a 7 digit no is there whose sum is 59 the chance that this no is divisible by 11 is
The test for a number to be divisible by 11 is that the sum of the digits in the even positions must be equal to the sum of the digits in the odd positions, or to differ from it by a multiple of 11. A 7-digit number looks like $\displaystyle \bullet \circ \bullet \circ \bullet \circ \bullet$. You want the sum of the digits in the $\displaystyle \bullet$ positions to differ from the sum of the digits in the $\displaystyle \circ$ positions by a multiple of 11 (the two sums cannot be equal, because the sum of the two of them must be 57, which is an odd number).
Bearing in mind that the sum of the three $\displaystyle \circ$ digits cannot be more than 27, and the sum of the four $\displaystyle \bullet$ digits cannot be more than 36, you should be able to convince yourself that the $\displaystyle \circ$ digits must have sum 23, and the $\displaystyle \bullet$ digits must have sum 34.
So the $\displaystyle \circ$ digits could be 689, 779 or 788 (any other possibilities?), and the $\displaystyle \bullet$ digits must be 9988 or 9997. Now you have to work out how many ways those combinations of digits can be fitted into the available slots in the 7-digit number.
Finally, you need to find out how many numbers there are altogether that have digital sum 57.
Edit. I misread 59 as 57. But the above method should give you some idea how to attack this problem.
Hello, prasum!
This is an awful problem!
I see no approach but brute-force Listing.
How many 7-digit numbers are there whose sum of digits is 59
and the number is divisible by 11?
There only 5 sets of seven digits whose sum is 59: .$\displaystyle \begin{array}{cc}(1) & 9999995 \\ (2) & 9999986 \\ (3) & 9999977 \\ (4) & 9999887 \\ (5) & 9998888 \end{array}$
To be divisible by 11, we must be able to partition the digits into two sets,
. . $\displaystyle \{a,b,c,d\}$ and $\displaystyle \{e,f,g\}$ so that:.$\displaystyle (a+b+c+d) - (e+f+g) \:=\:11k$
$\displaystyle \text{For case }(1)\;9999995\text{, it is impossible to do so.}$
$\displaystyle \text{For case }(2)\;9999986\text{, we have: }\,\{9998\}\text{ and }\{996\}\;\;(a)$
$\displaystyle \text{For case }(3)\;9999977\text{, we have: }\,\{9997\}\text{ and }\{997\}\;\;(b)$
$\displaystyle \text{For case }(4)\;9999887\text{, we have: }\,\{9998\}\text{ and }\{987\}\;\;(c)$
$\displaystyle \text{For case }(5)\;9998888\text{, we have: }\,\{9998\}\text{ and }\{888\}\;\;(d)$
We will form seven-digit numbers of the form:
. . . . . $\displaystyle \begin{array}{ccccccc} \_ & \_ & \_ & \_ & \_ & \_ & \_ \\ o & e & o & e & o & e & o \end{array}$
where the digits of the 4-element set are placed in the "odd" blanks (o)
. . and the digits of the 3-element set are placed in the "even" blanks (e).
Case (a): $\displaystyle \{9998\}\;\{996\}$
There are 4 choices for placing the "8".
There are 3 choices for placing the "6".
. . There are: $\displaystyle 4\cdot3 \,=\,12$ numbers.
Case (b): $\displaystyle \{9997\}\;\{997\}$
There are 4 choices for placing the "7".
There are 3 choices for placing the "7",
. . There are: $\displaystyle 4\cdot3 \,=\,12$ numbers.
Case (c): $\displaystyle \{9998\}\;\{987\}$
There are 4 choices for placing the "8".
There are $\displaystyle 3!$ choices for placing the "987".
. . There are: $\displaystyle 4\cdot6 \,=\,24$ numbers.
Case (d): $\displaystyle \{9998\}\;\{888\}$
There are 4 choices for placing the "8".
There is 1 choice for placing the "888".
. . There are: $\displaystyle 4\cdot1 \,=\,4$ numbers.
Therefore, there are:.$\displaystyle 12 + 12 + 48 + 4 \:=\:76$ such numbers.
Did I miss any cases? . . . I don't know!
.
Now that I have read the question correctly , let's try again.
Total number of 7-digit numbers with digital sum 59. The possible combinations of digits are:
9 9 9 9 9 9 5 (7 combinations);
9 9 9 9 9 8 6 (42 combinations);
9 9 9 9 9 7 7 (21 combinations);
9 9 9 9 8 8 7 (105 combinations);
9 9 9 8 8 8 8 (35 combinations).
Total number of 7-digit numbers with digital sum 59 is 7+42+21+105+35 = 210.
Numbers in the above list that are divisible by 11. The odd-numbered digits (those labelled $\displaystyle \bullet$ in my previous comment) must add up to 35. They must be 9 9 9 8 (4 combinations).
The even-numbered digits (those labelled $\displaystyle \circ$ in my previous comment) must add up to 24. They could be:
9 9 6 (3 combinations);
9 8 7 (6 combinations);
8 8 8 (1 combination).
Total number of multiples of 11 is 4(3+6+1) = 40.
Thus the probability of a 7-digit number with digital sum 59 being divisible by 11 is 40/210 = 4/21.