1990 AIME Problems/Problem 13
Contents
Problem
Let . Given that
has 3817 digits and that its first (leftmost) digit is 9, how many elements of
have 9 as their leftmost digit?
Solution 1
Since has 3816 digits more than
,
numbers have 9 as their leftmost digits.
Solution 2
Let's divide all elements of T into sections. Each section ranges from to
And, each section must have 1 or 2 elements. So, let's consider both cases.
If a section has 1 element, we claim that the number doesn't have 9 as the leftmost digit. Let this element be and the section ranges from
to
. To the contrary, let's assume the number (
) does have 9 as the leftmost digit. Thus,
. But, if you divide both sides by 9, you get
, and because
, so we have another number (
) in the same section (
). Which is a contradiction to our assumption that the section only has 1 element. So in this case, the number doesn't have 9 as the leftmost digit.
If a section has 2 elements, we claim one has to have a 9 as the leftmost digit, one doesn't. Let the elements be and
, and the section ranges from
to
.
So We know
. From
, we know
, and since
. The number(
)'s leftmost digit must be 9, and the other number(
)'s leftmost digit is 1.
There are total 4001 elements in T and 3817 sections (each esction can only has one or two elements). The two elements section must have 9 as the left most digit. We can take 4001 elements, subtract 3817 elements that don't have 9 as the leftmost digit, and get numbers that have 9 as the leftmost digit.
- AlexLikeMath
See also
1990 AIME (Problems • Answer Key • Resources) | ||
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Followed by Problem 14 | |
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