Difference between revisions of "2009 AMC 10A Problems/Problem 5"
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Note that <math>111111111</math> has <math>9</math> digits. Since the only operation in long multiplication will be <math>1 \cdot 1</math> and the addition tower will be <math>9</math> stacks long, there is no way a <math>1</math> stack can be <math>10</math> long and carry over its value. Thus, it is sufficient to conclude that we will multiply 1 with 1 <math>9 \cdot 9= \boxed{\textbf{(E) }81}</math> times, with no <math>1</math> being carried over. | Note that <math>111111111</math> has <math>9</math> digits. Since the only operation in long multiplication will be <math>1 \cdot 1</math> and the addition tower will be <math>9</math> stacks long, there is no way a <math>1</math> stack can be <math>10</math> long and carry over its value. Thus, it is sufficient to conclude that we will multiply 1 with 1 <math>9 \cdot 9= \boxed{\textbf{(E) }81}</math> times, with no <math>1</math> being carried over. | ||
+ | |||
+ | ~ Wesserwes7254 | ||
==See also== | ==See also== | ||
{{AMC10 box|year=2009|ab=A|num-b=4|num-a=6}} | {{AMC10 box|year=2009|ab=A|num-b=4|num-a=6}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 11:03, 15 July 2024
Contents
[hide]Problem
What is the sum of the digits of the square of ?
Solution 1
Using the standard multiplication algorithm, whose digit sum is
Solution 2
We note that
,
,
,
and .
We can clearly see the pattern: If is , with ones (and for the sake of simplicity, assume that ), then the sum of the digits of is
Aha! We know that has digits, so its digit sum is .
Solution 3
We see that can be written as .
We can apply this strategy to find , as seen below.
The digit sum is thus .
Solution 4 (Confusing But Fast)
We note that the only way for the sum of two numbers' digits to not equal the sum of the first numbers' digits plus the second numbers' digits is for a place (i.e., tens place, ones place, etc.) to be carried over. If that was a little confusing, note that and too. The ones don't carry over to the tens and the tens don't carry over to the hundreds. However, and The ones carry over to the tens and the tens carry over to the hundreds.
Note that has digits. Since the only operation in long multiplication will be and the addition tower will be stacks long, there is no way a stack can be long and carry over its value. Thus, it is sufficient to conclude that we will multiply 1 with 1 times, with no being carried over.
~ Wesserwes7254
See also
2009 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 4 |
Followed by Problem 6 | |
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 |
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