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Difference between revisions of "1992 AHSME Problems/Problem 17"

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== Solution ==
 
== Solution ==
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=== Solution 1===
  
 
We can determine if our number is divisible by <math>3</math> or <math>9</math> by summing the digits. Looking at the one's place, we can start out with <math>0, 1, 2, 3, 4, 5, 6, 7, 8, 9</math> and continue cycling though the numbers from <math>0</math> through <math>9</math>. For each one of these cycles, we add <math>0 + 1 + ... + 9 = 45</math>. This is divisible by <math>9</math>, thus we can ignore the sum. However, this excludes <math>19</math>, <math>90</math>, <math>91</math> and <math>92</math>. These remaining units digits sum up to <math>9 + 1 + 2 = 12</math>, which means our units sum is <math>3 \pmod 9</math>. As for the tens digits, for <math>2, 3, 4, \cdots , 8</math> we have <math>10</math> sets of those: <cmath>\frac{8 \cdot 9}{2} - 1 = 35,</cmath> which is congruent to <math>8 \pmod 9</math>. We again have <math>19, 90, 91</math> and <math>92</math>, so we must add <cmath>1 + 9 \cdot 3 = 28</cmath> to our total. <math>28</math> is congruent to <math>1 \pmod 9</math>. Thus our sum is congruent to <math>3 \pmod 9</math>, and <math>k = 1  
 
We can determine if our number is divisible by <math>3</math> or <math>9</math> by summing the digits. Looking at the one's place, we can start out with <math>0, 1, 2, 3, 4, 5, 6, 7, 8, 9</math> and continue cycling though the numbers from <math>0</math> through <math>9</math>. For each one of these cycles, we add <math>0 + 1 + ... + 9 = 45</math>. This is divisible by <math>9</math>, thus we can ignore the sum. However, this excludes <math>19</math>, <math>90</math>, <math>91</math> and <math>92</math>. These remaining units digits sum up to <math>9 + 1 + 2 = 12</math>, which means our units sum is <math>3 \pmod 9</math>. As for the tens digits, for <math>2, 3, 4, \cdots , 8</math> we have <math>10</math> sets of those: <cmath>\frac{8 \cdot 9}{2} - 1 = 35,</cmath> which is congruent to <math>8 \pmod 9</math>. We again have <math>19, 90, 91</math> and <math>92</math>, so we must add <cmath>1 + 9 \cdot 3 = 28</cmath> to our total. <math>28</math> is congruent to <math>1 \pmod 9</math>. Thus our sum is congruent to <math>3 \pmod 9</math>, and <math>k = 1  
 
\implies \boxed{B}</math>.
 
\implies \boxed{B}</math>.
  
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== Alternate Solution ==
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==Solution 2==
  
−
Every number is congruent to its digit sum mod <math>9</math>, so <math>N</math> is congruent to <math>1+9+2+0+...+9+2</math> mod <math>9</math>, but applying the result in reverse, <math>1+9</math> is congruent to <math>19</math>, <math>2+0</math> is congruent to <math>20</math>, etc., so the sum just become <math>19+20+...+92</math> mod <math>9</math>. We can simplify this using the formula for the sum of an arithmetic series, giving <math>\frac{1}{2} \times 74 \times (19+92) = 37 \times 111</math>, which is congruent to <math>1 \times 3 = 3</math> mod <math>9</math>, as before.
+
As our first trial with 3, We can say that <math>N\equiv 1+9+2+0+2+1+2+2+...+9+1+9+2\pmod{3}</math>. Since if the sum of the digits of a number is divisible by 3, then the number is also divisible by 3, we reverse that logic to say that <math>N\equiv 19+20+21+22+...+91+92\pmod{3}</math>, and adding that up using the rainbow strategy, we get <math>N\equiv (111\times37)\pmod{3}</math>. We can see that since 3 divides 111, the original number is divisible by 3. But, since 111 only has one 3 and 37 has no 3's, it is not divisble by 9. Thus, the answer is <math>\boxed{B}</math>
 +
 
 +
~mathpro12345
 +
 
 +
papa I wrote this second one
  
 
== See also ==
 
== See also ==

Latest revision as of 12:19, 4 January 2021

Problem

The 2-digit integers from 19 to 92 are written consecutively to form the integer $N=192021\cdots9192$. Suppose that $3^k$ is the highest power of 3 that is a factor of $N$. What is $k$?

$\text{(A) } 0\quad \text{(B) } 1\quad \text{(C) } 2\quad \text{(D) } 3\quad \text{(E) more than } 3$

Solution

Solution 1

We can determine if our number is divisible by $3$ or $9$ by summing the digits. Looking at the one's place, we can start out with $0, 1, 2, 3, 4, 5, 6, 7, 8, 9$ and continue cycling though the numbers from $0$ through $9$. For each one of these cycles, we add $0 + 1 + ... + 9 = 45$. This is divisible by $9$, thus we can ignore the sum. However, this excludes $19$, $90$, $91$ and $92$. These remaining units digits sum up to $9 + 1 + 2 = 12$, which means our units sum is $3 \pmod 9$. As for the tens digits, for $2, 3, 4, \cdots , 8$ we have $10$ sets of those: \[\frac{8 \cdot 9}{2} - 1 = 35,\] which is congruent to $8 \pmod 9$. We again have $19, 90, 91$ and $92$, so we must add \[1 + 9 \cdot 3 = 28\] to our total. $28$ is congruent to $1 \pmod 9$. Thus our sum is congruent to $3 \pmod 9$, and $k = 1 \implies \boxed{B}$.

Solution 2

As our first trial with 3, We can say that $N\equiv 1+9+2+0+2+1+2+2+...+9+1+9+2\pmod{3}$. Since if the sum of the digits of a number is divisible by 3, then the number is also divisible by 3, we reverse that logic to say that $N\equiv 19+20+21+22+...+91+92\pmod{3}$, and adding that up using the rainbow strategy, we get $N\equiv (111\times37)\pmod{3}$. We can see that since 3 divides 111, the original number is divisible by 3. But, since 111 only has one 3 and 37 has no 3's, it is not divisble by 9. Thus, the answer is $\boxed{B}$

~mathpro12345

papa I wrote this second one

See also

1992 AHSME (Problems • Answer Key • Resources)
Preceded by
Problem 16 		Followed by
Problem 18
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 • 26 • 27 • 28 • 29 • 30
All AHSME Problems and Solutions

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