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
Magoosh Test Prep


Education is not the filling of a pail, but the lighting of a fire. — William Butler Yeats (1865 – 1939)

dnalost I am only getting ways. So total of 36 ways. I guess your one 1349 is not correct one

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

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.


Thanks,
GyM
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