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parveen110: Senior | Next Rank: 100 Posts; Posts: 91; Joined: Fri Jan 17, 2014 7:34 am; Thanked: 7 times

by parveen110 » Wed Jun 18, 2014 3:21 am

How many 4 digit numbers divisible by 5 can be formed using digits 0,1,2,3,4,5,6 and 6?
a. 220
b. 249
c. 432
d. 216
e. 288

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Source: Beat The GMAT — Problem Solving | Wed Jun 18, 2014 3:21 am

GMATinsight: Legendary Member; Posts: 1100; Joined: Sat May 10, 2014 11:34 pmFollowed by: 24 members; Location: New Delhi, India; Thanked: 205 times

by GMATinsight » Wed Jun 18, 2014 3:45 am

I hope you have mentioned digit "6" two times by mistakes and since there is not restriction so we will assume that repetition of digits is allowed.

To make number divisible by 5, the unit digit should be either 0 or 5 only

Case 1: If unit digit is fixed as zero

Then the choices at the remaining three places are 6 x 7 x 7 = 294

Case 2: If unit digit is fixed as Five

Then the choices at the remaining three places are 6 x 7 x 7 = 294

Total Such number = 294+294 = 588

But since that isn't the available choice therefore Assuming that the repetition of digits is not allowed

Case 1: If unit digit is fixed as zero

Then the choices at the remaining three places are 6 x 5 x 4 = 120

Case 2: If unit digit is fixed as Five

Then the choices at the remaining three places are 5 x 5 x 4 = 100

Total Such number = 120+100 = 220 ANSWER OPTION A

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GMATinsight: Legendary Member; Posts: 1100; Joined: Sat May 10, 2014 11:34 pmFollowed by: 24 members; Location: New Delhi, India; Thanked: 205 times

by GMATinsight » Wed Jun 18, 2014 3:47 am

I request you to post the right questions on the forum as it may cause the waste of your time and the waste of time and efforts of individuals posting the answers.

One Suggestion -- Refrain from the sources that offer such questions with numerous mistakes and refer to the appropriate source of material for your GMAT Preparation.

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GMATGuruNY: GMAT Instructor; Posts: 15539; Joined: Tue May 25, 2010 12:04 pmFollowed by: 1906 members; Location: New York, NY; Thanked: 13060 times; GMAT Score:790

by GMATGuruNY » Wed Jun 18, 2014 12:09 pm

To clarify the intent of the problem, I've added the portion in red.

parveen110 wrote:How many 4 digit numbers divisible by 5 can be formed using the digits 0, 1, 2, 3, 4, 5, and 6, if no digit can be repeated?
a. 220
b. 249
c. 432
d. 216
e. 288

Alternate approach:

Good = total - bad.

Total:
Number of options for the units digit = 2. (Must be 0 or 5.)
Number of options for the thousands digit = 6. (Any of the remaining 6 digits.)
Number of options for the hundreds digit = 5. (Any of the 5 remaining digits.)
Number of options for the tens digit = 4. (Any of the 4 remaining digits.).
To combine these options, we multiply:
2*6*5*4 = 240.

Bad:
From the 240 options above, a BAD option is as follows:
0XX5.
Number of options for the hundreds digit = 5. (Any of the 7 digits but 0 or 5.)
Number of options for the tens digit = 4. (Any of the 6 remaining digits but 0 or 5.)
To combine these options, we multiply:
5*4 = 20.

Good:
240-20 = 220.

The correct answer is A.

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parveen110: Senior | Next Rank: 100 Posts; Posts: 91; Joined: Fri Jan 17, 2014 7:34 am; Thanked: 7 times

by parveen110 » Fri Jun 20, 2014 2:13 am

parveen110 wrote:How many 4 digit numbers divisible by 5 can be formed using digits 0,1,2,3,4,5,6 and 6?
a. 220
b. 249
c. 432
d. 216
e. 288

The intent of this question is that one has to choose only from digits 0,1,2,3,4,5,6 and 6(yes, 6 occurs twice) and make a resulting four-digit number divisible by 5 also.

For e.g.: 6645 is acceptable since 6 and 6 are from the digits given in the question but 4535 is not as 5 is repeated twice.

Solution:

Case I:

# of numbers with one 6 or no 6, divisible by 5:

Numbers ending in 5= 5*5*4 = 100(Since 0 can't be placed at the left end)
Numbers ending in 0= 6*5*4 = 120

Total # of cases for numbers with one 6 or no 6, divisible by 5= 220

Case II:

# of numbers with two 6's divisible by 5:

a. Numbers ending with 0= 5C1*3!/2!= 15
b. Numbers ending with 5, not including zero: 4C1*3!/2! = 12
add numbers including zero (i.e. 6605 and 6065) = 2

Total # of cases for numbers with two 6's, divisible by 5= 15+12+2= 29

Possible cases with four digit numbers divisible by 5 that can be formed using digits 0,1,2,3,4,5,6 and 6 are 220+29 = 249

I haven't solved it very systematically, if anyone has a better approach, would love to know about it.

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GMATGuruNY: GMAT Instructor; Posts: 15539; Joined: Tue May 25, 2010 12:04 pmFollowed by: 1906 members; Location: New York, NY; Thanked: 13060 times; GMAT Score:790

by GMATGuruNY » Fri Jun 20, 2014 4:45 am

parveen110 wrote:
The intent of this question is that one has to choose only from digits 0,1,2,3,4,5,6 and 6(yes, 6 occurs twice) and make a resulting four-digit number divisible by 5 also.

I haven't solved it very systematically, if anyone has a better approach, would love to know about it.

To the 220 options counted in my post above, add the following:

Case 2: 6 appears twice
Number of options for the units digit = 2. (0 or 5.)
Number of positions available for the NON-6 digit = 3. (Any of the first 3 positions in the integer.)
Number of options for the non-6 digit = 5. (Any of the six remaining digits other than 6.)
To combine these options, we multiply:
2*3*5 = 30.
Of these 30 options, one case -- 0665 -- is not allowed.
Subtracting the one bad case, we get:
30-1 = 29.

Adding these 29 options to the 220 options counted in my post above, we get:
220+29 = 249.

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GMATinsight: Legendary Member; Posts: 1100; Joined: Sat May 10, 2014 11:34 pmFollowed by: 24 members; Location: New Delhi, India; Thanked: 205 times

by GMATinsight » Fri Jun 20, 2014 5:22 am

To make number divisible by 5, the unit digit should be either 0 or 5 only

Case 1: If unit digit is fixed as zero

With two "6"s The choice to fill the remaining three digits = 3C2 x 5 = 15
With all remaining 3 digits different, The choice to fill the remaining three digits = 6 x 5 x 4= 120

Total Such cases = 120+15 = 135

Case 2: If unit digit is fixed as Five

With two "6"s and one "0" The choice to fill the remaining three digits = 2 [6605 or 6065]
With two "6"s and without "0" The choice to fill the remaining three digits = 3C2 x 4 = 12

With all remaining 3 digits different, and one "0" The choice to fill the remaining three digits = 2 (ways to place zero) x 5 x 4= 40

With all remaining 3 digits different, and without "0" The choice to fill the remaining three digits = 3 x 5 x 4= 60

Total Such cases = 2+12+40+60 = 114

Total numbers = 135+114 = 249 ANSWER OPTION B

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