Difference between revisions of "2006 AIME I Problems/Problem 3"
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== Problem == | == Problem == | ||
− | Find the least positive integer such that when its leftmost digit is deleted, the resulting integer is 1/ | + | Find the least [[positive]] [[integer]] such that when its leftmost [[digit]] is deleted, the resulting integer is <math>\frac{1}{29}</math> of the original integer. |
− | == | + | == Solutions == |
− | |||
+ | === 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> | ||
+ | |||
+ | === Solution 2 === | ||
+ | |||
+ | 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>. | ||
+ | |||
+ | 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> | ||
+ | |||
+ | == Solution 3 (Quick) == | ||
+ | |||
+ | 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 = 729.</math> | ||
+ | |||
+ | ~Dhillonr25 | ||
+ | |||
+ | (Note that the quick checking of six numbers was possible thanks to AIME problems having answers less than 1000). | ||
== See also == | == See also == | ||
− | |||
− | |||
− | |||
* [[Number Theory]] | * [[Number Theory]] | ||
+ | {{AIME box|year=2006|n=I|num-b=2|num-a=4}} | ||
[[Category:Intermediate Number Theory Problems]] | [[Category:Intermediate Number Theory Problems]] | ||
+ | {{MAA Notice}} |
Latest revision as of 03:53, 10 December 2022
Problem
Find the least positive integer such that when its leftmost digit is deleted, the resulting integer is of the original integer.
Solutions
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
.
Here's another way to finish using this solution. From the above, you have Divide by
, and you get
This means that
has to be divisible by
, and hence
Now, solve for
, which gives you
, giving you the number
Solution 3 (Quick)
Note that if we let the last digit be we must have
Thus we either have
which we can quickly check to be impossible (since the number after digit removal could be 10,20,30) or
Testing 5, 15, and 25 as the numbers after removal we find that our answer is clearly
~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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.