Difference between revisions of "2014 AMC 10A Problems/Problem 24"
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so that the <math>500,000</math>th number is the <math>506</math>th number on the <math>997</math>th row because <math>4+5+6+7......+999 = 499,494</math>. The last number of the <math>996</math>th row (when including the numbers skipped) is <math>499,494 + (1+2+3+4.....+996)= 996,000</math>, (we add the <math>1-996</math> because of the numbers we skip) so our answer is <math>996,000 + 506 = \boxed{\textbf{(A) }996,506}</math>. | so that the <math>500,000</math>th number is the <math>506</math>th number on the <math>997</math>th row because <math>4+5+6+7......+999 = 499,494</math>. The last number of the <math>996</math>th row (when including the numbers skipped) is <math>499,494 + (1+2+3+4.....+996)= 996,000</math>, (we add the <math>1-996</math> because of the numbers we skip) so our answer is <math>996,000 + 506 = \boxed{\textbf{(A) }996,506}</math>. | ||
+ | |||
+ | ===Note=== | ||
+ | One may also note we simply need to add the number of skipped numbers to <math>500,000</math> to get our answer. The number of skipped numbers is <math>\frac{996\cdot 997}{2}</math> which has a units digit of <math>6</math>. Looking at the answer choices, it becomes apparent that the answer is <math>\boxed{\textbf{(A) }996,506}</math> | ||
==Solution 2== | ==Solution 2== |
Revision as of 22:36, 30 October 2021
Contents
Problem
A sequence of natural numbers is constructed by listing the first , then skipping one, listing the next
, skipping
, listing
, skipping
, and, on the
th iteration, listing
and skipping
. The sequence begins
. What is the
th number in the sequence?
Solution 1
If we list the rows by iterations, then we get
etc.
so that the th number is the
th number on the
th row because
. The last number of the
th row (when including the numbers skipped) is
, (we add the
because of the numbers we skip) so our answer is
.
Note
One may also note we simply need to add the number of skipped numbers to to get our answer. The number of skipped numbers is
which has a units digit of
. Looking at the answer choices, it becomes apparent that the answer is
Solution 2
Let's start with natural numbers, with no skips in between.
All we need to do is count how many numbers are skipped, , and "push" (add on to)
according to however many numbers are skipped.
Clearly, . This means that the number of skipped number "blocks" in the sequence is
because we started counting from 4.
Therefore , and the answer is
.
Solution 3 (AOPS Video Transcript)
First, we group the numbers together in the following way: We quickly realize that the number of terms in the curly braces follow a pattern:
(where
is the
block. Now, we can tell that the last number in a curly brace will be the number of terms in the set added to the number of terms in all the previous sets. Luckily for us, odd numbers are easy to add. If we pretend that there was a
at the beginning, then the sum of all of the numbers before and including
is
. However, we have to subtract
which results in
. The amount of numbers in the parenthesis are the
triangular number or
. Next, we want to find the greatest
, where
. Simplifying, we get
. We realize that
results in a number just
greater than our target. Next, we square
:
. As we decrease
by
, we decrease the result of the equation by approximately
. In order to decrease by at least
, we have to decrease
times leading to
. We plug it in to
getting
. This is the last number in the
set. The number of terms used is
. We need to add
terms to get an answer of
.
~MathFun1000
Solution 4(Cheap)
We look at each option starting from . Clearly
where
is the number of the skipped numbers. So we have
. This factors as
. Since
is an integer the answer is
.
~coolmath_2018
Video Solution by Richard Rusczyk
https://www.youtube.com/watch?v=KfGtE4G6tBo
~ dolphin7
See Also
2014 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 23 |
Followed by Problem 25 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
All AMC 10 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.