Addition Problem
A B + C D = A A A where A B and C D are two-digit numbers and A A A is a three-digit number; A , B , C , and D are distinct positive integers. In the addition problem above, what is the value of D ?
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3 solutions
if A=1,B=8,C=9 and D=3 then AB+CD=AAA SO D=3
AAA is automatically 111, so A = 1 (B,C, & D)'s possible values are 2,3,4,5,6,7,8, & 9 here's what i found: AB + CD = AAA 18 + 93 = 111
so D is equal to 3!
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THE ONLY POSSIBLE VALUE THAT GIVEN TO A IS 1, SO AB+CD=111 , FROM GIVEN OPTION D CAN TAKE EITHER 3 OR 9 .AND IF WE PUT D=9 IN THE ABOVE EQ. THEN WE CANN'T GET L.H.S =111 SO ONLY OPTION REMAINS FOR THE VALUE OF D IS 3.SO THE EQ, WILL BE 18+93=111 ...........................:)
18 + 93 = 111 ... so the answer could have been 3 .... But, in this case , B + CD = 8 + 93 = 101 (now, applying brute force, we get, D=3,4,5,6,7,8)
18+93=111,,as we know A=1is only possibility.. So taking A=1 and we need B+D=11 soB=8,D=3..@B+D=11. And C=4
why the answer cannot be determined? here the answer is D=3
B can not be 9, C already equals 9
S i n c e t w o , 2 _ d i g i t n u m b e r s a d d t o t h i r d d i g i t , t h e t h i r d d i g i t m u s t b e o n l y 1 . S o A = 1 .
T h u s t h e s u m i s 1 1 1 . W e s h o u l d g e t 1 a t u n i t p l a c e . T h a t i s B + D = 1 1 , w e c a n h a v e
8 + 3 = 1 1 O R 2 + 9 = 1 1 . S o D c a n b e 3 O R 9 . C a n n o t b e d e t e r m i n e d .
The equation should be B + C D = 1 0 1 instead of B + D = 1 1 ?
You should further state explicit examples of different cases which yield a different value of D . The reason is that just because something is (currently) possible, doesn't mean that it can actually occur.
FYI - To type equations in Latex, you just need to add \ ( \ ) (no spaces) around your math code.
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We actually have B + C D = 1 0 1 . The fact B + D = 1 1 follows from the fact that the unit digit of 1 1 1 is 1 and B , D are the only digits responsible for it on the LHS. We have that B + D = 1 and B + D = 2 1 are impossible, thus B + D = 1 1 . This yields C = 9 , which means B + D = 1 1 , B , D = 1 , 9 is an equivalent problem to the one originally posted. Since it has multiple solutions, namely 6, so does the original problem.
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Thanks for adding the explanation!
Thank you for explaining my step very clearly.
Thank you.
To me with given details A ,B <c & D being distict positive nos,AAA will be 111 which can be by additions of AB =48 + CD= 63,can be 42 + 69 therefore D can be either 3 or 9..these are the options to choose.
K.K.GARG,India
The sum of 2 two digit numbers can't be greater than 198 (99+99), therefore;
A = 1
1B + CD = 111
So;
B + CD = 101
As the sum of 2 single-digit number can't be greater than 18,
C = 9
So;
B + 9D = 101
B + D = 11
Two unknowns can't be conclusively found from a single equation, meaning we don't have enough information to solve this problem.
AB + CD < 99 + 99 < 200 => AAA = 111
=> B+D = 1 or 11 =>
B + D = (3,8) (4,7) (5,6) (9,2) =>
So D cannot be determined