How many even 3 digit integers greater than 700 with distinct non zero

Question

How many even 3 digit integers greater than 700 with distinct non zero digits are there ?

A. 729
B. 243
C. 108
D. 88
E. 77

Bunuel · Accepted Answer

Notice that we are told that no digit in these numbers could be zero. So, we have only 9 digits to use for  XYZ , where  Z  is an even number.

If the first digit (X) is 7 or 9 ( 2 values ) then the third digit (Z) can take all  4 values  (2, 4, 6, or 8) and the second digit (Y) can take  7 values  (9 minus two digits we already used). So, for this case we have 2*4*7=56 numbers;

If the first digit (X) is 8 ( 1 value ) then the third digit (Z) can take only  3 values  (2, 4, or 6,) and the second digit (Y) can take  7 values  (9 minus two digits we already used). So, for this case we have 1*3*7=21 numbers;

Total: 56+21=77 numbers.

Answer: E.

Hope it's clear.

hardworker_indian · Answer

Exactly, the numbers need to be distinct. My answer:
The below are no. of ways of filling units digit, tenth digit and hundredth digit, in that order.

701-799 = 4.7.1 = 28
800-899 = 3.7.1 = 21
900-999 = 4.7.1 = 28
Total = 77

E

cyberjadugar · Answer

Hi,

My soltuion is as follows:
You have to find the even (3 digit) numbers greater than 700, with non-zero digits as well as distict digits. So, available digits would be (1 to 9)
Even numbers starting with 7 = 1*7*4 = 28 (Hundredth digit is 7 - so, only 1 choice, unit digit can be (2, 4, 6, 8). Now tens digit will not be 7 & a digit chosen at units place - 7 possibilities)
Even numbers starting with 8 = 1*7*3 = 21
Even numbers starting with 9 = 1*7*4 = 28
Total numbers = 77

Regards,

venksune · Answer

1. Number of 3 digit even numbers >than 700 and <1000 = 149. We can use Arthmetic progression to find 149. How? 702 is the first even number and 998 is the last even number. The difference is 2 between the numbers. So, we have 702 + (n-1)2 = 998 or n=149.

2. Number of 3 digit even numbers that need to eliminated = 41, How? 702, 704, 706, 708, 710 => 5 numbers x 3 (for 70x, 80x and 90x series)= 15. Also, 720, 730...790 - 8 numbers x 3 = 24. Also 800 and 900 - 2 numbers. So, 15 + 24 + 2 = 41.

3. So net result is 149-41 = 108. 

4. C is the answer.

Alex_NL · Answer

I have C as well.

the only hundred digit that are possible are 7, 8 and 9 (3)
the only ten digit that are possible are 1, 2, 3, 4, 5, 6, 7, 8, 9 (9)
the only unit digit that are possible are 2, 4, 6, 8 (4)

what I did is: 3 * 9 * 4 = 108

With a lot of practise these questions are better to understand.

Regards,

Alex

gmat800 · Answer

How many even 3 digit integers greater than 700 with distinct non zero digits are there ? 
A 729 
B 243 
C 108 
D 88 
E 77

I agree with ur explanation but using another method i m getting D...plz tell me wat is going wrong

the no.s are even, distinct n each place has a non-zero no.....

If the 1st hundreds digit is 7.......units can be (2,4,6,8)...i.e 4 possibilities....n tens can be any of the remaining 8 digits....(7 is in hundredes n one even no. occupies tens place)....so total options 8*4=32

for 8 in hundreds........8*3 options.....coz 8 in hundreds is already an even no.....so units can be any of the 3 remaining even....so that gives 24

for 9 in hundreds...again 8*4=32


so 32+24+32=88.........

although it seems long and may not seem to be a GMAT question which takes less time........it took no more than 30 sec to solve....

why this difference in answers....plz clarify!!!!

twixt · Answer

OA is E

Please note that these 3 digits have to be distinct so 771 for instance is not valid

It took me 10 minutes after 2 mistakes :

even integers in the range : 149
0 as last digit (including 00) = 29
0 as tens digit = 12 (702 704 706 708, ...)
Distinct digits :
7>9 : hundreds digit
1>9 : tens digit
2- 4-6-8 :units digit

=31 

Result = 149-29-12-31=77

hb · Answer

I am clear on the explanation provided by Bunuel for this question. However I am trying to understand the original answer provided using Arithmetic Progression. Where does the ( n-1)*2 come from ? Please explain and clarify.

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KarishmaB · Answer

First note that the explanation you are referring to has incorrect answer. Also, this method of solving is extremely treacherous - there are far too many limitations on the acceptable numbers to use this method. You must follow Bunuel's method for this question.

As for (n - 1)*2, we are given this particular sequence: 702, 704, 706, ..... 996, 998
We need to find how many numbers there are here. One way of doing that is using arithmetic progression concept. 
nth term = First term + (n - 1)*Common difference
(last term) 998 = 702 + (n - 1)*2
Here, n will be the total number of terms.

FZ · Answer

AbhiJ · Answer

XYZ

No of ways to enter hundreds digit = 3 ( 7, 8, 9 ; Because the number has to be more than 700)

No of ways to enter units digit = 2, 4, 6, 8 (We cannot enter 0 as all digits are non zero, and the numbers are to be even so 1, 3, 5, 7, 9 are out)

No of ways to enter tenth digit = 10 - 3 (All digits are to be distinct. Hence we cannot have 0 and the two digits used in units and hundreds place).

Hence total number of digits =  3 * 7 * 4  = 84.

Wait if we have 8 on the units and hundreds place then we are repeating digits. Hence we must remove them as per our initial conditions that all digits are distinct.

No of such numbers

No of ways to enter hundreds digit = 1 (As were considering 8 in units and hundreds place)

No of ways to enter tenth digit = 7 (All digits are to be distinct. Hence we cannot have 0 and the two digits used in units and hundreds place).

No of ways to enter tenth digit = 1 (As were considering 8 in units and hundreds place)
 
No of such numbers =  1 * 7 * 1 = 7

Answer =  84 - 7 = 77

12Sums · Answer

Hi Bunuel,

If the first digit (X) is 7 or 9 ( 2 values ) then the third digit (Z) can take all  4 values  (2, 4, 6, or 8) and the second digit (Y) can take  7 values  (9 minus two digits we already used).

Here what about 2 as second digit. For eg: 722. In this case second digit can't take 2. How are these numbers (such as 744,766,788,922,944) removed from 2*4*7 = 56 numbers?

If the first digit (X) is 8 ( 1 value ) then the third digit (Z) can take only  3 values  (2, 4, or 6,) and the second digit (Y) can take  7 values  (9 minus two digits we already used).

Here i didn't get the last part "(9 minus two digits we already used)" . Which 2 digits are we talking about? Here also how are we not considering 822,844 etc using the above approach?

Bunuel · Answer

We are told that X, Y and Z must be distinct non zero digits, thus XYZ cannot be 8 22  or 8 44 .

As for your other question:  9 minus two digits we already used  means that for Y we cannot use a digit we used for X and a digit we used for Z, so we can use only 9-2=7 values of it.

Hope it's clear.

sanguine2016 · Answer

Hi,

How many even 3 digit integers greater than 700 with distinct non zero digits are there ?

Now, we have to find 3 digit  even integers (Last digit 2,4,6 or 8) greater than 700 (i.e., first digit has to be 7,8 or 9).

So,
1.) No. of integers greater than 700 & less than 799, satisfying above mentioned conditions are:
==> 7 _ _
==> 7 _ (2,4,6,8)
==> 1 * 4 * 7 = 28 (4 digits--2,4,6, or 8 can be used to fill up the units digit and the tens place can be filled by remaining 7 digits as digits are distinct.).

2.) No. of integers greater than 800 & less than 899, satisfying above mentioned conditions are:
==> 8 _ _
==> 8 _ (2,4,6)
==> 1 * 3 * 7 = 21 (3 digits--2,4 or 6 can be used to fill up the units digit and the tens place can be filled by remaining 7 digits as digits are distinct.).

3.) No. of integers greater than 900 & less than 999, satisfying above mentioned conditions are:
==> 9 _ _
==> 9 _ (2,4,6,8)
==> 1 * 4 * 7 = 28 (4 digits--2,4,6, or 8 can be used to fill up the units digit and the tens place can be filled by remaining 7 digits as digits are distinct.).

So, adding them all we get total integers as : 28 + 21 + 21 = 77. 
So, (E).

I hope that the explanation is clear.

GMATinsight · Answer

There are only Three possible cases

In each case Hundreds places is fixed so we fill the unit's place first in order to make the number even and at last we fill the Ten's digit place

Case 1 : 7 __ __   When Hundreds digit is 7
  7    _7 choices (remaining 7 out of 9 non-zero digits)_ * _4 choices (2,4,6,8)_  = 28 Numbers

Case 2 : 8 __ __   When Hundreds digit is 8
  8    _7 choices (remaining 7 out of 9 non-zero digits)_ * _3 choices (2,4,6)_  = 21 Numbers

Case 3 : 9 __ __   When Hundreds digit is 9
  9    _7 choices (remaining 7 out of 9 non-zero digits)_ * _4 choices (2,4,6,8)_  = 28 Numbers

Total Such Numbers = 28 + 21 + 28 = 77 Numbers

Answer: Option E

SVaidyaraman · Answer

Let us try for an approximate answer since the values in the choices are far apart by relaxing some constraints.
 
The last digit can change in 4 ways
The middle digit can change in 8 ways
The first digit can change in 3 ways

This is 4*8*3 = 96 only even with relaxation of constraints. Choices A,B and C can be eliminated

There are some more numbers which will not be included.

when first digit is 7, second digit cannot be 7. So the possibilities reduce to 4*8*2 + 4*7*1=92
Similarly when first digit is 9, second digit cannot be 9 and the possibilities reduce by 4 more = 88.

The number of possibilities will be less than 88 because when first digit is 8, there are other impossible numbers.

So now we can pick the answer as E

spence11 · Answer

Quickly decide to use brute force. Count the number of matching cases in each group of ten numbers.

70 => 0
71 => 4
72 => 3
73 => 4
74 => 3
75 => 4
76 => 3
77 => 0
78 => 3
79 => 4
80 => 0
81 => 3
82 => 2
83 => 3
84 => 2
85 => 3
86 => 2
87 => 3
88 => 0
89 => 3
90 => 0
91 => 4
92 => 3
93 => 4
94 => 3
95 => 4
96 => 3
97 => 4
98 => 3
99 => 0

4*8 + 3*13 + 2*3 = 77

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SVaidyaraman · Answer

These types can be handled by considering the number of possibilities for each digit. Start with the one with the most constraints. 
1. Last digit can be 2,4,6,8 when the first digit is 7 or 9 and 2,4,6 when the first digit is 8. So there are 4 possibilities in the first case and 3 possibilities in the second case for the last digit.
2. Middle digit cannot be one of the first and last digits and cannot be 0. So there are 7 possibilities
3. When first digit is 7 or 9, then we have (4*7) + (4*7)=56
4. When first digit is 8, we have 3*7=21
5. Total is (3)+(4)= 77

spy2206 · Answer

carcass   Bunuel 
 "distinct non zero digits" - please explain this. does it mean three different digits in the number which is not zero or different numbers with no zero?

How many even 3 digit integers greater than 700 with distinct non zero

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