1Sir is the answer (12C2)3 x 9C2 ?
What is the number of pairs of 2 consecutve 4 digit integers which can be added without any carry?
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9 Answers
1i'm just getting a hint that the sum of a particular digit of the first no. and another digit of the 2nd no. must not be ≥ 10. Right ?
62yes absolutely
to make things more intersting, we can even pose a question, what is the number of consecutive numbers that we can add so that there is no carry..
remember,
10, 11
11, 12
10, 11, 12
will all be different cases :)
62doesnt seem so!
what is ur logic?
got to sleep.. gnite.. will check this tomorrow.. unless someone replies before that :)
1for the last digit the combinations of consecutive nos which add up to something less than 10 are 5,4;4,3;3,2;2,1;1,0. .so last places of the nos can be filled up in 5 ways.
second places and third places of both nos shud be equal.equal nos which add up to less than 10 are 4,4;3,3;2,2;1,1;0,0. so these two places can be filled up in 5 ways
.first place can be filled up in 4 ways excluding 0
so total no of pairs are 4*5*5*5=500
1sir is it
(11C2)3*92
a b c d
+
l m n o
___________
(a+l) (b+m) (c+n) (o+d)
now
(o+d) ≤9
o+d+k =9
no.of solution is 11C2
similarlty for others three
so
(11C2)3.92
62jagaran.. you have done a good job but missed the complete cases :)
@akari : the question is consecutive numbers