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2013 AIME I Problems/Problem 2

Problem

Find the number of five-digit positive integers, $n$, that satisfy the following conditions:

    (a) the number $n$ is divisible by $5,$
    (b) the first and last digits of $n$ are equal, and
    (c) the sum of the digits of $n$ is divisible by $5.$


Solution 1

The number takes a form of $\overline{5xyz5}$, in which $5|(x+y+z)$. Let $x$ and $y$ be arbitrary digits. For each pair of $x,y$, there are exactly two values of $z$ that satisfy the condition of $5|(x+y+z)$. Therefore, the answer is $10\times10\times2=\boxed{200}$

Solution 2 (casework)

We know the number will take the form $\overline{5xyz5}$ because of the first two conditions. The third condition means that $5|(x+y+z)$, where $x,y,z$ are nonnegative integers less than $10$. Let's split the problem into cases, where each case represents a possible sum of $x,y,z$.

1. If $x+y+z=0$, we only have $x=0,y=0,z=0$, so only $1$.

2. If $x+y+z=5$, we use stars & bars. We have $5$ stars and $3-1=2$ bars, so this case has $7\choose{2}$$= 21$.

3. If $x+y+z=10$, we use similar logic. We have $10$ stars and $2$ bars, so $12\choose{2}$$= 66$. However, $x,y,z$ must be less than $10$. Three of our order pairs have $10: (10,0,0), (0,10,0), (0,0,10)$. Therefore, this case has $66-3=63$.

4. If $x+y+z=15$, it gets more complicated. Using the same system as used previously would be too complicated. But remember that this case is equivalentto if we wanted to choose $x+y+z=12$. We know this because you can take any ordered pair satisfying $x+y+z=15,$ subtract each variable from $9$, and get an ordered pair satisfying $x+y+z=12$. For example, take $(5,6,4)$, which satisfies $x+y+z=15.$ Its corresponding ordered pair would be $(4,3,5)$, which satisfies $x+y+z=12$. Let's proceed to calculating. Applying stars and bars, we get $14\choose{2}$$= 91$, but we have to subtract the subcases including $10,11,$ or $12$ because $x,y,z$ must all be one-digit integers. There are $3$ cases with a $12$, $6$ cases with an $11$, and $3+6$ cases with a $10$. So this case has $91-18=73$.

5. If $x+y+z=20$, we can just calculate the ways to get $x+y+z=7$. Applying stars & bars, we get $9\choose{2}$$= 36$.

6. If $x+y+z=25$, we can just calculate the ways to get $x+y+z=2.$ Applying stars & bars, we get $4\choose{2}$$= 6$.

Therefore, our answer is $1+21+63+73+36+6 = \boxed{200}$.

Note: For case 4, the subcases that must be excluded are $(12,0,0), (11,1,0), (10,2,0), (10,1,1)$ and each of their respective permutations. Those $4$ ordered pairs have $3,6,6,3$ permutations respectively, which is why $18$ ordered pairs must be subtracted from $91$.

~lprado

Video Solution

https://www.youtube.com/watch?v=kz3ZX4PT-_0 ~Shreyas S

See also

2013 AIME I (Problems • Answer Key • Resources)
Preceded by
Problem 1 		Followed by
Problem 3
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions

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