MHB Finding ? to Make A Multiple of 7, 11, and 13

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To find the value of "?" that makes A a multiple of 7, 11, and 13, the equation A = 22?33 must be analyzed. Each condition requires testing values from 0 to 9 for "?" to see which results in A being divisible by the specified numbers. The participants discuss the calculations and possible values that satisfy each condition. Ultimately, the goal is to determine the correct digit that fits all three criteria. The discussion emphasizes the importance of systematic testing and divisibility rules.
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A=$\underbrace{22--22}\,?\,\underbrace{33--33}$
$\,\,\,\,\,\, \,\,\,\,\, \,\,\, \,\, \,\, 50$$\,\,\,\,\,\,\,\, \,\,\,\,\, \,\,\,\,\,\, \,\, 50$
find $?$
(1) to make A a multiple of $7$
(2) to make A a multiple of $11$
(3) to make A a multiple of $13$
 
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Albert said:
A=$\underbrace{22--22}\,?\,\underbrace{33--33}$
$\,\,\,\,\,\, \,\,\,\,\, \,\,\, \,\, \,\, 50$$\,\,\,\,\,\,\,\, \,\,\,\,\, \,\,\,\,\,\, \,\, 50$
find $? (0\leq ? \leq 9)$
(1) to make A a multiple of $7$
(2) to make A a multiple of $11$
(3) to make A a multiple of $13$
hint :
$7|111111,\,\,11|111111, \, and \,\, 13|111111$
 
Albert said:
A=$\underbrace{22--22}\,?\,\underbrace{33--33}$
$\,\,\,\,\,\, \,\,\,\,\, \,\,\, \,\, \,\, 50$$\,\,\,\,\,\,\,\, \,\,\,\,\, \,\,\,\,\,\, \,\, 50$
find $?$
(1) to make A a multiple of $7$
(2) to make A a multiple of $11$
(3) to make A a multiple of $13$

we have xxxxxx is divisible by 1001 or 7 * 11 * 13
so we remove 48 2's from A that is subtract 48 $2's * 10^{53}$ and then subtract {3 ...3}( 48 3's) and divide by 10^{48} and we get 22?33 . we have 22?22 divisible by 22 subtracting get
?11 . we can check that(by putting from 0 to 9) ? is 0 to be divisible by 11. 5 to be divisible by 7 and 6 to be divisible by 13
 
kaliprasad said:
we have xxxxxx is divisible by 1001 or 7 * 11 * 13
so we remove 48 2's from A that is subtract 48 $2's * 10^{53}$ and then subtract {3 ...3}( 48 3's) and divide by 10^{48} and we get 22?33 . we have 22?22 divisible by 22 subtracting get
?11 . we can check that(by putting from 0 to 9) ? is 0 to be divisible by 11. 5 to be divisible by 7 and 6 to be divisible by 13
nice !
 
Seemingly by some mathematical coincidence, a hexagon of sides 2,2,7,7, 11, and 11 can be inscribed in a circle of radius 7. The other day I saw a math problem on line, which they said came from a Polish Olympiad, where you compute the length x of the 3rd side which is the same as the radius, so that the sides of length 2,x, and 11 are inscribed on the arc of a semi-circle. The law of cosines applied twice gives the answer for x of exactly 7, but the arithmetic is so complex that the...

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