Difference between revisions of "2012 AIME I Problems/Problem 1"

Revision as of 23:13, 8 February 2013

Problem 1

Find the number of positive integers with three not necessarily distinct digits, $abc$ , with $a \neq 0$ and $c \neq 0$ such that both $abc$ and $cba$ are multiples of $4$ .

Solutions

Solution 1

A positive integer is divisible by $4$ if and only if its last two digits are divisible by $4.$ For any value of $b$ , there are two possible values for $a$ and $c$ , since we find that if $b$ is even, $a$ and $c$ must be either $4$ or $8$ , and if $b$ is odd, $a$ and $c$ must be either $2$ or $6$ . There are thus $2 \cdot 2 = 4$ ways to choose $a$ and $c$ for each $b,$ and $10$ ways to choose $b$ since $b$ can be any digit. The final answer is then $4 \cdot 10 = \boxed{40.}$

Solution 2

A number is divisible by four if its last two digits are divisible by 4. Thus, we require that $10b + a$ and $10b + c$ are both divisible by $4$ . If $b$ is odd, then $a$ and $c$ must both be $2 \pmod 4$ meaning that $a$ and $c$ are $2$ or $6$ . If $b$ is even, then $a$ and $c$ must be $0 \pmod 4$ meaning that $a$ and $c$ are $4$ or $8$ . For each choice of $b$ there are $2$ choices for $a$ and $2$ for $c$ for a total of $10 \cdot 2 \cdot 2 = \boxed{40}$ numbers.

@@ Line 10: / Line 10: @@
 === Solution 2 ===
-A number is divisible by four if its last two digits are divisible by 4. Thus, we require that <math> 10b + a </math> and <math> 10b + c</math> are both divisible by <math> 4 </math>. If <math> b </math> is odd, then <math> a </math> and <math> c </math> must both be <math> 2 \pmod 4 </math> meaning that <math> a </math> and <math> c </math> are <math> 2 </math> or <math> 6 </math>. If <math> b </math> is even, then <math> a </math> and <math> c </math> must be <math> 0 \pmod 4 </math> meaning that <math> a </math> and <math> c </math> are <math> 4 </math> or <math> 8 </math>. For each choice of <math> b </math> there are <math> 2 </math> choices for <math> a </math> and <math> 2 </math> for <math> c </math> for a total of <math> 10 \cdot 2 \cdot 2 = \boxed{040} </math> numbers.
+A number is divisible by four if its last two digits are divisible by 4. Thus, we require that <math> 10b + a </math> and <math> 10b + c</math> are both divisible by <math> 4 </math>. If <math> b </math> is odd, then <math> a </math> and <math> c </math> must both be <math> 2 \pmod 4 </math> meaning that <math> a </math> and <math> c </math> are <math> 2 </math> or <math> 6 </math>. If <math> b </math> is even, then <math> a </math> and <math> c </math> must be <math> 0 \pmod 4 </math> meaning that <math> a </math> and <math> c </math> are <math> 4 </math> or <math> 8 </math>. For each choice of <math> b </math> there are <math> 2 </math> choices for <math> a </math> and <math> 2 </math> for <math> c </math> for a total of <math> 10 \cdot 2 \cdot 2 = \boxed{40} </math> numbers.
 == See also ==
 {{AIME box|year=2012|n=I|before=First Problem|num-a=2}}

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