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

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Question Stats: 46% (02:09) correct 54% (02:19) wrong based on 967 sessions

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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 V
Joined: 02 Sep 2009
Posts: 58402
Re: this is what it has come down to  [#permalink]

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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 $$222$$.

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

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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: 116
Re: this is what it has come down to  [#permalink]

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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
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GMAT 1: 750 Q50 V40 Re: this is what it has come down to  [#permalink]

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3
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 _________________
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Manager  Joined: 10 Aug 2009
Posts: 101
Re: Testing number properties  [#permalink]

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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]

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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.
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Re: If x represents the sum of all the positive three-digit  [#permalink]

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.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
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Re: If x represents the sum of all the positive three-digit  [#permalink]

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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.
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Re: this is what it has come down to  [#permalink]

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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 V
Joined: 02 Sep 2009
Posts: 58402
Re: this is what it has come down to  [#permalink]

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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.
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Re: If x represents the sum of all the positive three-digit  [#permalink]

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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
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Re: If x represents the sum of all the positive three-digit  [#permalink]

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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.
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Re: If x represents the sum of all the positive three-digit  [#permalink]

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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".
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Re: If x represents the sum of all the positive three-digit  [#permalink]

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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).
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Re: If x represents the sum of all the positive three-digit  [#permalink]

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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.
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If x represents the sum of all the positive three-digit  [#permalink]

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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.
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Re: If x represents the sum of all the positive three-digit  [#permalink]

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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
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Re: If x represents the sum of all the positive three-digit  [#permalink]

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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
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