R2I4D wrote:
Wow. I have no idea what is happening here. Would you (or someone else) mind explaining a bit more...perhaps expounding on your method a little bit?
Perhaps, the same deal, with a two digit number using 1 and 2, with digits allowed to repeat?
[Answer: 11+12+21+22 = 66]
How can I use xcusemeplz2009's method to arrive at the same answer?
let me try again
let the no. be xyz
Place value of no will be 100x+10y+Z.............eqn 1
TOTAL NO OF WAYS IN WHICH WE CAN MAKE A NO. WITH GIVEN CONDITION IS 3*3*3=27( as digits are getting repeated)
now at unit place i.e Z if we fix 1 , then for y we have 3 options and for x we have 3 options , so total number where unit digit is one can be formed in 3*3=9 ways
similarly for other digits at unit place can be done in 9 ways
hence at unit place a digit is getting repeated 9 times
and in the same manner in tenths place repetion for all digits will be 9 each
now the task is to find the sum
sum of digits at z place is 9*1+9*5+9*8 , WHICH
IS SAME FOR Y AND X
for that we need to put all the options in eqn 1 format
i.e 100*9[1+5+8] + 10*9[1+5+8] + 9*[1+5+8]=13986
simple way is to remeber the quick formula to find out the repetition
find out tot no. ways in which the no. can be formed and divide it by the no. of digit , then mutiply this factor with the summation of all the dig and the place value
for second ex : 1,2
two dig no with repetn can be formewd in 2*2=4 ways , repetion factor=4/2 =2 ( tot no. of ways / tot no of dig)
sum=10(place value) * 2 (rep factor) *(1+2 summation of dig)+2*3=60+6=66
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