a\) from \(0\) to \(9\): \(a=0,c=9\) (9000); \(a=1,c=6\) (6111); \(a=2,c=3\) (3222); \(a=3,c=9\) (9333); \(a=4,c=6\) (6444); \(a=5,c=3\) (3555); \(a=6,c=9\) (9666); \(a=7,c=6\) (6777); \(a=8,c=3\) (3888); and for \(a=9\) the only fitting \(c\) is \(9\), which is not allowed since \(c \neq a\), so \(a=9\) gives nothing.
That is \(9\) numbers.

Step 4: Add the two counts and compare with the choices.
Total numbers \( = 11+9 = 20\).
None of the offered counts, \(12\), \(16\), \(19\) or \(21\), equals \(20\), so the correct choice is "none of the above". \[ \boxed{20 \Rightarrow \text{None of the above}} \]