The sum of the digits of a four digit number is 32. What is the probab

Question

The sum of the digits of a four digit number is 32. What is the probability that such a number is divisible by 11?

A.  \frac{9}{35}

B.  \frac{1}{5}

C.  \frac{1}{7}

D.  \frac{4}{11}

E.  \frac{2}{9}

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GMATinsight · Answer

For Sum of four digits = 32
The set of digits should be either {8, 8, 8, 8} or {7, 7, 9, 9} or {7, 8, 8, 9}

Total Numbers formed by {8, 8, 8, 8} = 1 and that will be divisible by 11
Total Numbers formed by {7, 7, 9, 9}  = \frac{4!}{2!2!} = 6  and Four of them {7997, 7799, 9779, 9977} will be divisible by 11
Total Numbers formed by {6, 8, 9, 9}  = \frac{4!}{2!} = 12  and none of them will be divisible by 11
Total Numbers formed by {7, 8, 8, 9}  = \frac{4!}{2!} = 12  and Four of them {8789, 7898, 9878, 8987) will be divisible by 11
Total Numbers formed by {5, 9, 9, 9}  = \frac{4!}{3!} = 4  and None of them will be divisible by 11

Total Outcomes = 1+6+12+12+4 = 35
Total Numbers divisible by 11 = 1+4+4 = 9

Probability  = \frac{9}{35}

ANswer: Option A

AdjustedEbitdad · Answer

Why can't the digits be {5,9,9,9} or {6,9,9,8}

Bunuel · Answer

The numbers created with digits {5, 9, 9, 9} and {6, 9, 9, 8} are included in the total numbers whose sum of the digits is 32. However, these numbers will not be divisible by 11.

According to the divisibility rule for 11, a number is divisible by 11 if the difference between the sums of its alternating digits is either 0 or divisible by 11. For numbers created with the digits {5, 9, 9, 9} and {6, 9, 9, 8}, this condition is not met, so they will not be divisible by 11.

P.S. It's highly unlikely that you'd ever need the divisibility rule for 11 on the GMAT.

HarshavardhanR · Answer

https://gmatclub.com/forum/download/file.php?id=81270 Harsha GMAT-Club-Forum-6sxpvsnq.png

Thannuh · Answer

Can you share an even more detailed step-by-step thought process which helped you think of these 5 different number combinations. GMAT-Club-Forum-0rrdntnd.png

Kinshook · Answer

The sum of the digits of a four digit number is 32.

What is the probability that such a number is divisible by 11?

Case 1: 4-digit number = 8888; Divisible by 11 = 8888

Case 2: 4-digit number comprises of 7988; Total numbers = 4!/2! = 12; Divisible by 11 = {7898,9878,8789,8987}

Case 3: 4-digit number comprises of 7799; Total numbers = 4!/2!/2! = 6; Divisible by 11 = {7799,9977,7997,9779}

Case 4: 4-digit number comprises of 6998; Total numbers = 4!/2! = 12; Divisible by 11 = None

Case 5: 4-digit number comprises of 5999; Total numbers = 4!/3! = 4; Divisible by 11 = None

Total numbers = 1 + 12*2 + 6 + 4 = 35

Divisible by 11 = 1 + 4 + 4 = 9

The probability that such a number is divisible by 11 =  \frac{9}{35}

IMO A

The sum of the digits of a four digit number is 32. What is the probab

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The sum of the digits of a four digit number is 32. What is the probab

Prep Toolkit

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