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```
if n=774958p96q and n is divisible by 11, then possible ordered pairs (p,q) are
if n=774958p96q and n is divisible by 11, then possible ordered pairs (p,q) are

```
8 months ago

Arun
25768 Points
```							Dear student For divisibility by 11 (Sum of digits at odd places) - (sum of digits at even places ) should be divisible by 11 (33+q) - (22+p)  = 11 +(q-p) should be divisible by 11 Hence p = q Hence (1,1) (2,2)(3,3).....(9,9) can exist
```
8 months ago
Vikas TU
14149 Points
```							To solve this, you need to know the divisibility rules of 8 and 9 Divisibility rule for 8If the sum of the digits of the number is divisible by 8, then the number is divisible by 8                                                              => 7+7+4+9+5+8+P+9+6+Q is divisible by 8 => 55 + P + Q is divisible by 8 Divisibility rule for 9For a number to be divisible by 9, the last digit should be divisible by 9=> Last digit of the number should be 0 or 9=> Q = 0 or 9Now, if Q = 0=> 55 + P should be divisible by 8=> P = 1 or 9 or 17 or 25=> There are four possible pairs of {P, Q}If Q = 9=> 55 + P + Q should be divisible by 8=>64+ P should be divisble by 8=> P = 0 or 8 or 16=> There are three possible pairs of {P, Q}Total possible pairs of {P, Q} are 4+3 = 7Those pairs are {1, 0} {9, 0} {17, 0} {25, 0} {0, 9} {8,9} {16,9}
```
8 months ago
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