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Difference between revisions of "2017 AMC 10A Problems/Problem 25"

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Revision as of 17:39, 8 February 2017

Let the three-digit number be $ACB$:

If a number is divisible by 11, then the difference between the sums of alternating digits is a multiple of 11.

There are two cases: $A+B=C$ $A+B=C+11$

We now proceed to break down the cases.

$\textbf{Case 1}$: $A+B=C$. This has $18+45+33+21+9=126$ cases.


$\textbf{Part 1}$: $B=0$ $A=C$, this case results in 110, 220, 330...990. There are two ways to arrange the digits in each of those numbers. $2 \cdot 9 = 18$

$\textbf{Part 2}$: $B>0$ $B=1, A=C+1$, this case results in 121, 231,... 891. There are $6$ ways to arrange the digits in all of those number except the first, and 3 ways ways for the first. This leads to $45$ cases.

$\textbf{Part 3}$: $B=2, A=C+2$, this case results in 242, 352,... 792. There are $6$ ways to arrange the digits in all of those number except the first, and 3 ways ways for the first. This leads to $33$ cases.

$\textbf{Part 4}$: $B=3, A=C+3$, this case results in 363, 473,...693. There are $6$ ways to arrange the digits in all of those number except the first, and 3 ways ways for the first. This leads to $21$ cases.

$\textbf{Part 5}$: $B=4, A=C+4$, this case results in 484 and 594. There are $6$ ways to arrange the digits in all of those number except the first, and 3 ways ways for the first. This leads to $9$ cases.



$\textbf{Case 2}$: $A+B=C+11$.


$\textbf{Part 1}$: $C=0, A+B=11$, this cases results in 209, 308, ...506. There are $4$ ways to arrange each of those cases. This leads to $16$ cases.

$\textbf{Part 2}$: $C=1, A+B=12$, this cases results in 319, 418, ...616. There are $6$ ways to arrange each of those cases, except the last. This leads to $21$ cases.

$\textbf{Part 3}$: $C=2, A+B=13$, this cases results in 429, 528, ...617. There are $6$ ways to arrange each of those cases. This leads to $18$ cases.

... If you continue this counting, you receive $16+21+18+15+12+9+6+3=100$ cases.

$100+126=\boxed{(A) 226}$

See Also

2017 AMC 10A (Problems • Answer Key • Resources)
Preceded by
Problem 8 		Followed by
Problem 10
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 10 Problems and Solutions

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