Difference between revisions of "2006 AIME I Problems/Problem 3"
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== Solution == | == Solution == | ||
+ | == Solution 1 == | ||
Suppose the original number is <math>N = \overline{a_na_{n-1}\ldots a_1a_0},</math> where the <math>a_i</math> are digits and the first digit, <math>a_n,</math> is nonzero. Then the number we create is <math>N_0 = \overline{a_{n-1}\ldots a_1a_0},</math> so <cmath>N = 29N_0.</cmath> But <math>N</math> is <math>N_0</math> with the digit <math>a_n</math> added to the left, so <math>N = N_0 + a_n \cdot 10^n.</math> Thus, <cmath>N_0 + a_n\cdot 10^n = 29N_0</cmath> <cmath>a_n \cdot 10^n = 28N_0.</cmath> The right-hand side of this equation is divisible by seven, so the left-hand side must also be divisible by seven. The number <math>10^n</math> is never divisible by <math>7,</math> so <math>a_n</math> must be divisible by <math>7.</math> But <math>a_n</math> is a nonzero digit, so the only possibility is <math>a_n = 7.</math> This gives <cmath>7 \cdot 10^n = 28N_0</cmath> or <cmath>10^n = 4N_0.</cmath> Now, we want to minimize ''both'' <math>n</math> and <math>N_0,</math> so we take <math>N_0 = 25</math> and <math>n = 2.</math> Then <cmath>N = 7 \cdot 10^2 + 25 = \boxed{725},</cmath> and indeed, <math>725 = 29 \cdot 25.</math> <math>\square</math> | Suppose the original number is <math>N = \overline{a_na_{n-1}\ldots a_1a_0},</math> where the <math>a_i</math> are digits and the first digit, <math>a_n,</math> is nonzero. Then the number we create is <math>N_0 = \overline{a_{n-1}\ldots a_1a_0},</math> so <cmath>N = 29N_0.</cmath> But <math>N</math> is <math>N_0</math> with the digit <math>a_n</math> added to the left, so <math>N = N_0 + a_n \cdot 10^n.</math> Thus, <cmath>N_0 + a_n\cdot 10^n = 29N_0</cmath> <cmath>a_n \cdot 10^n = 28N_0.</cmath> The right-hand side of this equation is divisible by seven, so the left-hand side must also be divisible by seven. The number <math>10^n</math> is never divisible by <math>7,</math> so <math>a_n</math> must be divisible by <math>7.</math> But <math>a_n</math> is a nonzero digit, so the only possibility is <math>a_n = 7.</math> This gives <cmath>7 \cdot 10^n = 28N_0</cmath> or <cmath>10^n = 4N_0.</cmath> Now, we want to minimize ''both'' <math>n</math> and <math>N_0,</math> so we take <math>N_0 = 25</math> and <math>n = 2.</math> Then <cmath>N = 7 \cdot 10^2 + 25 = \boxed{725},</cmath> and indeed, <math>725 = 29 \cdot 25.</math> <math>\square</math> | ||
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+ | ==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. \\ | ||
+ | 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>. | ||
==Solution 2 == | ==Solution 2 == |
Revision as of 17:40, 25 June 2017
Problem
Find the least positive integer such that when its leftmost digit is deleted, the resulting integer is of the original integer.
Solution
Solution 1
Suppose the original number is where the are digits and the first digit, is nonzero. Then the number we create is so But is with the digit added to the left, so Thus, The right-hand side of this equation is divisible by seven, so the left-hand side must also be divisible by seven. The number is never divisible by so must be divisible by But is a nonzero digit, so the only possibility is This gives or Now, we want to minimize both and so we take and Then and indeed,
Solution 2
Let be the required number, and be with the first digit deleted. Now, we know that (because this is an AIME problem). Thus, has or digits. Checking the other cases, we see that it must have digits. \\ Let , so . Thus, . By the constraints of the problem, we see that , so Now, we subtract and divide to get Clearly, must be a multiple of because both and are multiples of . Thus, . Now, we plug that into the equation: By the same line of reasoning as earlier, . We again plug that into the equation to get Now, since , , and , our number .
Solution 2
Let be the required number, and be with the first digit deleted. Now, we know that (because this is an AIME problem). Thus, has or digits. Checking the other cases, we see that it must have digits. \\ Let , so . Thus, . By the constraints of the problem, we see that , so Now, we subtract and divide to get Clearly, must be a multiple of because both and are multiples of . Thus, . Now, we plug that into the equation: By the same line of reasoning as earlier, . We again plug that into the equation to get Now, since , , and , our number .
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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