2013 AMC 10B Problems/Problem 25

The following problem is from both the 2013 AMC 12B #23 and 2013 AMC 10B #25, so both problems redirect to this page.

Problem

Bernardo chooses a three-digit positive integer $N$ and writes both its base-5 and base-6 representations on a blackboard. Later LeRoy sees the two numbers Bernardo has written. Treating the two numbers as base-10 integers, he adds them to obtain an integer $S$ . For example, if $N = 749$ , Bernardo writes the numbers $10,\!444$ and $3,\!245$ , and LeRoy obtains the sum $S = 13,\!689$ . For how many choices of $N$ are the two rightmost digits of $S$ , in order, the same as those of $2N$ ?

$\textbf{(A)}\ 5 \qquad\textbf{(B)}\ 10 \qquad\textbf{(C)}\ 15 \qquad\textbf{(D)}\ 20 \qquad\textbf{(E)}\ 25$

Solution 1

First, we can examine the units digits of the number base 5 and base 6 and eliminate some possibilities.

Say that $N \equiv a \pmod{6}$

also that $N \equiv b \pmod{5}$

Substituting these equations into the question and setting the units digits of $2N$ and $S$ equal to each other, it can be seen that $b < 5$ (because otherwise $a$ and $b$ will have different parities), and thus $a=b$ . $N \equiv a \pmod{6}$ , $N \equiv a \pmod{5}$ , $\implies N=a \pmod{30}$ , $0 \le a \le 4$

Therefore, $N$ can be written as $30x+y$ and $2N$ can be written as $60x+2y$

Just keep in mind that $y$ can be one of five choices: $0, 1, 2, 3,$ or $4$ , ; Also, we have already found which digits of $y$ will add up into the units digits of $2N$ .

Now, examine the tens digit, $x$ by using $\mod{25}$ and $\mod{36}$ to find the tens digit (units digits can be disregarded because $y=0,1,2,3,4$ will always work) Then we take $N=30x+y$ $\mod{25}$ and $\mod{36}$ to find the last two digits in the base $5$ and $6$ representation. $N \equiv 30x \pmod{36}$ $N \equiv 30x \equiv 5x \pmod{25}$ Both of those must add up to $2N\equiv60x \pmod{100}$

( $33 \ge x \ge 4$ )

Now, since $y=0,1,2,3,4$ will always work if $x$ works, then we can treat $x$ as a units digit instead of a tens digit in the respective bases and decrease the mods so that $x$ is now the units digit. $N \equiv 5x \pmod{6}$ $N \equiv 6x \equiv x \pmod{5}$ $2N\equiv 6x \pmod{10}$

Say that $x=5m+n$ (m is between 0-6, n is 0-4 because of constraints on x) Then

$N \equiv 5m+n \pmod{5}$ $N \equiv 25m+5n \pmod{6}$ $2N\equiv30m + 6n \pmod{10}$

and this simplifies to

$N \equiv n \pmod{5}$ $N \equiv m+5n \pmod{6}$ $2N\equiv 6n \pmod{10}$

From careful inspection, this is true when

$n=0, m=6$

$n=1, m=6$

$n=2, m=2$

$n=3, m=2$

$n=4, m=4$

This gives you $5$ choices for $x$ , and $5$ choices for $y$ , so the answer is $5* 5 = \boxed{\textbf{(E) }25}$

Solution 2

Notice that there are exactly $1000-100=900=5^2\cdot 6^2$ possible values of $N$ . This means, in $100\le N\le 999$ , every possible combination of $2$ digits will happen exactly once. We know that $N=900,901,902,903,904$ work because $900\equiv\dots00_5\equiv\dots00_6$ .

We know for sure that the units digit will add perfectly every $30$ added or subtracted, because $\text{lcm }(5, 6)=30$ . So we only have to care about cases of $N$ every $30$ subtracted. In each case, $2N$ subtracts $6$ /adds $4$ , $N_5$ subtracts $1$ and $N_6$ adds $1$ for the $10$ 's digit.

$\textbf{5 }\textcolor{red}{\text{ 0}}\text{ 4 3 2 1 0 }\textcolor{red}{\text{4}}\text{ 3 2 1 0 4 3 2 1 0 4 }\textcolor{red}{\text{3 2}}\text{ 1 0 4 3 2 1 0 4 3 2 }\textcolor{red}{\text{1}}$

$\textbf{6 }\textcolor{red}{\text{ 0}}\text{ 1 2 3 4 5 }\textcolor{red}{\text{0}}\text{ 1 2 3 4 5 0 1 2 3 4 }\textcolor{red}{\text{5 0}}\text{ 1 2 3 4 5 0 1 2 3 4 }\textcolor{red}{\text{5}}$

$\textbf{10}\textcolor{red}{\text{ 0}}\text{ 4 8 2 6 0 }\textcolor{red}{\text{4}}\text{ 8 2 6 0 4 8 2 6 0 4 }\textcolor{red}{\text{8 2}}\text{ 6 0 4 8 2 6 0 4 8 2 }\textcolor{red}{\text{6}}$

As we can see, there are $5$ cases, including the original, that work. These are highlighted in $\textcolor{red}{\text{red}}$ . So, thus, there are $5$ possibilities for each case, and $5\cdot 5=\boxed{\textbf{(E) }25}$ .

Solution 3

Notice that $N_5$ ranges from $3$ to $5$ digits and $N_6$ ranges from $3$ to $4$ digits.

So we can let $N_5 = a_1b_1c_1d_1e_1$ and $N_6 = a_2b_2c_2d_2$ . $S = 10000a_1 + 1000(b_1 + a_2) + 100(c_1 + b_2) + 10(d_1 + c_2) + e_1 + d_2.$ To make it easier to work with mod 100, we will let $N = 625a_1 + 125b_1 + 25c_1 + 5d_1 + e_1$ here. So we have:

$2N \equiv S \pmod{100} \implies 2(625a_1 + 125b_1 + 25c_1 + 5d_1 + e_1) \equiv 10(d_1 + c_2) + e_1 + d_2 \pmod{100}.$

$50(a_1 + b_1 + c_1) + \cancel{10d_1} + 2e_1 \equiv \cancel{10d_1} + 10c_2 + e_1 + d_2 \pmod{100}.$

$50(a_1 + b_1 + c_1) + e_1 \equiv 10c_2 + d_2 \pmod{100}.$

Now note that $50(a_1 + b_1 + c_1)$ and $10c_2$ will always end with zeros, so the only way for this expression to be true is if $e_1 = d_2.$

$50(a_1 + b_1 + c_1) + \cancel{e_1} \equiv 10c_2 + \cancel{d_2} \pmod{100}$

$5(a_1 + b_1 + c_1) \equiv c_2 \pmod{10}.$

So $c_2$ can either be $0$ or $5$ .

Now we can express $N$ in its base-six and base-five forms and set them equal to each other:

$625a_1 + 125b_1 + 25c_1 + 5d_1 + \cancel{d_2} = 216a_2 + 36b_2 + 6c_2 + \cancel{d_2}$

$5(125a_1 + 25b_1 + 5c_1 + d_1) = 6c_2 + 36(b_2 + 6a_2).$

Since we established $5 \mid c_2$ (remember, $c_2$ can either be $0$ or $5$ ), for the right hand side to be a multiple of $5$ , $5 \mid b_2 + 6a_2.$ So we have seven possible pairs for $a_2$ and $b_2$ :

$(a_2, b_2) \rightarrow (5,0), (4,1), (3,2), (2,3), (1,4), (0, 5), (0,0).$ But if we had $(0,0),$ then $N_6$ would only have two digits, and $N$ would not be a three digit number. And if we had $a_2 = 5,$ then $N$ would exceed 1000 and also not be a three digit number. So we actually have five valid pairs: $(4,1), (3,2), (2,3), (1,4), (0, 5).$ Since a number in one base can be expressed in the other base in exactly one way, and the value of $c_2$ is dependent on $a_1, b_1,$ and $c_1,$ there is one solution $(a_1, b_1, c_1, d_1)$ for every $(a_2, b_2)$ in

$N_5 = 625a_1 + 125b_1 + 25c_1 + 5d_1 + d_2 = N_6 = 216a_2 + 36b_2 + 6c_2 + d_2.$

And for every pair we have, there are five choices for $e_1 = d_2$ (any integer from $0 - 4$ ). So the answer is $5 \cdot 5 = \boxed{25}.$

~ grogg007, Nafer

Solution 4

Observe that the maximum possible value of the sum of the last two digits of the base $5$ number and the base $6$ number is $44+55=99$ . Let $N \equiv a \pmod {25}$ and $N \equiv b \pmod {36}$ .

If $a < \frac{25}{2}$ , $2N \equiv 2a \pmod {25}$ and if $a > \frac{25}{2}$ , $2N \equiv 2a - 25 \pmod {25}$ .

Using the same logic for $b$ , if $b < 18$ , $2N \equiv 2b \pmod {36}$ , and in the other case $2N \equiv 2b - 36 \pmod {36}$ .

We can do four cases:

Case 1: $a + b = 2a - 25 + 2b - 36 \implies a + b = 61$ .

For this case, there is trivially only one possible solution, $(a, b) = (25, 36)$ , which is equivalent to $(a, b) = (0, 0)$ .

Case 2: $a + b = 2a - 25 + 2b \implies a + b = 25$ .

Note that in this case, $a \geq 13$ must hold, and $b < 18$ must hold. We find the possible ordered pairs to be: $(13, 12), (14, 11), (15, 10), ..., (24, 1)$ for a total of $12$ ordered pairs.

Case 3: $a + b = 2a + 2b - 36 \implies a + b = 36$ .

Note that in this case, $b \geq 18$ must hold, and $a < 13$ must hold. We find the possible ordered pairs to be: $(24, 12), (25, 11), (26, 10), ..., (35, 1)$ for a total of $12$ ordered pairs.

Case 4: $a + b = 2a + 2b$ .

Trivially no solutions except $(a, b) = (0, 0)$ , which matches the solution in Case 1, which makes this an overcount.

By CRT, each solution $(a, b)$ corresponds exactly one positive integer in a set of exactly $\text{lcm} (25, 36) = 900$ consecutive positive integers, and since there are $900$ positive integers between $100$ and $999$ , our induction is complete, and our answer is $1 + 12 + 12 = \boxed{\textbf{(E) }25}$ .

~ fidgetboss_4000

Video Solution

https://www.youtube.com/watch?v=tgCK-H5jsOE&t=497s

2013 AMC 12B (Problems • Answer Key • Resources)
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2013 AMC 10B (Problems • Answer Key • Resources)
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