Difference between revisions of "2024 AIME II Problems/Problem 7"

Latest revision as of 20:13, 31 May 2024

Problem

Let $N$ be the greatest four-digit positive integer with the property that whenever one of its digits is changed to $1$ , the resulting number is divisible by $7$ . Let $Q$ and $R$ be the quotient and remainder, respectively, when $N$ is divided by $1000$ . Find $Q+R$ .

Solution 1

We note that by changing a digit to $1$ for the number $\overline{abcd}$ , we are subtracting the number by either $1000(a-1)$ , $100(b-1)$ , $10(c-1)$ , or $d-1$ . Thus, $1000a + 100b + 10c + d \equiv 1000(a-1) \equiv 100(b-1) \equiv 10(c-1) \equiv d-1 \pmod{7}$ . We can casework on $a$ backwards, finding the maximum value.

(Note that computing $1000 \equiv 6 \pmod{7}, 100 \equiv 2 \pmod{7}, 10 \equiv 3 \pmod{7}$ greatly simplifies computation).

Applying casework on $a$ , we can eventually obtain a working value of $\overline{abcd} = 5694 \implies \boxed{699}$ . ~akliu

Solution 2

Let our four digit number be $abcd$ . Replacing digits with 1, we get the following equations:

$1000+100b+10c+d \equiv 0 \pmod{7}$

$1000a+100+10c+d \equiv 0 \pmod{7}$

$1000a+100b+10+d \equiv 0 \pmod{7}$

$1000a+100b+10c+1 \equiv 0 \pmod{7}$

Reducing, we get

$6+2b+3c+d \equiv 0 \pmod{7}$ $(1)$

$6a+2+3c+d \equiv 0 \pmod{7}$ $(2)$

$6a+2b+3+d \equiv 0 \pmod{7}$ $(3)$

$6a+2b+3c+1 \equiv 0 \pmod{7}$ $(4)$

Subtracting $(2)-(1), (3)-(2), (4)-(3), (4)-(1)$ , we get:

$3a-b \equiv 2 \pmod{7}$

$2b-3c \equiv 6 \pmod{7}$

$3c-d \equiv 2 \pmod{7}$

$6a-d \equiv 5 \pmod{7}$

For the largest 4 digit number, we test values for a starting with 9. When a is 9, b is 4, c is 3, and d is 7. However, when switching the digits with 1, we quickly notice this doesnt work. Once we get to a=5, we get b=6,c=9,and d=4. Adding 694 with 5, we get $\boxed{699}$ -westwoodmonster

Solution 3

Let our four digit number be $\overline{abcd}$ . Replacing digits with $1$ , we get the following equations:

$1000+100b+10c+d \equiv 0 \pmod{7}$

$1000a+100+10c+d \equiv 0 \pmod{7}$

$1000a+100b+10+d \equiv 0 \pmod{7}$

$1000a+100b+10c+1 \equiv 0 \pmod{7}$

Add the equations together, we get:

$3000a+300b+30c+3d+1111 \equiv 0 \pmod{7}$

And since the remainder of 1111 divided by 7 is 5, we get:

$3\overline{abcd} \equiv 2 \pmod{7}$

Which gives us:

$\overline{abcd} \equiv 3 \pmod{7}$

And since we know that changing each digit into $1$ will make $\overline{abcd}$ divisible by $7$ , we get that $d-1$ , $10c-10$ , $100b-100$ , and $1000a-1000$ all have a remainder of $3$ when divided by $7$ . Thus, we get $a=5$ , $b=6$ , $c=9$ , and $d=4$ . Thus, we get $5694$ as $\overline{abcd}$ , and the answer is $694+5=\boxed{699}$ .

~Callisto531

Solution 4

Let our four digit number be $abcd$ . Replacing digits with 1, we get the following equations:

$1000+100b+10c+d \equiv 0 \pmod{7}$

$1000a+100+10c+d \equiv 0 \pmod{7}$

$1000a+100b+10+d \equiv 0 \pmod{7}$

$1000a+100b+10c+1 \equiv 0 \pmod{7}$

Then, we let x, y, z, t be the smallest whole number satisfying the following equations:

$1000a \equiv x \pmod{7}$

$100b \equiv y \pmod{7}$

$10a \equiv z \pmod{7}$

$d \equiv t \pmod{7}$

Since 1000, 100, 10, and 1 have a remainder of 6, 2, 3, and 1 when divided by 7, we can get the equations of:

(1): $6+y+z+t \equiv 0 \pmod{7}$

(2): $x+2+z+t \equiv 0 \pmod{7}$

(3): $x+y+3+t \equiv 0 \pmod{7}$

(4): $x+y+z+1 \equiv 0 \pmod{7}$

Add (1), (2), (3) together, we get:

$2x+2y+2z+3t+11 \equiv 0 \pmod{7}$

We can transform this equation to:

$2(x+y+z+1)+3t+9 \equiv 0 \pmod{7}$

Since, according to (4), $x+y+z+1$ has a remainder of 0 when divided by 7, we get:

$3t+9 \equiv 0 \pmod{7}$

And because t is 0 to 6 due to it being a remainder when divided by 7, we use casework and determine that t is 4.

Using the same methods of simplification, we get that x=2, y=5, and z=6, which means that 1000a, 100b, 10c, and d has a remainder of 2, 5, 6, and 4, respectively. Since a, b, c, and d is the largest possible number between 0 to 9, we use casework to determine the answer is a=5, b=6, c=9, and d=4, which gives us an answer of $5+694=\boxed{699}$

~Callisto531 and his dad

Video Solution

https://youtu.be/DxBjaFJneBs

~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)

@@ Line 1: / Line 1: @@
-Let x+5y = cos(10). If 2+7y = 14cos(x), then what is the value of tan(xy)?
+==Problem==
+Let <math>N</math> be the greatest four-digit positive integer with the property that whenever one of its digits is changed to <math>1</math>, the resulting number is divisible by <math>7</math>. Let <math>Q</math> and <math>R</math> be the quotient and remainder, respectively, when <math>N</math> is divided by <math>1000</math>. Find <math>Q+R</math>.
+==Solution 1==
+We note that by changing a digit to <math>1</math> for the number <math>\overline{abcd}</math>, we are subtracting the number by either <math>1000(a-1)</math>, <math>100(b-1)</math>, <math>10(c-1)</math>, or <math>d-1</math>. Thus, <math>1000a + 100b + 10c + d \equiv 1000(a-1) \equiv 100(b-1) \equiv 10(c-1) \equiv d-1 \pmod{7}</math>. We can casework on <math>a</math> backwards, finding the maximum value.
+(Note that computing <math>1000 \equiv 6 \pmod{7}, 100 \equiv 2 \pmod{7}, 10 \equiv 3 \pmod{7}</math> greatly simplifies computation).
+Applying casework on <math>a</math>, we can eventually obtain a working value of <math>\overline{abcd} = 5694 \implies \boxed{699}</math>. ~akliu
+==Solution 2==
+Let our four digit number be <math>abcd</math>. Replacing digits with 1, we get the following equations:
+<math>1000+100b+10c+d \equiv 0 \pmod{7}</math>
+<math>1000a+100+10c+d \equiv 0 \pmod{7}</math>
+<math>1000a+100b+10+d \equiv 0 \pmod{7}</math>
+<math>1000a+100b+10c+1 \equiv 0 \pmod{7}</math>
+Reducing, we get
+<math>6+2b+3c+d \equiv 0 \pmod{7}</math>  <math>(1)</math>
+<math>6a+2+3c+d \equiv 0 \pmod{7}</math>   <math>(2)</math>
+<math>6a+2b+3+d \equiv 0 \pmod{7}</math>     <math>(3)</math>
+<math>6a+2b+3c+1 \equiv 0 \pmod{7}</math>    <math>(4)</math>
+Subtracting <math>(2)-(1), (3)-(2), (4)-(3), (4)-(1)</math>, we get:
+<math>3a-b \equiv 2 \pmod{7}</math>
+<math>2b-3c \equiv 6 \pmod{7}</math>
+<math>3c-d \equiv 2 \pmod{7}</math>
+<math>6a-d \equiv 5 \pmod{7}</math>
+For the largest 4 digit number, we test values for a starting with 9. When a is 9, b is 4, c is 3, and d is 7.  However, when switching the digits with 1, we quickly notice this doesnt work. Once we get to a=5, we get b=6,c=9,and d=4. Adding 694 with 5, we get <math>\boxed{699}</math> -westwoodmonster
+==Solution 3==
+Let our four digit number be <math>\overline{abcd}</math>. Replacing digits with <math>1</math>, we get the following equations:
+<math>1000+100b+10c+d \equiv 0 \pmod{7}</math>
+<math>1000a+100+10c+d \equiv 0 \pmod{7}</math>
+<math>1000a+100b+10+d \equiv 0 \pmod{7}</math>
+<math>1000a+100b+10c+1 \equiv 0 \pmod{7}</math>
+Add the equations together, we get:
+<math>3000a+300b+30c+3d+1111 \equiv 0 \pmod{7}</math>
+And since the remainder of 1111 divided by 7 is 5, we get:
+<math>3\overline{abcd} \equiv 2 \pmod{7}</math>
+Which gives us:
+<math>\overline{abcd} \equiv 3 \pmod{7}</math>
+And since we know that changing each digit into <math>1</math> will make <math>\overline{abcd}</math> divisible by <math>7</math>, we get that <math>d-1</math>, <math>10c-10</math>, <math>100b-100</math>, and <math>1000a-1000</math> all have a remainder of <math>3</math> when divided by <math>7</math>. Thus, we get <math>a=5</math>, <math>b=6</math>, <math>c=9</math>, and <math>d=4</math>. Thus, we get <math>5694</math> as <math>\overline{abcd}</math>, and the answer is <math>694+5=\boxed{699}</math>.
+~Callisto531
+==Solution 4==
+Let our four digit number be <math>abcd</math>. Replacing digits with 1, we get the following equations:
+<math>1000+100b+10c+d \equiv 0 \pmod{7}</math>
+<math>1000a+100+10c+d \equiv 0 \pmod{7}</math>
+<math>1000a+100b+10+d \equiv 0 \pmod{7}</math>
+<math>1000a+100b+10c+1 \equiv 0 \pmod{7}</math>
+Then, we let x, y, z, t be the smallest whole number satisfying the following equations:
+<math>1000a \equiv x \pmod{7}</math>
+<math>100b \equiv y \pmod{7}</math>
+<math>10a \equiv z \pmod{7}</math>
+<math>d \equiv t \pmod{7}</math>
+Since 1000, 100, 10, and 1 have a remainder of 6, 2, 3, and 1 when divided by 7, we can get the equations of:
+(1): <math>6+y+z+t \equiv 0 \pmod{7}</math>
+(2): <math>x+2+z+t \equiv 0 \pmod{7}</math>
+(3): <math>x+y+3+t \equiv 0 \pmod{7}</math>
+(4): <math>x+y+z+1 \equiv 0 \pmod{7}</math>
+Add (1), (2), (3) together, we get:
+<math>2x+2y+2z+3t+11 \equiv 0 \pmod{7}</math>
+We can transform this equation to:
+<math>2(x+y+z+1)+3t+9 \equiv 0 \pmod{7}</math>
+Since, according to (4), <math>x+y+z+1</math> has a remainder of 0 when divided by 7, we get:
+<math>3t+9 \equiv 0 \pmod{7}</math>
+And because t is 0 to 6 due to it being a remainder when divided by 7, we use casework and determine that t is 4.
+Using the same methods of simplification, we get that x=2, y=5, and z=6, which means that 1000a, 100b, 10c, and d has a remainder of 2, 5, 6, and 4, respectively. Since a, b, c, and d is the largest possible number between 0 to 9, we use casework to determine the answer is a=5, b=6, c=9, and d=4, which gives us an answer of <math>5+694=\boxed{699}</math>
+~Callisto531 and his dad
+==Video Solution==
+https://youtu.be/DxBjaFJneBs
+~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
+==See also==
+{{AIME box|year=2024|num-b=6|num-a=8|n=II}}
+[[Category:Intermediate Number Theory Problems]]
+{{MAA Notice}}

2024 AIME II (Problems • Answer Key • Resources)
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Difference between revisions of "2024 AIME II Problems/Problem 7"

Latest revision as of 20:13, 31 May 2024

Contents

Problem

Solution 1

Solution 2

Solution 3

Solution 4

Video Solution

See also