Difference between revisions of "2017 AMC 10B Problems/Problem 23"
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− | ==Problem | + | ==Problem== |
Let <math>N=123456789101112\dots4344</math> be the <math>79</math>-digit number that is formed by writing the integers from <math>1</math> to <math>44</math> in order, one after the other. What is the remainder when <math>N</math> is divided by <math>45</math>? | Let <math>N=123456789101112\dots4344</math> be the <math>79</math>-digit number that is formed by writing the integers from <math>1</math> to <math>44</math> in order, one after the other. What is the remainder when <math>N</math> is divided by <math>45</math>? | ||
<math>\textbf{(A)}\ 1\qquad\textbf{(B)}\ 4\qquad\textbf{(C)}\ 9\qquad\textbf{(D)}\ 18\qquad\textbf{(E)}\ 44</math> | <math>\textbf{(A)}\ 1\qquad\textbf{(B)}\ 4\qquad\textbf{(C)}\ 9\qquad\textbf{(D)}\ 18\qquad\textbf{(E)}\ 44</math> | ||
− | |||
==Solution 1== | ==Solution 1== | ||
We only need to find the remainders of N when divided by 5 and 9 to determine the answer. | We only need to find the remainders of N when divided by 5 and 9 to determine the answer. | ||
By inspection, <math>N \equiv 4 \text{ (mod 5)}</math>. | By inspection, <math>N \equiv 4 \text{ (mod 5)}</math>. | ||
− | The remainder when <math>N</math> is divided by <math>9</math> is <math>1+2+3+4+ \cdots +1+0+1+1 +1+2 +\cdots + 4+3+4+4</math>, but since <math>10 \equiv 1 \text{ (mod 9)}</math>, we can also write this as <math>1+2+3 +\cdots +10+11+12+ \cdots 43 + 44 = \frac{44 \cdot 45}2 = 22 \cdot 45</math>, which has a remainder of 0 mod 9. Solving these modular congruence using | + | The remainder when <math>N</math> is divided by <math>9</math> is <math>1+2+3+4+ \cdots +1+0+1+1 +1+2 +\cdots + 4+3+4+4</math>, but since <math>10 \equiv 1 \text{ (mod 9)}</math>, we can also write this as <math>1+2+3 +\cdots +10+11+12+ \cdots 43 + 44 = \frac{44 \cdot 45}2 = 22 \cdot 45</math>, which has a remainder of 0 mod 9. Solving these modular congruence using the [[Chinese Remainder Theorem]] we get the remainder to be <math>9 \pmod{45}</math>. Therefore, the answer is <math>\boxed{\textbf{(C) } 9}</math>. |
− | ==Alternative Ending to Solution 1== | + | ===Alternative Ending to Solution 1=== |
− | Once we find our 2 modular congruences, we can narrow our options down to <math>{C}</math> and <math>{D}</math> because the remainder when <math>N</math> is divided by <math>45</math> should be a multiple of 9 by our modular congruence that states <math>N</math> has a remainder of <math>0</math> when divided by <math>9</math>. Also, our other modular congruence states that the remainder when divided by <math>45</math> should have a remainder of <math>4</math> when divided by <math>5</math>. Out of options <math>C</math> and <math>D</math>, only <math>\boxed{\textbf{(C) } 9}</math> satisfies that the remainder when <math>N</math> is divided by | + | Once we find our 2 modular congruences, we can narrow our options down to <math>{C}</math> and <math>{D}</math> because the remainder when <math>N</math> is divided by <math>45</math> should be a multiple of 9 by our modular congruence that states <math>N</math> has a remainder of <math>0</math> when divided by <math>9</math>. Also, our other modular congruence states that the remainder when divided by <math>45</math> should have a remainder of <math>4</math> when divided by <math>5</math>. Out of options <math>C</math> and <math>D</math>, only <math>\boxed{\textbf{(C) } 9}</math> satisfies that the remainder when <math>N</math> is divided by <math>45 \equiv 4 \text{ (mod 5)}</math>. |
==Solution 2== | ==Solution 2== | ||
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<cmath>10(1 + 2 + \cdots + 43) + 44 \equiv 10 \left (\frac{43 \cdot 44}2 \right ) + 44 \equiv</cmath> | <cmath>10(1 + 2 + \cdots + 43) + 44 \equiv 10 \left (\frac{43 \cdot 44}2 \right ) + 44 \equiv</cmath> | ||
<cmath>10 \left (\frac{-2 \cdot -1}2 \right ) - 1 \equiv \boxed{\textbf{(C) } 9} \pmod{45}</cmath> | <cmath>10 \left (\frac{-2 \cdot -1}2 \right ) - 1 \equiv \boxed{\textbf{(C) } 9} \pmod{45}</cmath> | ||
+ | |||
+ | ==Solution 3 (Clever way using divisibility rules)== | ||
+ | We know that <math>45 = 5 \cdot 9</math>, so we can apply our restrictions to that. We know that the units digit must be <math>5</math> or <math>0</math>, and the digits must add up to a multiple of <math>9</math>. <math>1+2+3+4+\cdots + 44 = \frac{44 \cdot 45}{2}</math>. We can quickly see this is a multiple of <math>9</math> because <math>\frac{44}{2} \cdot 45 = 22 \cdot 45</math>. We know <math>123 \ldots 4344</math> is not a multiple of <math>5</math> because the units digit isn't <math>5</math> or <math>0</math>. We can just subtract by 9 until we get a number whose units digit is 5 or 0. | ||
+ | |||
+ | We have <math>123 \ldots 4344</math> is divisible by <math>9</math>, so we can subtract by <math>9</math> to get <math>123 \ldots 4335</math> and we know that this is divisible by 5. So our answer is <math>\boxed{\textbf{(C) } 9}</math> | ||
+ | |||
+ | ~Arcticturn | ||
==See Also== | ==See Also== | ||
{{AMC10 box|year=2017|ab=B|num-b=22|num-a=24}} | {{AMC10 box|year=2017|ab=B|num-b=22|num-a=24}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 16:39, 29 September 2023
Contents
Problem
Let be the
-digit number that is formed by writing the integers from
to
in order, one after the other. What is the remainder when
is divided by
?
Solution 1
We only need to find the remainders of N when divided by 5 and 9 to determine the answer.
By inspection, .
The remainder when
is divided by
is
, but since
, we can also write this as
, which has a remainder of 0 mod 9. Solving these modular congruence using the Chinese Remainder Theorem we get the remainder to be
. Therefore, the answer is
.
Alternative Ending to Solution 1
Once we find our 2 modular congruences, we can narrow our options down to and
because the remainder when
is divided by
should be a multiple of 9 by our modular congruence that states
has a remainder of
when divided by
. Also, our other modular congruence states that the remainder when divided by
should have a remainder of
when divided by
. Out of options
and
, only
satisfies that the remainder when
is divided by
.
Solution 2
Realize that for all positive integers
.
Apply this on the expanded form of :
Solution 3 (Clever way using divisibility rules)
We know that , so we can apply our restrictions to that. We know that the units digit must be
or
, and the digits must add up to a multiple of
.
. We can quickly see this is a multiple of
because
. We know
is not a multiple of
because the units digit isn't
or
. We can just subtract by 9 until we get a number whose units digit is 5 or 0.
We have is divisible by
, so we can subtract by
to get
and we know that this is divisible by 5. So our answer is
~Arcticturn
See Also
2017 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 22 |
Followed by Problem 24 | |
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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