Difference between revisions of "2006 AIME I Problems/Problem 3"
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== Solution == | == Solution == | ||
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+ | 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 1 == | == Solution 1 == |
Revision as of 17:42, 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 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
.
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.