CA CS CIMA Prakash Somani (Landmark Group) (23502 Points)
12 September 2008Find the values of each of the alphabets.
N o o n
s o o n +
m o o n
----------
j u n e
hint: each one of alphabets stands for distinct number between 0 and 9
:)cheers:)
Amresh Choudhary
(M.Sc. (Maths) MBA LLB (DU))
(694 Points)
Replied 12 September 2008
| Originally posted by :CA Prakash Somani | ||
| " | Find the values of each of the alphabets. N o o n s o o n + m o o n ---------- j u n e hint: each one of alphabets stands for distinct number between 0 and 9 :)cheers:) |
" |
n o o n
s o o n
m o o n
---------------
j u n e
---------------
step1:
when u add n, n ,n you got e
so e must me a multiple of 3
from 1 to 9, multiple of 3 are 3,6,9
if e = 3, n = 1,
if e = 6, n = 2,
if e = 9, n = 3.
step2:
when u add o,o,o you got n. tricky part is here n is just
the units place, if o+o+o is 24 then n = 4 and 2 is send
as carry... i know u know additions........... :)
assuming e = 3, n = 1 ,
when u add o,o,o you must again get multiple of 3,
but here they can be greater than 9, but less than 27,
since, 'o' can not be greater than 9.
so possible multiples are 3,6,9,12,15,18,21,24,27.
since we are assuming e = 3, n = 1,
to get 1 in n's place of "june", only possible multiple of
3 is 21. satisfies, n = 1 and 2 is send as carry.
to get 21, o must be 7, bcaz 7+7+7 = 21
so now we got, n = 1, e = 3, o = 7.
step3:
we got so far, n = 1, e = 3, o = 7.
here again, we add, o+o+o+2(carry from 21) and we get 'u'
and carry is forwarded.
since o = 7, 21+2 = 23, which makes u = 3 and 2 as
carry,
buttttttttt, already we have 'e' = 3 ,so 'u' can not be
equal to 3,
so our assumption in step 2, e = 3, n = 1 is false not
we have to try with remaining two assumptions.... dont
sleep...... look at the remaining part tooo.
other two assumptions are
if e = 6, n = 2,
if e = 9, n = 3.
now
step 4:
assuming if e = 6, n = 2,
when u add o,o,o you must again get multiple of 3,
but here they can be greater than 9, but less than 27,
since, 'o' can not be greater than 9.
so possible multiples are 3,6,9,12,15,18,21,24,27.
since we are assuming e = 6, n = 2,,
to get 2 in n's place of "june", only possible multiple of
3 is 12. satisfies, n = 2 and 1 is send as carry.
to get 12, o must be 4, bcaz 4+4+4 = 12,
o = 4, n = 2, e = 6
step5:
so far, we got o = 4, n = 2, e = 6
here again, we add, o+o+o+1(carry from 12) and we get 'u'
and the new carry is forwarded.
since o = 4, 12+1 = 13, which makes u = 3 and 1 as
carry,
so, u =3, o = 4, n = 2, e = 6
all are unique so no problem... we can go further and you
dont yawnnn
step 6:
till now we got u =3, o = 4, n = 2, e = 6
here, n+s+m+1(carry from 13) is j
so, 2+s+m+1 = j
from1 to 9, the reamining numbers are 1,5,7,8,9. since
2,3,4,6 are used by n,u,o,e respectively
here, 3+s+m = j, so j is greater than 3,
so the possible numbers for ' j ' are now, 5,7,8,9.
assume j = 5,
3+s+m =5, which means, s = m =1, uniquness is not
achieved, so j != 5,
assume j = 7,
3+s+m = 7, means s+m =4
here, possbile values of s and m to satisfy s+m =4, are
1+3 =4 not possible since u = 3
2+2 = 4 not possible since both cant be equal
3+1 =4 not possible since u = 3
assume j = 8
3+s+m = 8, means s+m = 5
possbile values of s and m to satisfy s+m =5, are
1+4 = 5 not possible since o = 4
2+3= 5 not possible since u = 3
3+2 = 5 not possible since u = 3
4+1 + 5 not possible since o = 4
therefore j = 9
3+s+m = 9
s+m = 6
s = 1 and m =5 or s = 5 and m = 1
n o o n
2 4 4 2
s o o n
1 4 4 2
m o o n
5 4 4 2
---------------
--------------
j u n e
---------------
9 3 2 6
-------------
-------------
CA CS CIMA Prakash Somani
(Landmark Group)
(23502 Points)
Replied 15 September 2008
I have two answers...
First one If we use N and n are same than the answer is ....
u=3, o=4, N=n=2, e=6, s=5, m=0, j=8.
If we use N and n are different then the answer is....
u=3, 0=4, n=2, e=6, N=0, s=7, m=1, j=9.
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