divisible by 111

Art of Problem Solving

AoPS Online

Middle School Math Grades 5-8, Ages 10-13, MATHCOUNTS, AMC 8

Grades 5-8, Ages 10-13, MATHCOUNTS, AMC 8

3 M G

BBookmark VNew Topic kLocked

Middle School Math Grades 5-8, Ages 10-13, MATHCOUNTS, AMC 8

Grades 5-8, Ages 10-13, MATHCOUNTS, AMC 8

3 M G

BBookmark VNew Topic kLocked

No tags match your search

M

divisible by 111

G H J

Reveal topic tags

L

G H BBookmark kLocked kLocked NReply

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

38 posts

#1 h Apr 1, 2025, 3:16 PM

Y by

How many 6-digit natural numbers (with distinct digits) can be formed using the digits 2, 3, 4, 5, 6, and 7 that are divisible by 111?

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

38 posts

#2 h Apr 7, 2025, 10:57 AM

Y by

aria123 wrote:

How many 6-digit natural numbers (with distinct digits) can be formed using the digits 2, 3, 4, 5, 6, and 7 that are divisible by 111?

Please help me!!!

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

297 posts

#3 h Apr 7, 2025, 11:22 AM • 1 Y

Y by aria123

Use the fact that 1000 is 1 mod 111 to come up with a divisibility rule(kind of like proving the divisibility rules for 9 and 11).

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

24 posts

#4 h Apr 7, 2025, 1:26 PM • 1 Y

Y by aria123

For divisibility by 111 check divisibility by 3 and 37 as 3×37 =111

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

2479 posts

#5 h Apr 7, 2025, 1:34 PM • 1 Y

Y by aria123

ethan2011 wrote:

Use the fact that 1000 is 1 mod 111 to come up with a divisibility rule(kind of like proving the divisibility rules for 9 and 11).

orz!!!

if num is abcdef then 111|abcdef implies 111|(abc+def) so we just test!

cases on abc+def:
444: 1
555: 8
666: 27
777: 64
888: 125
999: 216
1110: 125
1221: 64
1332: 27
1443: 8
1554: 1

add them up to get (1+...+216)+(125+...+1)=(1+...+6)^2+(1+...+5)^2=21^2+15^2=441+225=666!

edit: if someone says that the ans is 666 not 666! im going to lose it, i was using ! to show excitement not factorial

This post has been edited 1 time. Last edited by fruitmonster97, Apr 7, 2025, 1:35 PM

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

38 posts

#6 h Apr 15, 2025, 9:18 AM

Y by

fruitmonster97 wrote:

ethan2011 wrote:

Use the fact that 1000 is 1 mod 111 to come up with a divisibility rule(kind of like proving the divisibility rules for 9 and 11).

orz!!!

if num is abcdef then 111|abcdef implies 111|(abc+def) so we just test!

cases on abc+def:
444: 1
555: 8
666: 27
777: 64
888: 125
999: 216
1110: 125
1221: 64
1332: 27
1443: 8
1554: 1

add them up to get (1+...+216)+(125+...+1)=(1+...+6)^2+(1+...+5)^2=21^2+15^2=441+225=666!

edit: if someone says that the ans is 666 not 666! im going to lose it, i was using ! to show excitement not factorial

Thank you! But a, b, c, d, e, f are distinct and in {2, 3, 4, 5, 6, 7} so the answer is 48.

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

1345 posts

#7 h Apr 18, 2025, 2:43 AM • 1 Y

Y by aria123

the number is divisible by 111 iff the first three digits + the last three digits is divisible by 111. this sum is divisible by 9, so trying 333, 666, 999, 1332, 1665 gives that 999 is the only possible sum, and this is achieved when the first and fourth digits add to 9, the second and fifth digits add to 9, and the third and sixth digits add to 9. thus the answer is 3!*2^3=48.

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

38 posts

#8 h Apr 20, 2025, 3:06 AM

Y by

vincentwant wrote:

the number is divisible by 111 iff the first three digits + the last three digits is divisible by 111. this sum is divisible by 9, so trying 333, 666, 999, 1332, 1665 gives that 999 is the only possible sum, and this is achieved when the first and fourth digits add to 9, the second and fifth digits add to 9, and the third and sixth digits add to 9. thus the answer is 3!*2^3=48.

Thank you!!!

Z K Y

N Quick Reply

G

H

=

© 2025 AoPS Incorporated

a