# sum of digits of a 5 digit number is 41. find the probability that such a number is divisible by 11?

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## Given that the sum of digits of a 5 digit number is 41 . Find the probability that such a number is divisible by 11 ?

Given that the sum of digits of a 5 digit number is 41 . Find the probability that such a number is divisible by 11 ?

Byju's Answer Standard XII Mathematics Axiomatic Approach Given that th... Question

Given that the sum of digits of a 5 digit number is 41. Find the probability that such a number is divisible by 11?

A 12/35 B 23/35 C 29/35 D 6/35 E None of these Open in App Solution

The correct option is **A**

6/35

In order to get the sum as 41, the following 5 digit combinations exist:

99995 → number of 5 digit numbers = 5

99986 → number of 5 digit numbers = 20

99977 → number of 5 digit numbers = 10

99887 → number of 5 digit numbers = 30

98888 → number of 5 digit numbers = 5

Now, 70 such numbers exist.

Now for a 5 digit number of form (pqrst) to be divisible by 11,

(p+r+t)−(q+s) = 11, also (p+r+t)+(q+s) = 41

∴

p+r+t = 26, q+s = 15

Note that, (p+r+t)−(q+s) = 22 is not possible as it will give fractional values of (p+r+t) and (q+s).

Also, (p+r+t)−(q+s) = 33 is not possible as it will give (p+r+t) = 37 and we know p,r,t are digits.

(p,r,t) = (9,9,8) and (q,s) = (8,7) -------- (i)

or (p,r,t) = (9,9,8) and (q,s) = (9,6) -------- (ii)

Using 1 s t

equation we can construct 6 numbers.

Using 2 n d

equation we can construct 6 numbers.

∴

Number of favourable cases = 12.

Hence, required probability =

12 70 = 6 35 Suggest Corrections 0

SIMILAR QUESTIONS

**Q.**

Given that the sum of digits of a 5 digit number is 41. Find the probability that such a number is divisible by 11?

**Q.**The sum of the digits of a seven digit number is

59

.Find the probability that this number is divisible by

11 .

**Q.**Given a number 12345678901234567890..... up to 500 digits. Find the smallest number n that should be added to the number such that the sum is exactly divisible by 11.

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1 , 2 , 3 , 4 , 5 , 6 , 7 , 8 , 9

, written in random order, the probability that the number is divisible by

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**Q.**A

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digit number is formed by the digits

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without repetition. Find the probability that the number is divisible by

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## Sum of the digit of a five number is 41 find the probability that such a number is divisible by

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Previous Question: Next Question: Salad problem.. Four girls Robin, Mandy, Stacy, Erica of four families Miller, Jacob, Flure and Clark prepare four salads using the fruits Apples cherries bananas, grapes .Each girls uses 3 fruits in her salad. No body have the same combination. 1 )Robin not a Miller girl uses apples. 2) Miller and Mandy uses apples and cherries. 3) Clark uses cherries and grapes but Flure uses only one of them. 4)Erica is not Clark nor Flure. htt Then 4 questions asked. 1. Which is robins family: a. Miller b. Jacob c.Flure d.Clark 2. Which fruit is not used by Mandy? aa a. Cherries b. Grapes c. Apples d. Bananas 3. Which is the combination by Erica? a. Apples, cherries, Bananas b. Apples, Cherries, Grapes c. Apples, Grapes, Bananas d. Cherries, Grapes, Bananas 4. Which is robin's fruit combination? a. Apples, cherries, Bananas b. Apples, Cherries, Grapes c. Apples, Grapes, Bananas d. Cherries, Grapes, Bananas

## combinatorics

Problem Statement:- Consider the collection of all $5$-digit numbers such that the sum of digits of each number is $43$. A number is selected at random from the collection. Find the probabilit...

Number of5 5

-digit numbers which are divisible by

11 11

and whose digits add up to

43 43 Ask Question Asked 6 years ago

Modified 6 years ago

Viewed 7k times 9

**Problem Statement:-**

Consider the collection of all

5 5

-digit numbers such that the sum of digits of each number is

43 43

. A number is selected at random from the collection. Find the probability that the number is divisible by

11 11 .

**My Solution:-**

Let the five-digit numbers be represented by

x 1 x 2 x 3 x 4 x 5 x1x2x3x4x5 .

As the digits of these numbers should add up to

43 43 , we have x 1 + x 2 + x 3 + x 4 + x 5 =43 (where 1≤ x i ≤9, x i ∈N)

(where 1≤xi≤9, xi∈N)x1+x2+x3+x4+x5=43

So, the number of such numbers is equal to the coefficient of

x 43 x43 in ( 1− x 10 1−x ) 5 =15 (1−x101−x)5=15

As this is a listable amount of numbers so lets just list them

99997 99979 99799 97999 79999 99988 99889 98899 88999 99898 98989 89899 98998 89989 89998

999979997999799979997999999988998899889988999998989898989899989988998989998

Its easy to spot the numbers divisible by 11 which are

97999,99979,98989 97999,99979,98989

So, the required probability comes out to be

3 15 = 1 5 315=15

The number of numbers in this list turned out to be small hence just listing them turned out to be good enough for finding the numbers divisible by

11 11

, but what if the sum of the digits of the number turns out to be such that the number of numbers satisfying the condition turns out to be too big. In that case, how am I supposed to find the number of numbers which are divisible by

11 11

without the help of a computer but just with pen and paper.

combinatoricselementary-number-theory

Share

edited Mar 17, 2017 at 17:59

amWhy 1

asked Mar 17, 2017 at 17:47

user350331 2,5804 4 gold badges 19 19 silver badges 38 38 bronze badges

How did you get to the line: number of such number is equal to the coefficient of x^43 in ((1-x^10)/(1-x))^5 ? –

Rangan Aryan Mar 7, 2021 at 8:52 Add a comment

## 2 Answers

4

As you note, the requirement for a digit sum of

43 43

restricts the possible digit choices quite sharply.

In general numbers divisible by

11 11

have alternating digit sums (eg. adding

a a s and b b s from abababa abababa

) that differ by a multiple of

11 11

(which could be zero, if sums are equal). This is due to powers of

10 10 alternating between −1 −1 and 1mod11 1mod11 .

In this case since the total digit sum is odd, we must have alternating sums that differ by

11 11 , that is 16 16 and 27 27

for the two- and three-digit sums respectively. Clearly this gives us something of the form

9□9□9 9◻9◻9

and the options for the intervening digits adding to

16 16 are (7,9),(8,8),(9,7) (7,9),(8,8),(9,7) as you found.

By request in comments:

If the digit sum were

36 36

, all other conditions unchanged, we would have to have the digit sums equal

→(18,18) →(18,18) (as a difference of 22→(29,7) 22→(29,7)

is not feasible). Then there is only one option for the two-digit set

(9,9) (9,9)

and we can find the number of divisions on the three-digit set by a small inclusion-exclusion to partition the

18 18

into three valid digits. First there are two options that include a zero,

(99099,99990) (99099,99990)

after which we can take the partitions as being non-zero. Then without constraint on the upper size of partition, the options would be

( 17 2 ) (172)

, and removing the cases with a digit greater than

9 9 reduces this by 3⋅( 8 2 ) 3⋅(82) , giving 2+136−84=54 2+136−84=54 options. Share

edited Mar 17, 2017 at 18:39

answered Mar 17, 2017 at 17:59

Joffan 39.1k5 5 gold badges 43 43 silver badges 84 84 bronze badges

I do know about the divisibility rule of 11 and that is how I was quickly able to determine the numbers that were divisible by 11 in the list. Can you show how to find the number of numbers divisible by 11 if 36 was to be the sum of the digits. –

user350331

Mar 17, 2017 at 18:09

Add a comment 2

what if the sum of the digits of the number turns out to be such that the number of numbers satisfying the condition turns out to be too big. In that case, how am I supposed to find the number of numbers which are divisible by

Guys, does anyone know the answer?