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Difference between revisions of "2006 AIME I Problems/Problem 3"

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=== Solution 2 ===
 
=== Solution 2 ===
  
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Let <math>N</math> be the required number, and <math>N'</math> be <math>N</math> with the first digit deleted. Now, we know that <math>N<1000</math> (because this is an AIME problem). Thus, <math>N</math> has <math>1,</math> <math>2</math> or <math>3</math> digits. Checking the other cases, we see that it must have <math>3</math> digits. \\
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Let <math>N</math> be the required number, and <math>N'</math> be <math>N</math> with the first digit deleted. Now, we know that <math>N<1000</math> (because this is an AIME problem). Thus, <math>N</math> has <math>1,</math> <math>2</math> or <math>3</math> digits. Checking the other cases, we see that it must have <math>3</math> digits.
 
Let <math>N=\overline{abc}</math>, so <math>N=100a+10b+c</math>. Thus, <math>N'=\overline{bc}=10b+c</math>. By the constraints of the problem, we see that <math>N=29N'</math>, so <cmath>100a+10b+c=29(10b+c).</cmath> Now, we subtract and divide to get <cmath>100a=28(10b+c)</cmath> <cmath>25a=70b+7c.</cmath> Clearly, <math>c</math> must be a multiple of <math>5</math> because both <math>25a</math> and <math>70b</math> are multiples of <math>5</math>. Thus, <math>c=5</math>. Now, we plug that into the equation: <cmath>25a=70b+7(5)</cmath> <cmath>25a=70b+35</cmath> <cmath>5a=14b+7.</cmath> By the same line of reasoning as earlier, <math>a=7</math>. We again plug that into the equation to get <cmath>35=14b+7</cmath> <cmath>b=2.</cmath> Now, since <math>a=7</math>, <math>b=2</math>, and <math>c=5</math>, our number <math>N=100a+10b+c=\boxed{725}</math>.
 
Let <math>N=\overline{abc}</math>, so <math>N=100a+10b+c</math>. Thus, <math>N'=\overline{bc}=10b+c</math>. By the constraints of the problem, we see that <math>N=29N'</math>, so <cmath>100a+10b+c=29(10b+c).</cmath> Now, we subtract and divide to get <cmath>100a=28(10b+c)</cmath> <cmath>25a=70b+7c.</cmath> Clearly, <math>c</math> must be a multiple of <math>5</math> because both <math>25a</math> and <math>70b</math> are multiples of <math>5</math>. Thus, <math>c=5</math>. Now, we plug that into the equation: <cmath>25a=70b+7(5)</cmath> <cmath>25a=70b+35</cmath> <cmath>5a=14b+7.</cmath> By the same line of reasoning as earlier, <math>a=7</math>. We again plug that into the equation to get <cmath>35=14b+7</cmath> <cmath>b=2.</cmath> Now, since <math>a=7</math>, <math>b=2</math>, and <math>c=5</math>, our number <math>N=100a+10b+c=\boxed{725}</math>.
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Here's another way to finish using this solution. From the above, you have <cmath>100a = 28(10b + c).</cmath> Divide by <math>4</math>, and you get <cmath>25a = 7(10b + c).</cmath> This means that <math>25a</math> has to be divisible by <math>7</math>, and hence <math>a = 7.</math> Now, solve for <math>25 = 10b + c</math>, which gives you <math>a = 7, b = 2, c = 5</math>, giving you the number <math>\boxed{725}</math>
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== Solution 3 (Quick) ==
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Note that if we let the last digit be <math>c</math> we must have <math>9c \equiv c \pmod{10}.</math> Thus we either have <math>c=0</math> which we can quickly check to be impossible (since the number after digit removal could be 10,20,30) or <math>c=5.</math> Testing 5, 15, and 25 as the numbers after removal we find that our answer is clearly <math>29 \cdot 25 = 725.</math>
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~Dhillonr25
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(Note that the quick checking of six numbers was possible thanks to AIME problems having answers less than 1000).
  
 
== See also ==
 
== See also ==

Latest revision as of 21:47, 13 October 2024

Problem

Find the least positive integer such that when its leftmost digit is deleted, the resulting integer is $\frac{1}{29}$ of the original integer.

Solutions

Solution 1

Suppose the original number is $N = \overline{a_na_{n-1}\ldots a_1a_0},$ where the $a_i$ are digits and the first digit, $a_n,$ is nonzero. Then the number we create is $N_0 = \overline{a_{n-1}\ldots a_1a_0},$ so \[N = 29N_0.\] But $N$ is $N_0$ with the digit $a_n$ added to the left, so $N = N_0 + a_n \cdot 10^n.$ Thus, \[N_0 + a_n\cdot 10^n = 29N_0\] \[a_n \cdot 10^n = 28N_0.\] The right-hand side of this equation is divisible by seven, so the left-hand side must also be divisible by seven. The number $10^n$ is never divisible by $7,$ so $a_n$ must be divisible by $7.$ But $a_n$ is a nonzero digit, so the only possibility is $a_n = 7.$ This gives \[7 \cdot 10^n = 28N_0\] or \[10^n = 4N_0.\] Now, we want to minimize both $n$ and $N_0,$ so we take $N_0 = 25$ and $n = 2.$ Then \[N = 7 \cdot 10^2 + 25 = \boxed{725},\] and indeed, $725 = 29 \cdot 25.$ $\square$

Solution 2

Let $N$ be the required number, and $N'$ be $N$ with the first digit deleted. Now, we know that $N<1000$ (because this is an AIME problem). Thus, $N$ has $1,$ $2$ or $3$ digits. Checking the other cases, we see that it must have $3$ digits. Let $N=\overline{abc}$, so $N=100a+10b+c$. Thus, $N'=\overline{bc}=10b+c$. By the constraints of the problem, we see that $N=29N'$, so \[100a+10b+c=29(10b+c).\] Now, we subtract and divide to get \[100a=28(10b+c)\] \[25a=70b+7c.\] Clearly, $c$ must be a multiple of $5$ because both $25a$ and $70b$ are multiples of $5$. Thus, $c=5$. Now, we plug that into the equation: \[25a=70b+7(5)\] \[25a=70b+35\] \[5a=14b+7.\] By the same line of reasoning as earlier, $a=7$. We again plug that into the equation to get \[35=14b+7\] \[b=2.\] Now, since $a=7$, $b=2$, and $c=5$, our number $N=100a+10b+c=\boxed{725}$.

Here's another way to finish using this solution. From the above, you have \[100a = 28(10b + c).\] Divide by $4$, and you get \[25a = 7(10b + c).\] This means that $25a$ has to be divisible by $7$, and hence $a = 7.$ Now, solve for $25 = 10b + c$, which gives you $a = 7, b = 2, c = 5$, giving you the number $\boxed{725}$

Solution 3 (Quick)

Note that if we let the last digit be $c$ we must have $9c \equiv c \pmod{10}.$ Thus we either have $c=0$ which we can quickly check to be impossible (since the number after digit removal could be 10,20,30) or $c=5.$ Testing 5, 15, and 25 as the numbers after removal we find that our answer is clearly $29 \cdot 25 = 725.$

~Dhillonr25

(Note that the quick checking of six numbers was possible thanks to AIME problems having answers less than 1000).

See also

2006 AIME I (Problems • Answer Key • Resources)
Preceded by
Problem 2 		Followed by
Problem 4
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions

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