Difference between revisions of "2010 AMC 10A Problems/Problem 25"
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Note: We can go up to <math>167</math>, and then notice the pattern of units digits alternating between <math>3</math> and <math>7</math>, so we do not need to calculate <math>7223</math>. | Note: We can go up to <math>167</math>, and then notice the pattern of units digits alternating between <math>3</math> and <math>7</math>, so we do not need to calculate <math>7223</math>. | ||
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+ | ==Solution 2== | ||
+ | Notice that to get the previous term, we must add the smallest square number, (let's call it <math>n^2</math>) such that the sum is less than <math>(n+1)^2</math>. Otherwise, instead of subtracting <math>n^2</math> from the previous term, we're subtracting a greater square number. | ||
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+ | Remember that <math>(x+1)^2 = x^2 + x + (x+1)</math>. Recall that to find the previous term, we must add a square number such that it is less than the next square number. <math>a + n^2 < (n+1)^2</math>. For this to be true, <math>a < n + (n+1)</math>. What that means is that given a term <math>a</math>, we can find the previous term by adding <math>n^2</math> where <math>n > \frac {a-1}{2}</math>. | ||
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+ | For example, to find the term that precedes <math>167</math>, we know that <math>n>166/2 = 83</math>. Therefore, <math>n=84</math> and the previous term is <math>167 + 84^2 = 7223</math>. The last digit of <math>7223</math> is <math>3 \Rightarrow \boxed{\textbf{(B)}\ 3}</math> | ||
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+ | ~Zeric Hang | ||
== See also == | == See also == | ||
{{AMC10 box|year=2010|num-b=24|after=Last Question|ab=A}} | {{AMC10 box|year=2010|num-b=24|after=Last Question|ab=A}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 13:53, 2 July 2018
Contents
Problem
Jim starts with a positive integer and creates a sequence of numbers. Each successive number is obtained by subtracting the largest possible integer square less than or equal to the current number until zero is reached. For example, if Jim starts with
, then his sequence contains
numbers:
Let be the smallest number for which Jim’s sequence has
numbers. What is the units digit of
?
Solution
We can find the answer by working backwards. We begin with on the bottom row, then the
goes to the right of the equal's sign in the row above. We find the smallest value
for which
and
, which is
.
We repeat the same procedure except with for the next row and
for the row after that. However, at the fourth row, we see that solving
yields
, in which case it would be incorrect since
is not the greatest perfect square less than or equal to
. So we make it a
and solve
. We continue on using this same method where we increase the perfect square until
can be made bigger than it. When we repeat this until we have
rows, we get:
Hence the solution is the last digit of , which is
.
Note: We can go up to , and then notice the pattern of units digits alternating between
and
, so we do not need to calculate
.
Solution 2
Notice that to get the previous term, we must add the smallest square number, (let's call it ) such that the sum is less than
. Otherwise, instead of subtracting
from the previous term, we're subtracting a greater square number.
Remember that . Recall that to find the previous term, we must add a square number such that it is less than the next square number.
. For this to be true,
. What that means is that given a term
, we can find the previous term by adding
where
.
For example, to find the term that precedes , we know that
. Therefore,
and the previous term is
. The last digit of
is
~Zeric Hang
See also
2010 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 24 |
Followed by Last Question | |
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.