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mikemcgarry
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Let abcd be a general four-digit number and all the digits are non-zer 27 Feb 2017, 12:43
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95% (hard)

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38% (02:57) correct 62% (02:34) wrong based on 345 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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Let abcd be a general four-digit number and all the digits are non-zer Updated on: 05 Mar 2017, 21:37
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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mikemcgarry
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.
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Let abcd be a general four-digit number and all the digits are non-zer 28 Feb 2017, 19:44
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dnalost
mikemcgarry
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
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Let abcd be a general four-digit number and all the digits are non-zer 28 Feb 2017, 20:30
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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.
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Let abcd be a general four-digit number and all the digits are non-zer 01 Mar 2017, 01:21
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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.


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Let abcd be a general four-digit number and all the digits are non-zer 05 Mar 2017, 21:13
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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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Let abcd be a general four-digit number and all the digits are non-zer 03 Jul 2018, 10:27
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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


Answer E.


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Let abcd be a general four-digit number and all the digits are non-zer 05 Jul 2018, 23:34
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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?
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Let abcd be a general four-digit number and all the digits are non-zer 26 Mar 2023, 21:28
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mikemcgarry
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 :-)

Focus on first 3 digits. The last digit will automatically be the sum of first 3.
Since all digits are distinct, the smallest sum of first 3 will be 1 + 2 + 3 = 6.
Since their sum must be a digit, it cannot be more than 9.

The possible combinations with 1 and 2 are:
1, 2, 3
1, 2, 4
1, 2, 5
1, 2, 6 (sum = 9)

The possible combinations with 1 and 3 are:
1, 3, 4
1, 3, 5 (sum = )

There are no possible combinations with 1 and 4 (or higher) since 1, 4, 5 will give 10 as last digit which is not possible.

The possible combinations with 2 and 3 are:
2, 3, 4 (already sum is 9)

Hence there are 7 combinations each of which can be arranged in 3! ways (since each arrangement is a different number)
Total such numbers = 7 * 3! = 42

Answer (E)
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Let abcd be a general four-digit number and all the digits are non-zer 04 Jun 2024, 12:36
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mikemcgarry
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 :-)
­a,b,c,d all have to be distinct. Furthermore, a+b+c=d but not necessarily in the order a,b,c so they can be arranged in 3!=6 ways.
The possible sets of values that a,b, and c can take so that d is less than or equal to 9 are (1,2,3), (1,2,4), (1,2,5), (1,2,6), (1,3,4), (1,3,5), (2,3,4) i.e. 7 sets.
Therefore 7*6!=42. Option (E) is correct.
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Let abcd be a general four-digit number and all the digits are non-zer 22 Jun 2025, 07:35
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