Number 3: My fav number
Q1. What is the probability of presence of the digit 3 in a 3 digit number?
Q2. What is the reminder when 3^33 is divided by 333
Q3. A trader makes a profit of 3% by selling milk at Rs. 33/lit which contains 333.33 ml of water. If the price of milk increases by 33% then by what percentage the trade needs to increase the amount of water in milk so as to get the same profit with out increasing the price.
Q4. How many times 3 occurs from 3 to 3333?
Q5. If 33 men and 333 children working together can build 33.33% of “TEEN MAHAL” in 3 days then in how many days only the children can complete the rest of the building work? (The efficiency of a child is 33.33% of a man)
Q6. 3 equilateral triangular shape pieces are cut from a square tin plate of side 3m. What is the ratio of the area of remaining portion of the tin plate to the original area of the plate?
Q7. In how many ways 333 chocolates can be distributed in a class of 33 students where no student gets same number of chocolates.
OSO points: 66
Location:
Comments : 10
a. number of 3 digit numbers containg only one 3 - i.e XY3/X3Y/3XY
8 x 9 x 3 =216 = number of 3's
OSO bhai i think u r wrong in first step
3XY=1*9*9=81
X3Y=8*1*9=72
XY3=8*9*1=72
So total number in first case=225
me right or wrong tell me?
Little Star
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Q2. What is the reminder when 3^33 is divided by 333
ans is 270
Little Star
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Question 1: What is the probability of presence of the digit 3 in a three digit number?
Solution:
Altogether there are 9 x 10 x 10 = 900 three digit numbers.
Case 1: When there is only one three in the number, then the number of such numbers is:
For 3 in the Unit's Place: 8 x 9 x 1 = 72
For 3 in the Ten's Place: 8 x 1 x 9 = 72
For 3 in the Hundred's Place: 1 x 9 x 9 = 81
Hence there are: 225 numbers consisting of only one 3 in them.
Case 2: When there are two threes in the number, then the number of such numbers is:
For 3 in the Unit's and Ten's Place: 8 x 1 x 1 = 8
For 3 in the Ten's and Hundred's Place: 1 x 1 x 9 = 9
For 3 in the Hundred's and Unit's Place: 1 x 9 x 1 = 9
Hence there are: 26 numbers consisting of two threes in them.
Case 3: When there are three threes in the number, then the number of such numbers is only 1 and that is 333.
Hence there are a total of 252 numbers out of 900 numbers consisting of at least one three in them.
Hence the required probability is: 252/900 = 0.28
Thank You
Ravi Raja
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Question 2: What is the reminder when 3^33 is divided by 333?
Solution:
First we cancel the common factor 9 from both the numbers and get:
3^31/37
[{(3^5)^6}(3)]/37
= [{(16)^6}(3)]/37
= (2^24)(3)/37
= [{(2^5)^4}(2^4)(3)]/37
= (5^4)(2^4)(3)/37
= (33)(11)/37
= 30
Hence the required remainder is: 30 x 9 = 270
Thak You
Ravi Raja
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Total number of 3 digit numbers = 900
number of digits = 2700
Number of 3's in all three digit numbers :
a. number of 3 digit numbers containg only one 3 - i.e XY3/X3Y/3XY
8 x 9 x 3 =216 = number of 3's
b. number of 3 digit numbers containg only two 3 - i.e X33/3X3/33X
8 + 9 + 9 = 26 number of 3's = 52
b. number of 3 digit numbers containg three 3 - i.e 333
is only 1 number of 3's = 3
Total number of 3 = 271
Required probability = 271/2700
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