consider a no 235 where last digit is the sum of first two digit no eg- 2 3 5 how

bwtween 100 and 200 there are 9 such numbers
101,112,123,134,145,156,167,178,189
in the same way btwn 200 and 300
202,213,224,235,246,257,268,279 =8 numbers totally
so series continues as 9+8+7+6+5+4+3+2+1=45 numbers
9 years agoHelpfull: Yes(26) No(0)

45

total combinatioons using 0 are 9
total combinations using 1 as first digit are 8 similalry using 9 is 1
9+8+7+6+5+4+3+2+1=45
9 years agoHelpfull: Yes(19) No(1)
189/819
279/729
369/639
459/549
909
for 9 it gives 9 cases
for 8 it gives 8 cases & so on....
that is why 9+8+7+...1=45
9 years agoHelpfull: Yes(4) No(0)
ans : 45
we have digits 0-9
for 1 combinations are 001 only like that
for 2, 002,112
for 3, 003,123,213
for 4, 004,224,234,324 so like that we have
for 5, 5 combinations etc add all these combinations
1+2+3+4+5+6+7+8+9=45
9 years agoHelpfull: Yes(2) No(0)
using 1 at hundredth place,we can get 7 no
using 2 at hundredth place,we can get 6 no
similarly 5,4,4,3,2,1 no.
total=(7+6+5+4+4+3+2+1)=32
9 years agoHelpfull: Yes(0) No(6)
45
9+8+7+6+5+4+3+2+1=45
9 years agoHelpfull: Yes(0) No(0)
Fady Bro:-
9+8+7+6+5+4+3+2+1 ;)
9 years agoHelpfull: Yes(0) No(2)

consider a no 235, where last digit is the sum of first two digit no eg- 2+3=5 , how many such 3-digit no are there ???