--How --> 3-digit value
+Much --> 4-digit value
Power --> 5-digit value
*********************************
w+h=r, o+c=e, h+u=w, _+m=o,

Note here "m" is added with no alphabet... So if i take "M" value between (0-8), it ll not produce a remainder... So it ll again produce a 4-digit no.... But our final result should be a 5-digit value...
If 3-digit + 4-digit = 5-digit means, 1st no. of 4-digit no. should start with "9"...Nd obviously the 1st no. of 5-digit ll be "1" nd 2nd no. ll be "0"...
So here P=1...M=9... O=0...
So Definitely P=1, m=9, O=0

So remaining numbers we can use is {2,3,4,5,6,7,8}
So Take "R=2" first...
Possible values of (W,H) ll be (4,8), (5,7), (7,5) (8,4)
When v take (W,H) = (4,8) , we cannot get (H+U) =w...
So take (W,H) = (5,7)... So W=5, H=7, R=2 carry 1...
Already O=0... So possible value of C ={3,4,6,8}...
When C=3, C={carry+0+3}=4... So E=4...
H is already 7... Possible values of U={6,8}...
We need to get H+U=W... (ie) Result should be 5... So U should be 8... 7+8=5 carry 1...
M is already 9... So {carry + 9} = 0 carry 1...
nd hence P=1 (carry of prev result)

HOW ==> 705
MUCH ==> 9837
POWER ==> 10542

705
(+)9837
-------------------
10542
------------------
P+O+W+E+R = 1+0+5+4+2 = 12

Ans : 12

@SARASWATHY can u please give some materials on this type problms..and please provides some links..

803
9468
------------
10371
.....first three digit can be obtained by the concept described by sarshwathy...and remaining can be done by selecting proper digits...though the addition me produce differ result ...but p+o+w+e+r...well always be 12

ans is 12 :)

H O W
7 0 5
M U C H
+ 9 8 3 7
______________
P O W E R
1 0 5 4 2

THEN P+O+W+E+R=1+0+5+4+2=12

HOW ==> 805
MUCH ==> 9718
POWER ==> 10523

705
(+)9837
-------------------
10542
------------------
P+O+W+E+R = 1+0+5+2+3 = 11

how + much = POWER.
p=1,
o=0,
m=9 obvious.
now. if w= 5, h=7, so it gives exact answer. 10542

·····h·o·w
+·m·u·c·h
--------------
p··o·w·e·r

Since there are no numbers above the p, it must be a 1. Since
there is only one number above the m, it must be a 9 and o must be a 0. So, so far we have
p = 1, o = 0, m = 9

For the o + c = m column to work out, w+h must be greater than 9 so that a 1 gets carried into the next column. So it must be true that
1+c = e.

It must also be true that
h + u = 10 + w and
w + h = 10 + r

So 2h + u = 10 + w + h = 20 + r.
That is u - r = 20 - 2h

If h=8, then u-r = 4
If h=7, then u-r = 6
If h=6, then u-r = 8

There are no other possibilities.

So the following combinations are possible for o,p,r,h,u,and m.

0·1·2·3·4·5·6·7·8·9
o·p·r ···w···u···h·m
o·p····r ···w··u·h·m
o·p·r·······w· h·u·m

Since e = c+1, we need room
for e followed by c. The only
possibility from the above is

0123456789
oprcew·hum

So

how
much
-----
power

becomes

705
9837
-----
10542

p+o+w+e+r = 12