GMAT Question of the Day - Daily to your Mailbox; hard ones only

 It is currently 06 Dec 2019, 01:42

GMAT Club Daily Prep

Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

If x represents the sum of all the positive three-digit

Author Message
TAGS:

Hide Tags

Intern
Joined: 02 Oct 2009
Posts: 15
If x represents the sum of all the positive three-digit  [#permalink]

Show Tags

Updated on: 30 Jul 2012, 04:29
6
66
00:00

Difficulty:

95% (hard)

Question Stats:

47% (02:10) correct 53% (02:18) wrong based on 985 sessions

HideShow timer Statistics

If x represents the sum of all the positive three-digit numbers that can be constructed using each of the distinct nonzero digits a, b, and c exactly once, what is the largest integer by which x must be divisible?

(A) 3
(B) 6
(C) 11
(D) 22
(E) 222

Originally posted by rvthryet on 13 Nov 2009, 20:35.
Last edited by Bunuel on 30 Jul 2012, 04:29, edited 2 times in total.
Math Expert
Joined: 02 Sep 2009
Posts: 59561
Re: this is what it has come down to  [#permalink]

Show Tags

13 Nov 2009, 21:34
49
22
rvthryet wrote:
If x represents the sum of all the positive three-digit numbers that can be constructed using each of the distinct nonzero digits a, b, and c exactly once, what is the largest integer by which x must be divisible?
(A) 3
(B) 6
(C) 11
(D) 22
(E) 222

I have never really understood the thinking behind this...
OA E

Using THREE non-zero digits a,b,c only, we can construct 3!=6 numbers: abc, acb, bac, bca, cab, cba. Their sum will be:

$$x=(100a+10b+c)+(100a+10c+b)+(100b+10a+c)+(100b+10c+a)+(100c+10a+b)+(100c+10b+a)=$$
$$=200*(a+b+c)+20*(a+b+c)+2*(a+b+c)=$$
$$=222*(a+b+c)$$

Largest integer by which x MUST be divisible is $$222$$.

General Discussion
Manager
Joined: 05 Jun 2009
Posts: 65
Re: this is what it has come down to  [#permalink]

Show Tags

13 Nov 2009, 20:40
where did this question come from wow I have like no idea where to begin I would assume 123 and 987 which are two combinations are both both divisible by 3 as the GCD so 3?
A?
Manager
Joined: 11 Sep 2009
Posts: 115
Re: this is what it has come down to  [#permalink]

Show Tags

13 Nov 2009, 21:47
Bunuel wrote:
rvthryet wrote:
If x represents the sum of all the positive three-digit numbers that can be constructed using each of the distinct
nonzero digits a, b, and c exactly once, what is the largest integer by which x must be divisible?
(A) 3
(B) 6
(C) 11
(D) 22
(E) 222

I have never really understood the thinking behind this...
OA E

Using THREE non-zero digits a,b,c only, we can construct 3!=6 numbers: abc, acb, bac, bca, cab, cba. Their sum would be:

$$x=(100a+10b+c)+(100a+10c+b)+(100b+10a+c)+(100b+10c+a)+(100c+10a+b)+(100c+10b+a)=$$
$$=200*(a+b+c)+20*(a+b+c)+2*(a+b+c)=$$
$$=222*(a+b+c)$$

Largest integer by which x MUST be divisible is $$222$$.

Good explanation, exactly how I solved it. I love questions with elegant solutions like this. +1
CEO
Joined: 17 Nov 2007
Posts: 2996
Concentration: Entrepreneurship, Other
Schools: Chicago (Booth) - Class of 2011
GMAT 1: 750 Q50 V40
Re: this is what it has come down to  [#permalink]

Show Tags

13 Nov 2009, 21:59
4
1
1
We can also solve this one without math using symmetry: hundreds, tens and units are symmetric, so sum can be written as (y)*111. We need to check that y is even. For example, for fixed a at hundred position, there is two bc,cb combinations. Therefore, a is included twice (even number of times) into sum of hundreds. So, it is 222

By the way, it is the first time when I add something after Bunuel
_________________
HOT! GMAT Club Forum 2020 | GMAT ToolKit 2 (iOS) - The OFFICIAL GMAT CLUB PREP APPs, must-have apps especially if you aim at 700+
Manager
Joined: 10 Aug 2009
Posts: 99

Show Tags

03 Mar 2010, 04:26
1
E

Maybe there is a faster way to do it but I did it like this:

How many ways can you arrange abc?
abc
acb
bac
bca
cab
cba

which are equivalent to:
100a + 10b + c
100a + 10c + b
100b + 10a + c
100b + 10c + a
100c + 10a + b
100c + 10b + a

if you add them all together you get 222a + 222b + 222c
Intern
Joined: 03 Dec 2010
Posts: 21
Re: If x represents the sum of all the positive three-digit  [#permalink]

Show Tags

31 Mar 2012, 03:27
To Bunuel,

I've gone thorugh ur notes for each Quant topic and I try to solve topic wise questions from gmatclub. Sometimes I'm not able to figure out how to start with the problem, or I should say how to apply the properties learned since, the techniques you give in your solution for a given problem are not there in properties or formulaes. What do you recommend ? I plan to give my Gmat nxt mnth end. This Tuesday, Veritas prep test I took I scored 600, Q44, verbal 33.

Kindly assist.
Thanks.
Manager
Joined: 26 Jul 2011
Posts: 77
Location: India
WE: Marketing (Manufacturing)
Re: If x represents the sum of all the positive three-digit  [#permalink]

Show Tags

30 Jul 2012, 04:26
.Though I was able to solve it (in a random way), but was unable to come up with a concrete approach.
@NickK kudos for that perfect one. This is how I did.....

The question asked for the largest divisor and thus we need to form 6 largest number that could be made using 3 distinct nonzero digits....987+978+897+879+798+789 = 5328...start from the largest number provided in the answer..222 divides 5328 completely hence is the answer
Manager
Joined: 23 May 2013
Posts: 92
Re: If x represents the sum of all the positive three-digit  [#permalink]

Show Tags

01 Oct 2013, 06:40
ratinarace wrote:
.Though I was able to solve it (in a random way), but was unable to come up with a concrete approach.
@NickK kudos for that perfect one. This is how I did.....

The question asked for the largest divisor and thus we need to form 6 largest number that could be made using 3 distinct nonzero digits....987+978+897+879+798+789 = 5328...start from the largest number provided in the answer..222 divides 5328 completely hence is the answer

Agree, substitution works the best for 'must be true' problems.
Manager
Joined: 29 Aug 2013
Posts: 69
Location: United States
GMAT 1: 590 Q41 V29
GMAT 2: 540 Q44 V20
GPA: 3.5
WE: Programming (Computer Software)
Re: this is what it has come down to  [#permalink]

Show Tags

02 Oct 2013, 00:27
Bunuel wrote:
rvthryet wrote:
If x represents the sum of all the positive three-digit numbers that can be constructed using each of the distinct nonzero digits a, b, and c exactly once, what is the largest integer by which x must be divisible?
(A) 3
(B) 6
(C) 11
(D) 22
(E) 222

I have never really understood the thinking behind this...
OA E

Using THREE non-zero digits a,b,c only, we can construct 3!=6 numbers: abc, acb, bac, bca, cab, cba. Their sum will be:

$$x=(100a+10b+c)+(100a+10c+b)+(100b+10a+c)+(100b+10c+a)+(100c+10a+b)+(100c+10b+a)=$$
$$=200*(a+b+c)+20*(a+b+c)+2*(a+b+c)=$$
$$=222*(a+b+c)$$

Largest integer by which x MUST be divisible is $$222$$.

Hi Bunuel,
Can you please explain me what will be the value of "x" in this question. If it were asked what is the value of x?

Thanks!
Math Expert
Joined: 02 Sep 2009
Posts: 59561
Re: this is what it has come down to  [#permalink]

Show Tags

02 Oct 2013, 03:12
shameekv wrote:
Bunuel wrote:
rvthryet wrote:
If x represents the sum of all the positive three-digit numbers that can be constructed using each of the distinct nonzero digits a, b, and c exactly once, what is the largest integer by which x must be divisible?
(A) 3
(B) 6
(C) 11
(D) 22
(E) 222

I have never really understood the thinking behind this...
OA E

Using THREE non-zero digits a,b,c only, we can construct 3!=6 numbers: abc, acb, bac, bca, cab, cba. Their sum will be:

$$x=(100a+10b+c)+(100a+10c+b)+(100b+10a+c)+(100b+10c+a)+(100c+10a+b)+(100c+10b+a)=$$
$$=200*(a+b+c)+20*(a+b+c)+2*(a+b+c)=$$
$$=222*(a+b+c)$$

Largest integer by which x MUST be divisible is $$222$$.

Hi Bunuel,
Can you please explain me what will be the value of "x" in this question. If it were asked what is the value of x?

Thanks!

We cannot say what x is.

If a, b, and c, are 1, 2, and 3 respectively, then x = 123 + 132 + 213 + 231 + 312 + 321 = 1,332 = 6*222 (the least possible value of x).
...
If a, b, and c, are 7, 8, and 9 respectively, then x = 789 + 798 + 879 + 897 + 978 + 987 = 5,328 = 24*222 (the greatest possible value of x).

Hope it helps.
Manager
Joined: 29 Aug 2013
Posts: 69
Location: United States
GMAT 1: 590 Q41 V29
GMAT 2: 540 Q44 V20
GPA: 3.5
WE: Programming (Computer Software)
Re: If x represents the sum of all the positive three-digit  [#permalink]

Show Tags

02 Oct 2013, 03:27
Hi Bunuel,

Thanks for the clarification. I thought it is the sum of all such 3-digit numbers that have distinct numbers.

What in the case "x is the sum of all the 3-digit numbers that have distinct numbers". How do you calculate the value of x in such case. I tried many things but couldn't work it out.

I saw such type of question recently where x was required to be calculated but the digits could be repeated and that made it simple. But I couldn't figure out with this restriction. Could you please help me out on that?

Thanks,
Shameek
Intern
Joined: 08 Oct 2012
Posts: 1
Location: United States
Concentration: General Management, Technology
GPA: 2.3
WE: Engineering (Computer Software)
Re: If x represents the sum of all the positive three-digit  [#permalink]

Show Tags

26 Oct 2014, 12:52
1
Shamee, to solve the problem in a simpler manner why don't you assume the numbers a, b and c to be 1, 2 and 3 respectively?

Thus, the distinct numbers that can be formed would be -
123
132
213
231
312
321

If you sum these up you get a total of 1332.

Then proceed to plug in the answer options to find the greatest number that divides 1332.

From the options -
(A) 3 - Yes
(B) 6 - Yes
(C) 11 - No
(D) 22 - No
(E) 222 - Yes

Clearly, since 222 is the greatest, E is the right option.
Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 9849
Location: Pune, India
Re: If x represents the sum of all the positive three-digit  [#permalink]

Show Tags

18 Jan 2016, 22:29
1
pritishpratap wrote:
Shamee, to solve the problem in a simpler manner why don't you assume the numbers a, b and c to be 1, 2 and 3 respectively?

Thus, the distinct numbers that can be formed would be -
123
132
213
231
312
321

If you sum these up you get a total of 1332.

Then proceed to plug in the answer options to find the greatest number that divides 1332.

From the options -
(A) 3 - Yes
(B) 6 - Yes
(C) 11 - No
(D) 22 - No
(E) 222 - Yes

Clearly, since 222 is the greatest, E is the right option.

Here is the catch in "assuming values" in this question:
The question is a "must be true" question. How do you know that what holds for values 1, 2 and 3 will be true for values say 2, 3 and 7 too? What if sum of numbers formed by 2, 3 and 7 is not divisible by 222? You do need to apply logic to confirm "must be true".
_________________
Karishma
Veritas Prep GMAT Instructor

Senior SC Moderator
Joined: 14 Nov 2016
Posts: 1347
Location: Malaysia
Re: If x represents the sum of all the positive three-digit  [#permalink]

Show Tags

31 Mar 2017, 00:56
1
rvthryet wrote:
If x represents the sum of all the positive three-digit numbers that can be constructed using each of the distinct nonzero digits a, b, and c exactly once, what is the largest integer by which x must be divisible?

(A) 3
(B) 6
(C) 11
(D) 22
(E) 222

Bunuel, This question has been wrongly tagged. The original source is Manhattan Prep, Challenge Problems (2002, December 2, Three-Digit Divisibility).
Attachments

Untitled.jpg [ 64.06 KiB | Viewed 13126 times ]

_________________
"Be challenged at EVERY MOMENT."

“Strength doesn’t come from what you can do. It comes from overcoming the things you once thought you couldn’t.”

"Each stage of the journey is crucial to attaining new heights of knowledge."

Math Expert
Joined: 02 Sep 2009
Posts: 59561
Re: If x represents the sum of all the positive three-digit  [#permalink]

Show Tags

31 Mar 2017, 01:50
ziyuen wrote:
rvthryet wrote:
If x represents the sum of all the positive three-digit numbers that can be constructed using each of the distinct nonzero digits a, b, and c exactly once, what is the largest integer by which x must be divisible?

(A) 3
(B) 6
(C) 11
(D) 22
(E) 222

Bunuel, This question has been wrongly tagged. The original source is Manhattan Prep, Challenge Problems (2002, December 2, Three-Digit Divisibility).

Edited. Thank you.
Manager
Joined: 30 Dec 2016
Posts: 230
GMAT 1: 650 Q42 V37
GPA: 4
If x represents the sum of all the positive three-digit  [#permalink]

Show Tags

09 Oct 2017, 03:10
I solved it in a bit different way. Not sure if this is correct.

There are 3 numbers a, b, c so there can be 6 arrangements of these numbers. So, 6 possible numbers are there ( just to be sure i am not missing)

abc
+acb
+bac
+bca
+cba
+cab
6a+6b+6c

Now if we factor out 6 --> 6( a+b+c ) from this we know the answer must be a multiple of 6. It can not be 6 as a+b+c wpuld yield some integer and 6*someinteger > 6 .
So the only possible outcome is 222 which is a multiple of 6 other than Choice B.
Manager
Joined: 02 Jul 2016
Posts: 134
Location: India
GMAT 1: 650 Q49 V28
GPA: 4
Re: If x represents the sum of all the positive three-digit  [#permalink]

Show Tags

22 Aug 2018, 06:30
1
Bunuel wrote:
rvthryet wrote:
If x represents the sum of all the positive three-digit numbers that can be constructed using each of the distinct nonzero digits a, b, and c exactly once, what is the largest integer by which x must be divisible?
(A) 3
(B) 6
(C) 11
(D) 22
(E) 222

I have never really understood the thinking behind this...
OA E

Using THREE non-zero digits a,b,c only, we can construct 3!=6 numbers: abc, acb, bac, bca, cab, cba. Their sum will be:

$$x=(100a+10b+c)+(100a+10c+b)+(100b+10a+c)+(100b+10c+a)+(100c+10a+b)+(100c+10b+a)=$$
$$=200*(a+b+c)+20*(a+b+c)+2*(a+b+c)=$$
$$=222*(a+b+c)$$

Largest integer by which x MUST be divisible is $$222$$.

I tried solving it using the formula
(n-1)!*(sum of the digits)*(111…..n times)

n =3
sum of the digits = a+b+c
so
(3-1)! *(a+b+c)*111
222*(a+b+c)

Hence clearly the number will be divisible by 222
CrackVerbal Quant Expert
Joined: 12 Apr 2019
Posts: 312
Re: If x represents the sum of all the positive three-digit  [#permalink]

Show Tags

06 Sep 2019, 07:46
1
In my humble opinion, although this is not exactly a problem that is tested extensively on the GMAT, there’s no harm in solving questions like these to strengthen your concepts of Permutations with numbers.

If a, b and c are the digits from which we have to form 3-digit numbers such that all the digits are non-zero and distinct, then, the number of ways of doing that is given by,

(a+b+c) * (3-1)! * (111).

In general, if there are ‘n’ distinct digits using which we have to form ‘n’ digit numbers where the digits are non-zero and distinct, then the number of ways of doing this is given by,
(Sum of the n digits) *(n-1)! *(1111….. n times).

The reason for this is as follows:
When a comes in the units place, its place value will be a. There will be (n-1)! numbers where a will be the units digit. Therefore, the sum of these values will be a * (n-1)!

When a comes in the tens place, its value will be 10a. There will be (n-1)! numbers where a will be the tens digit. The sum of these values will be 10a * (n-1)!

When a comes in the hundreds place, its value will be 100a. There will be (n-1)! numbers where a will be the hundreds digit. The sum of these values will be 100a * (n-1)!.

The sum of all these values will be (n-1)! * a (100 + 10 + 1) = (n-1)! * a * 111.

Similarly, for the other digits, b, c, d and so on, the sum of the values can be worked out as shown above.

In our case, since the sum total is (a+b+c) * 2! * 111 which is nothing but (a+b+c) * 222, we can say that the sum will be definitely divisible by 222.
The correct answer option is E.

Hope this helps!
_________________
Re: If x represents the sum of all the positive three-digit   [#permalink] 06 Sep 2019, 07:46
Display posts from previous: Sort by