ABA and BAB are not multiplication, they just denote digits.
I realized that ABA plus BAB will result in the last 3 digits being repeated, so do I find all multiple of 74 that have 3 repeating digits?
How many pairs of digits A,B such that ABA+BAB is divisible by 74?
0
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algebra-precalculus
modular-arithmetic
divisibility
2 Answers
1
ABA+BAB=111A+111B=111(A+B)=3x37(A+B) which divides 74 when A+B is even
where A,B is an element of {1,2,3,4,5,6,7,8,9}
so (A,B)=(1,1) (1,3) (1,5) (1,7) (1,9) (3,...) (5,...) (7,...) (9,...) so there are 25 pairs where both A and B are odd
There is also (2,2) (2,4) (2,6) (2,8) (4,..) (6,..) (8,..) which is another 16 solutions
Total =41 pairs
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0Are we supposed to count ordered pairs, or unordered ones? – 2017-02-19
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0We are supposed to count the pairs I'd say A=1 B=3 is a different pair to A=3 B=1 – 2017-02-19
2
Render the sum as $111\times(a+b)=3 \times 37\times (a+b)$. What property must the factor $a+b$ have to complete divisibility by $2\times 37=74$? Once you know this it is easy to count the favored pairs.