# If 1111=R, 2222=T, 3333=E, 4444=N Then 5555=? [CLOSED]

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If 1111=R, 2222=T, 3333=E, 4444=N Then 5555=? [CLOSED]
closed with the note: BEST ANSWER PICKED
posted Jul 27, 2014

``````1111=4=fouR
2222=8=eighT
3333=12=twelvE
4444=16=sixteeN
5555=20=twentY
``````

By addition also we can solve the same

1*4=4 fou`R`
2*4=eigh`T`
3*4=twelv`E`
4*4=sixtee`N`
5*4=twent`Y`

answer Jul 31, 2014 by anonymous

1111=r
2222=t
3333=e
4444=n
5555=y

answer Aug 22, 2014 by anonymous
+1 vote

Y........................................................

+1 vote

5+5+5+5=20=twentY

+1 vote

1111=1+1+1+1=FOU(R)
similarly
,2222=EIGH(T)=T,
3333=TWELV(E)=E,
4444=SIXTEE(N)=N
Then 5555TWENT(Y)=Y.

+1 vote

ENTRY (jumbled words R T E N and Y)
Y is the missing alphabet.

answer Sep 28, 2015 by anonymous
+1 vote

if 1+1+1+1=4 in word FOUR last word id R
and 2+2+2+2=8 in word EIGHT LAST WORD IS T
3+3+3+3=12 IN WORD TWELVE LAST WORD IS E
4+4+4+4=16 IN WORD SIXTEEN LAST WORD IS N
5+5+5+5=20 IN WORD TWENTY LAST WORD IS Y

+1 vote

Y.
1111=4, whose last letter is R
2222=8, whose last letter is T
and so on.

Its Y .................so simple