2006 AIME I Problems/Problem 3
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,
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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