Difference between revisions of "2023 AMC 10B Problems/Problem 16"
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== Solution 2 == | == Solution 2 == | ||
− | Note that you can obtain a downo by reversing an upno (like 235 is an upno, and you can obtain 532). So, we need to find the amount of downos that end with 0. We can use stars and bars to get: <cmath>\sum_{n=0}^9 \dbinom{9}{n} = \dbinom{9}{0} + \dbinom{9}{1} + \cdots + \dbinom{9}{9}</cmath> to get 512. However, 0 is not a valid case so we subtract 1. Our answer is 511. | + | Note that you can obtain a downo by reversing an upno (like <math>235</math> is an upno, and you can obtain <math>532</math>). So, we need to find the amount of downos that end with 0. We can use stars and bars to get: <cmath>\sum_{n=0}^9 \dbinom{9}{n} = \dbinom{9}{0} + \dbinom{9}{1} + \cdots + \dbinom{9}{9}</cmath> to get 512. However, 0 is not a valid case so we subtract 1. Our answer is 511. |
-aleyang | -aleyang | ||
-ap246(LaTeX) | -ap246(LaTeX) |
Revision as of 18:21, 15 November 2023
Contents
Problem
Define an to be a positive integer of 2 or more digits where the digits are strictly
increasing moving left to right. Similarly, define a
to be a positive integer
of 2 or more digits where the digits are strictly decreasing moving left to right. For
instance, the number 258 is an upno and 8620 is a downno. Let 𝑈 equal the total
number of
and let 𝑑 equal the total number of
. What is |𝑈 − 𝐷|?
Solution
is greater than
because
can't start with
.
So the differences are in the form
When x has length
we have
The number of possible
.
~Technodoggo
Solution
Since Upnos do not allow 0s to be in their first -- and any other -- digit, there will be no zeros in any digits of an Upno. Thus, Upnos only contain digits [1,2,3,4,5,6,7,8,9].
Upnos are 2 digits in minimum and 9 digits maximum (repetition is not allowed). Thus the total number of Upnos will be (9C2)+(9C3)+(9C4)+...+(9C9), since every selection of distinct numbers from the set [1,2,3,4,5,6,7,8,9] can be arranged so that it is an Upno. There will be (9C2) 2 digit Upnos, (9C3) 3 digit Upnos and so on.
Thus, the total number of Upnos will be (9C2)+(9C3)+(9C4)+...+(9C9) = 2^9-(9C0)-(9C1) = 512 - 10 = 502.
Notice that the same combination logic can be done for Downnos, but Downnos DO allow zeros to be in their last digit. Thus, there are 10 possible digits [0,1,2,3,4,5,6,7,8,9] for Downnos.
Therefore, it is visible that the total number of Downnos are (10C2)+(10C3)+(10C4)+...+(10C10) = 2^10-(10C0)-(10C10) = 1024 - 11 = 1013.
Thus abs(#Upno-#Downno) = abs(1013-502) = 511.
~yxyxyxcxcxcx
Solution 2
Note that you can obtain a downo by reversing an upno (like is an upno, and you can obtain
). So, we need to find the amount of downos that end with 0. We can use stars and bars to get:
to get 512. However, 0 is not a valid case so we subtract 1. Our answer is 511.
-aleyang
-ap246(LaTeX)