ABA and BAB

True or False

If A B A \overline{ABA} is a multiple of 7, then so is B A B \overline{BAB} .

Always true Always false Sometimes true, sometimes false

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3 solutions

Maria Kozlowska
Jan 11, 2017

A B A = 101 A + 10 B = 7 d \overline{ABA} =101A+10B =7d

B A B = 101 B + 10 A \overline{BAB}=101B+10A

A B A B A B = 101 ( A B ) + 10 ( B A ) = ( A B ) × 7 × 13 \overline{ABA} -\overline{BAB} =101(A-B)+10(B-A)=(A-B) \times 7 \times 13 .

B A B = A B A ( A B ) × 7 × 13 = 7 ( d 13 ( A B ) ) \overline{BAB} =\overline{ABA} - (A-B) \times 7 \times 13 = 7(d-13(A-B))

Note: The same condition holds for multiples of 13 13 . It works for numbers 494 , 585 , 676 , 767 , 858 , 949 494, 585, 676, 767,858,949 .

I love what you did here! I took the sum (and so we get ( A + B ) × 111 (A+B) \times 111 meaning that the condition holds for multiples of 3 and 37), but didn't think of taking the difference.

I wonder which other multiples this would hold for.

Chung Kevin - 4 years, 4 months ago

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I checked all the pairs. It works for 3 , 7 , 13 3, 7, 13 . It also works for 37 37 in numbers 111 , 222 , 333...999 111, 222, 333 ... 999 .

Maria Kozlowska - 4 years, 4 months ago

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The generalized question of n A B A n B A B n \mid \overline{ABA} \Leftrightarrow n \mid \overline{BAB} is pretty interesting. While I agree that n 111 × 91 = 10101 n \mid 111 \times 91 = 10101 is a sufficient condition, it is not immediately apparant to me why this is also a necessary condition.

Calvin Lin Staff - 4 years, 4 months ago

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@Calvin Lin The relation is symmetrical by definition.

Maria Kozlowska - 4 years, 4 months ago
Anirudh Sreekumar
Jan 14, 2017

A B A = 101 A + 10 B \overline{ABA}=101A+10B is a multiple of 7 7

thus 101 A + 10 B 91 A 101A+10B-91A is also a multiple of 7 7 ( 91 A (91 A is a multiple of 7 ) 7)

10 A + 10 B \Rightarrow 10A+10B is a multiple of 7 7

10 A + 10 B + 91 B \Rightarrow 10A+10B+91B is a multiple of 7 7 ( 91 B (91 B is a multiple of 7 ) 7)

101 B + 10 A = 100 B + 10 A + B = B A B \Rightarrow 101B+10A=100B+10A+B=\overline{BAB} is a multiple of 7 7

That's a sweet solution.

Chung Kevin - 4 years, 4 months ago

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thank you :)

Anirudh Sreekumar - 4 years, 4 months ago

For single (non-negative) digits A , B A,B and some positive integer n n we have that

100 A + 10 B + A = 101 A + 10 B = 7 n 100A + 10B + A = 101A + 10B = 7n \Longrightarrow

3 A + 3 B = 7 n 98 A 7 B = 7 ( n 14 A B ) = 7 m 3A + 3B = 7n - 98A - 7B = 7*(n - 14A - B) = 7m for some integer m m .

Now B A B = 100 B + 10 A + B = 7 ( 14 B + A ) + 3 ( B + A ) = 7 k + 7 m = 7 d \overline{BAB} = 100B + 10A + B = 7*(14B + A) + 3*(B + A) = 7k + 7m = 7d

for some integer d d , and thus 7 A B A 7 B A B \boxed{7 | \overline{ABA} \Longrightarrow 7 | \overline{BAB}} .

Comment: Note that 3 ( A + B ) = 7 m 3(A + B) = 7m implies that 7 ( A + B ) 7|(A + B) , so the 3-digit number pairs ( A B A , B A B (\overline{ABA}, \overline{BAB} ) we are looking at are ( 161 , 616 ) , ( 252 , 525 ) , ( 343 , 434 ) , ( 595 , 959 ) , ( 686 , 868 ) (161,616), (252,525), (343,434), (595,959), (686,868) and of course ( 777 , 777 ) (777,777) .

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