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# Let abcd be a general four-digit number and all the digits are non-zer

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Magoosh GMAT Instructor
Joined: 28 Dec 2011
Posts: 4476
Let abcd be a general four-digit number and all the digits are non-zer  [#permalink]

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27 Feb 2017, 12:43
3
6
00:00

Difficulty:

95% (hard)

Question Stats:

30% (02:52) correct 70% (02:39) wrong based on 146 sessions

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Let abcd be a general four-digit number and all the digits are non-zero. How many four-digits numbers abcd exist such that the four digits are all distinct and such that
a + b + c = d?

(A) 6
(B) 7
(C) 24
(D) 36
(E) 42

This is the first of a set of 15 challenging math questions. To see the whole set, as well as the OE for this question, see:
Challenging GMAT Math Practice Questions

Mike

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Mike McGarry
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Let abcd be a general four-digit number and all the digits are non-zer  [#permalink]

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Updated on: 05 Mar 2017, 21:37
2
1
mikemcgarry wrote:
Let abcd be a general four-digit number and all the digits are non-zero. How many four-digits numbers abcd exist such that the four digits are all distinct and such that
a + b + c = d?

(A) 6
(B) 7
(C) 24
(D) 36
(E) 42

This is the first of a set of 15 challenging math questions. To see the whole set, as well as the OE for this question, see:
Challenging GMAT Math Practice Questions

Mike

There are 7 combinations:
abcd
1236
2349
1359
1348
1247
1258
1269
Each can be rearranged in 6 ways. Total 6*7 = 42 .
C.

Originally posted by dnalost on 27 Feb 2017, 13:39.
Last edited by dnalost on 05 Mar 2017, 21:37, edited 1 time in total.
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Re: Let abcd be a general four-digit number and all the digits are non-zer  [#permalink]

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28 Feb 2017, 19:44
dnalost wrote:
mikemcgarry wrote:
Let abcd be a general four-digit number and all the digits are non-zero. How many four-digits numbers abcd exist such that the four digits are all distinct and such that
a + b + c = d?

(A) 6
(B) 7
(C) 24
(D) 36
(E) 42

This is the first of a set of 15 challenging math questions. To see the whole set, as well as the OE for this question, see:
Challenging GMAT Math Practice Questions

Mike

There are 7 combinations:
abcd
1236
2349
1349
1348
1247
1258
1269
Each can be rearranged in 6 ways. Total 6*7 = 42 .
C.

dnalost I am only getting ways. So total of 36 ways. I guess your one 1349 is not correct one
Intern
Joined: 06 Oct 2015
Posts: 45
Re: Let abcd be a general four-digit number and all the digits are non-zer  [#permalink]

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28 Feb 2017, 20:30
Could someone help explain this to me, I'm getting 12. The way I thought about it was finding the smallest value for D possible first given a,b,and c have to be unique. This would mean D=1+2+3; D=6.

You can arrange the 1,2,and 3 six different ways (6!) so that's six cases where D=6.

The next smallest value for D is 9 (2+3+4). You can arrange the 2,3,and 4 six ways again so you are at 12 numbers.
Intern
Joined: 17 Nov 2014
Posts: 4
Re: Let abcd be a general four-digit number and all the digits are non-zer  [#permalink]

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01 Mar 2017, 01:21
1+2+3=6 (1236)
Arranging 1,2,3 so six combinations.

1+2+4=7 (1247) arranging 1,2,4 six combinations

1+2+5 = 8 (1258) again 6 combinations

1+2+6 = 9 (1269) again 6 combinations

1+3+4 =8 (1348) ; 6 combinations

1+3+5 =9 (1359) ; 6 combinations

2+3+4 =9 (2349); 6 combinations

Hence , total 7 × 6 = 42 combinations of such 4 digit nos.

Sent from my Moto G (4) using GMAT Club Forum mobile app
Intern
Joined: 16 Jan 2014
Posts: 3
Let abcd be a general four-digit number and all the digits are non-zer  [#permalink]

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05 Mar 2017, 21:13
1
if d is 3 abc will have to be (111)- >> not possible as number are distinct
if d is 4 abc will have to be 112 or some other combination with 0 -->> Both not possible as neither 0 not repetition is allowed

Like this if we see unique combinations are possible only when d is
6,7,8,9

when d = 6 abc could be (123) -> can be arranged in 3! ways so total 6 ways
when d is 7 abc could be (124) -> can be arranged in 3! ways so total 6 ways
when d is 8 abc could be
(125) -> can be arranged in 3! ways so total 6 ways
(134)-> can be arranged in 3! ways so total 6 ways
when d is 9 abc could be
126 -> arranged in 3! ways = 6 ways
135 -> arranged in 3! ways = 6 ways
234 -> arranged in 3! ways = 6 ways

If we sum all these we get 42 combinations in total
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Re: Let abcd be a general four-digit number and all the digits are non-zer  [#permalink]

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03 Jul 2018, 10:27
Given abcd is a 4 digit number, such that a,b,c,d are distinct non zero digits & a + b + c = d

Hence lets start from the max value that d can take, d = 9, hence we have {a,b,c} = {6,2,1} or {5,3,1} or {4,3,2}

each combination of {a,b,c} can be arranged among themselves in 3! = 6 ways.

Therefore for d = 9, we have 6 * 3 = 18 ways

Similarly for d =8, we have {a,b,c} = {5,2,1} or {4,3,1}, hence 12 ways

for d = 7, we have {a,b,c} = {4,2,1}, hence 6 ways

& for d = 6, we have {a,b,c} = {3,2,1}, hence 6 ways.

we cannot take d < 6, since we don't have {a,b,c} such that they satisfy a+b+c = d

Hence Total # of 4 digit numbers required = 18 + 12 + 6 + 6 = 42

Thanks,
GyM
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Re: Let abcd be a general four-digit number and all the digits are non-zer  [#permalink]

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05 Jul 2018, 23:34
Is there any way to solve it without having to list the possible digits?
For sum =6 and 7, we have 1 combination (and its arrangements), for sum= 8 we have 2 and for sum=9 we have 3.
Is that a rule to predict the number of combinations for each sum, so we can just multiply them by their arrangements of 3!, without having to figure out which they are?
mikemcgarry
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Re: Let abcd be a general four-digit number and all the digits are non-zer  [#permalink]

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02 Dec 2019, 14:29
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Re: Let abcd be a general four-digit number and all the digits are non-zer   [#permalink] 02 Dec 2019, 14:29
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