Problem Solving

vinay1983 wrote:I presume this will be easy to most of the "solving" people here

How many 4-digit positive integers do not contain the digit 3 or 6?

A. 2401
B. 3584
C. 4096
D. 5040
E. 7200

Take the task of building 4-digit positive integers and break it into stages.

Stage 1: Choose a thousands digit
This can be 1,2,4,5,7,8,or 9, so we can complete stage 1 in 7 ways

Stage 2: Choose a hundreds digit
This can be 0,1,2,4,5,7,8,or 9, so we can complete stage 2 in 8 ways

Stage 3: Choose a tens digit
This can be 0,1,2,4,5,7,8,or 9, so we can complete stage 3 in 8 ways

Stage 4: Choose a units digit
This can be 0,1,2,4,5,7,8,or 9, so we can complete stage 4 in 8 ways

By the Fundamental Counting Principle (FCP) we can complete all 4 stages (and thus build a 4-digit positive integer) in (7)(8)(8)(8) ways

IMPORTANT: we don't really need to calculate the product (7)(8)(8)(8)
We can just recognize that the units digit will be 4. That is (7)(8)(8)(8) = ---4
Since answer choice B, is the only one with units digit 4, it must be correct.

Cheers,
Brent

Aside: For more information about the FCP, watch our free video: https://www.gmatprepnow.com/module/gmat-counting?id=775

Zoser wrote:What would the answer be if the question said 3 AND 6 instead of 3 or 6?

Do you mean to ask, "How many 4 digit positive integers do not contain the digit 3 AND 6 together (not necessarily in order)?"

If yes, then let's follow this.

# of 4 digit positive integers do not contain the digit 3 AND 6 together = Total # of 4 digit positive integers - Total # of 4 digit positive integers that contain the digit 3 AND 6 together

Total # of 4 digit positive integers = 9*10*10*10 = 9(10^3) = 9000

"Total # of 4 digit positive integers that contain the digit 3 AND 6 together" can be computed the following way.

A. For '3' appears before '6':

a. Say '3' is in the thousands place and '6' in the hundreds place.

3 6 _ _ : 1*1*10*10 = 100

b. Say '3' is in the thousands place and '6' in the tens place.

3 _ 6 _ : 1*10*1*10 = 100

c. Say '3' is in the thousands place and '6' in the unit place.

3 _ _ 6 : 1*10*10*1 = 100

-----------------------

d. Say '3' is in the hundreds place and '6' in the tens place.

_3 6 _ : 9*1*1*10 = 90

e. Say '3' is in the hundreds place and '6' in the unit place.

_3 _ 6 : 9*1*10*1 = 90

------------------------

f. Say '3' is in the tens place and '6' in the unit place.

_ _3 6 : 9*10*1*1 = 90

-------------------------

Total # in which '3' appears before '6' = 100 + 100+ 100 + 90 + 90 + 90 = 570;

Similarly, total # in which '6' appears before '3' = 570;

Total # of 4 digit positive integers that contain the digit 3 AND 6 together = 570 + 570 = 1140

Thus, # of 4 digit positive integers do not contain the digit 3 AND 6 together = 9000 - 1140 = 7860.

Hope this is clear.

Relevant book: Manhattan Review GMAT Number Properties Guide

-Jay
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