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

  • Thread starter Thread starter Albert1
  • Start date Start date
  • Tags Tags
    Multiple
Click For Summary
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.
Albert1
Messages
1,221
Reaction score
0
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 $13$
 
Last edited:
Mathematics news on Phys.org
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 !
 
Thread 'Erroneously  finding discrepancy in transpose rule'
Obviously, there is something elementary I am missing here. To form the transpose of a matrix, one exchanges rows and columns, so the transpose of a scalar, considered as (or isomorphic to) a one-entry matrix, should stay the same, including if the scalar is a complex number. On the other hand, in the isomorphism between the complex plane and the real plane, a complex number a+bi corresponds to a matrix in the real plane; taking the transpose we get which then corresponds to a-bi...

Similar threads

  • · Replies 3 ·
Replies
3
Views
940
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 3 ·
Replies
3
Views
1K
  • · Replies 3 ·
Replies
3
Views
1K
Replies
3
Views
943
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 11 ·
Replies
11
Views
3K
  • · Replies 12 ·
Replies
12
Views
2K