Difference between revisions of "2024 AIME II Problems/Problem 7"
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==Solution 3== | ==Solution 3== | ||
− | Let our four digit number be <math>abcd</math>. Replacing digits with 1, we get the following equations: | + | 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>1000+100b+10c+d \equiv 0 \pmod{7}</math> | ||
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And since the remainder of 1111 divided by 7 is 5, we get: | And since the remainder of 1111 divided by 7 is 5, we get: | ||
− | <math> | + | <math>3\overline{abcd} \equiv 2 \pmod{7}</math> |
Which gives us: | Which gives us: | ||
− | <math>abcd \equiv 3 \pmod{7}</math> | + | <math>\overline{abcd} \equiv 3 \pmod{7}</math> |
− | And since we know that changing each digit into 1 will make abcd divisible by 7, 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 3 when divided by 7. Thus, we get <math>a=5</math>, <math>b=6</math>, <math>c=9</math>, and <math>d=4</math>. Thus, we get 5694 as abcd, and the answer is <math>694+5=\boxed{699}</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 | ~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== | ==Video Solution== | ||
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{{AIME box|year=2024|num-b=6|num-a=8|n=II}} | {{AIME box|year=2024|num-b=6|num-a=8|n=II}} | ||
− | [[Category:]] | + | [[Category:Intermediate Number Theory Problems]] |
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 21:13, 31 May 2024
Problem
Let be the greatest four-digit positive integer with the property that whenever one of its digits is changed to
, the resulting number is divisible by
. Let
and
be the quotient and remainder, respectively, when
is divided by
. Find
.
Solution 1
We note that by changing a digit to for the number
, we are subtracting the number by either
,
,
, or
. Thus,
. We can casework on
backwards, finding the maximum value.
(Note that computing greatly simplifies computation).
Applying casework on , we can eventually obtain a working value of
. ~akliu
Solution 2
Let our four digit number be . Replacing digits with 1, we get the following equations:
Reducing, we get
Subtracting , we get:
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 -westwoodmonster
Solution 3
Let our four digit number be . Replacing digits with
, we get the following equations:
Add the equations together, we get:
And since the remainder of 1111 divided by 7 is 5, we get:
Which gives us:
And since we know that changing each digit into will make
divisible by
, we get that
,
,
, and
all have a remainder of
when divided by
. Thus, we get
,
,
, and
. Thus, we get
as
, and the answer is
.
~Callisto531
Solution 4
Let our four digit number be . Replacing digits with 1, we get the following equations:
Then, we let x, y, z, t be the smallest whole number satisfying the following equations:
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):
(2):
(3):
(4):
Add (1), (2), (3) together, we get:
We can transform this equation to:
Since, according to (4), has a remainder of 0 when divided by 7, we get:
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
~Callisto531 and his dad
Video Solution
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
See also
2024 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 6 |
Followed by Problem 8 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.