Difference between revisions of "2009 AMC 10A Problems/Problem 5"
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==Problem== | ==Problem== | ||
− | <!-- don't remove the following tag, for PoTW on the Wiki front page--><onlyinclude>What is the sum of the digits of the square of <math>\text | + | <!-- don't remove the following tag, for PoTW on the Wiki front page--><onlyinclude>What is the sum of the digits of the square of <math>\text 111111111</math>?<!-- don't remove the following tag, for PoTW on the Wiki front page--></onlyinclude> |
<math>\mathrm{(A)}\ 18\qquad\mathrm{(B)}\ 27\qquad\mathrm{(C)}\ 45\qquad\mathrm{(D)}\ 63\qquad\mathrm{(E)}\ 81</math> | <math>\mathrm{(A)}\ 18\qquad\mathrm{(B)}\ 27\qquad\mathrm{(C)}\ 45\qquad\mathrm{(D)}\ 63\qquad\mathrm{(E)}\ 81</math> | ||
==Solution 1== | ==Solution 1== | ||
− | Using the standard multiplication algorithm, <math>111,111,111^2=12,345,678,987,654,321,</math> whose digit sum is <math> | + | Using the standard multiplication algorithm, <math>111,111,111^2=12,345,678,987,654,321,</math> whose digit sum is <math>\boxed{(E)\text{ }81.}</math> |
− | |||
==Solution 2== | ==Solution 2== | ||
− | + | We note that | |
− | <math> | + | <math>1^2 = 1</math>, |
− | + | <math>11^2 = 121</math>, | |
+ | |||
+ | <math>111^2 = 12321</math>, | ||
+ | |||
+ | and <math>1,111^2 = 1234321</math>. | ||
+ | |||
+ | We can clearly see the pattern: If <math>X</math> is <math>111\cdots111</math>, with <math>n</math> ones (and for the sake of simplicity, assume that <math>n<10</math>), then the sum of the digits of <math>X^2</math> is | ||
+ | |||
+ | <math>1+2+3+4+5\cdots n+(n-1)+(n-2)\cdots+1</math> | ||
+ | |||
+ | <math>=(1+2+3\cdots n)+(1+2+3+\cdots n-1)</math> | ||
+ | |||
+ | <math>=\dfrac{n(n+1)}{2}+\dfrac{(n-1)n}{2}</math> | ||
+ | |||
+ | <math>=\dfrac{n(n+1+n-1)}{2}=\dfrac{2n^2}{2}=n^2.</math> | ||
+ | |||
+ | Aha! We know that <math>111,111,111</math> has <math>9</math> digits, so its digit sum is <math>9^2=\boxed{81(E)}</math>. | ||
==Solution 3== | ==Solution 3== | ||
− | + | We see that <math>111^2</math> can be written as <math>111(100+10+1)=11100+1110+111=12321</math>. | |
− | <math>111 | + | |
+ | We can apply this strategy to find <math>111,111,111^2</math>, as seen below. | ||
+ | |||
+ | <math>111111111^2=111111111(100000000+10000000\cdots+10+1)</math> | ||
− | <math> | + | <math>=11111111100000000+1111111110000000+\cdots+111111111</math> |
− | |||
− | |||
− | = | + | <math>=12,345,678,987,654,321</math> |
− | |||
− | <math>1 | + | The digit sum is thus <math>1+2+3+4+5+6+7+8+9+8+7+6+5+4+3+2+1=81 \boxed{(E)}</math>. |
− | <math> | + | ==Solution 4 (Confusing But Fast)== |
+ | We note that the only way for the sum of multiple numbers' digits to not equal the sum of the digits after they are all added is for a place (i.e., tens place, ones place, etc.) to be carried over. If that was a little confusing, note that <math>63+22=85,</math> and <math>8+5=13.</math> <math>6+3+2+2=13,</math> too. The ones don't carry over to the tens and the tens don't carry over to the hundreds. However, <math>63+37=100,</math> and <math>6+3+3+7=19 \neq 1.</math> The ones carry over to the tens and the tens carry over to the hundreds, meaning it loses <math>2(9-1)=18</math> of its digit sum. | ||
− | |||
− | <math>1, | + | If that made sense, 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 <math>1</math> with <math>1</math> <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}} |
Latest revision as of 20:00, 16 July 2024
Contents
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 multiple numbers' digits to not equal the sum of the digits after they are all added 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, meaning it loses
of its digit sum.
If that made sense, 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
with
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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