## Interview Question

Quantitative Developer Interview

-Atlanta, GA

# How many numbers between 1 and 1000 contain a 3?

4

I would go with elimination of one character from a base 10 numbering system gives you a base 9 numbering system. 9^3 = 729 permutations of a base 9 numbering system (a system with no number 3) with 3 digits, since 10^3 = 1000; 10^3 - 9^3 = 271 thus 271 numbers have a 3 in them.

DareN on

4

There are only 19 numbers between 1 and 100 containing '3': 3, 13, 23, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 43, 53, 63, 73, 83, 93. The same logic holds for 101-200, 301-400, 401-500, 501-600, 601-700, 701-800, 801-900, and 901-1000. That is 19 * 8 = 152. There are 20 numbers between 201-300 that contain a 3: 203, 213, 223, 230, 231, 232, 233, 234, 235, 236, 237, 238, 239, 243, 253, 263, 273, 283, 293, 300. That brings the sum up to 152 + 20 = 172. Finally, 99 numbers from 301-400 contain a 3. The final sum is 172+99=271. Of course, there are much more intelligent ways to count. For example, instead of counting how many numbers contain a 3, count how many do NOT contain a 3. That means that there are 9 possibilities for the 1st number (0-9 except 3), 9 possibilities for the 2nd number, and 9 possibilities for the 3rd number. Thus, there are 9 * 9 * 9 = 729 that do not contain a 3 which means that there are 1000 - 729 = 271 numbers that do contain a 3.

Interview Candidate on

2

Thanks Candidate, or more simply For 300 to 399 = 100 numbers For other x00 to x99 = 19 numbers each = 19 x 9 = 171 numbers Total of 271 numbers

A Learner on

0

http://brainteaserbible.com/interview-brainteaser-numbers-between-0-and-1000-at-least-one-5 basically the same thing

JDizz on

1

1000 does not contain a 3. So, count the number of 3 digit numbers without a three. There are 9 choices for the first entry, 9 for the second, and 9 for the third. So, there are 729 numbers without a 3, and 1000-729 = 271 with a 3.

KP on

0

I have a much simpler and faster method: let A be the cardinality of numbers between 1 and 1000 that contain a 3 and let A' be the cardinality of numbers between 1 and 1000 that do not contain a 3. There are 3 digits that can take the form of (0,1,2,3,4,5,6,7,8,9), so 10 possibilities.. To obtain A' cardinality we have 9 possibilities because 3 is excluded so A' = 9^3 = 729. Hence, the amount of numbers that don't have a 3 from 1 to 1000 is 729 so to obtain the amount that does contain at least one 3 is : 1000 - 729 = 271

Alex R on

0

lol im in 7th grade and this question is easy to me. first you count the numbers which DON't have a three. there are 9 choices for the first digit, 9 for the second digit, and 9 for the third. You probably noticed that this counts 0 to 999 instead of 1 to 1000 but its okay because, we count the same amount of numbers. 9^3=729 -1000=|271| (i like using absolute value cause it makes me look cool, its just a way to show "difference" in math)

ethan on

0

3xx 100 a3x a is not 3 here to reduce duplication, 10*9 ab3 a , b are neither not 3 , 9*9 =271

Henry on

0

You can use this formula to work it out for any power of 10: Tn+1=9Tn+10^n. Tn being the number of threes in the numbers between 1 and the previous power of 10. Tn+1 is simply saying the number of threes in the next power (the one you are working out). The 10^n is the power of 10 that you add on. This is the previous power of ten. You must start off knowing that there is one 3 between 1 and 10 For 100: Tn+1=9*1+10^1=19 For 1000: Tn+1=9*19+10^2=271 For 10 000: Tn+1=9*271+10^3=3439

Anonymous on

0

1000-1-9^3+1 = 271

StanLXR on

0

It is an infinite number 3.1, 3.11, 3.111, 3.11111

Anonymous on

6

300 numbers contain a 3, but you counted numbers of the form x33, 3x3, and 33x *twice* so you must subtract them 300-30=270, but you subtracted 333 once too many, so add it back 300-30+1=271. Answer is 271.

Anonymous on

0

271 does not seem right . I think it is 111. Aren't there just 11 numbers containing 3 between 0 an 100, like 3,13,23 30,33 etc. So between 0 and 1000 there are 111; after adding in for 300.

Bob on

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