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PALINDROME becomes a pandigital number
Clash Royale CLAN TAG#URR8PPP
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Each of the letters of the word PALINDROME is assigned a different integer value between 0 and 9, for a total value for the word of 45. With the same value for each of the letters, the values of the letters in the eleven words below are all different. Moreover, the eleven words have been organized from left to right in increasing order of value (thus NAIL has the greatest value).
DRIP
PAIN
LIME
MEAL
DROP
DINE
RIND
LIAR
LORE
LEAD
NAIL
What was the value assigned to each of the letters of PALINDROME?
logical-deduction
edited 57 mins ago
gabbo1092
3,157532
asked 2 hours ago
Bernardo Recamán Santos
1,8971136
add a comment |
up vote
8
down vote
favorite
Each of the letters of the word PALINDROME is assigned a different integer value between 0 and 9, for a total value for the word of 45. With the same value for each of the letters, the values of the letters in the eleven words below are all different. Moreover, the eleven words have been organized from left to right in increasing order of value (thus NAIL has the greatest value).
DRIP
PAIN
LIME
MEAL
DROP
DINE
RIND
LIAR
LORE
LEAD
NAIL
What was the value assigned to each of the letters of PALINDROME?
logical-deduction
edited 57 mins ago
gabbo1092
3,157532
asked 2 hours ago
Bernardo Recamán Santos
1,8971136
add a comment |
up vote
8
down vote
favorite
up vote
8
down vote
favorite
Each of the letters of the word PALINDROME is assigned a different integer value between 0 and 9, for a total value for the word of 45. With the same value for each of the letters, the values of the letters in the eleven words below are all different. Moreover, the eleven words have been organized from left to right in increasing order of value (thus NAIL has the greatest value).
DRIP
PAIN
LIME
MEAL
DROP
DINE
RIND
LIAR
LORE
LEAD
NAIL
What was the value assigned to each of the letters of PALINDROME?
logical-deduction
edited 57 mins ago
gabbo1092
3,157532
asked 2 hours ago
Bernardo Recamán Santos
1,8971136
Each of the letters of the word PALINDROME is assigned a different integer value between 0 and 9, for a total value for the word of 45. With the same value for each of the letters, the values of the letters in the eleven words below are all different. Moreover, the eleven words have been organized from left to right in increasing order of value (thus NAIL has the greatest value).
DRIP
PAIN
LIME
MEAL
DROP
DINE
RIND
LIAR
LORE
LEAD
NAIL
What was the value assigned to each of the letters of PALINDROME?
logical-deduction
logical-deduction
edited 57 mins ago
gabbo1092
3,157532
asked 2 hours ago
Bernardo Recamán Santos
1,8971136
edited 57 mins ago
gabbo1092
3,157532
asked 2 hours ago
Bernardo Recamán Santos
1,8971136
edited 57 mins ago
gabbo1092
3,157532
edited 57 mins ago
gabbo1092
3,157532
edited 57 mins ago
gabbo1092
3,157532
3,157532
asked 2 hours ago
Bernardo Recamán Santos
1,8971136
asked 2 hours ago
Bernardo Recamán Santos
1,8971136
asked 2 hours ago
Bernardo Recamán Santos
1,8971136
1,8971136
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
up vote
4
down vote
One possible solution is
$$ P=0,, A=5, ,L=9, , I=4, ,N=6, , D=7, , R=3, , O=8, , M=1, , E=2 $$
Partial Reasoning
Each word takes a different value and the words are ordered left to right. This means, for example, that $$PAIN + 8 < NAIL Rightarrow P+8 < LRightarrow P=0, ,L=9$$ $$MEAL +5 < LEAD Rightarrow M + 5 < D Rightarrow D>6 Rightarrow DRIP > 9$$ In fact, since $E < R$ and $M < D-5$ the smallest value that $DRIP$ can attain is $13$. $$DROP > DRIP + 3 Rightarrow O > I + 3 Rightarrow I < 5$$ This means that $NAIL$ can be at most $28$. In fact, we also observe that since $D>6$ and $O > I+3$ this further restricts the maximum value of $NAIL$ to be $24$. Since the words all have distinct values and are ordered left to right, we must have $DRIP + 9 < NAIL$.
If we set $DRIP=13$, we quickly find that $NAIL < 23$ (a contradiction) so it must be that $DRIP=14$ which forces $NAIL=24$. From there, the value of each word is fully determined and the corresponding letters that work are $$ P=0,, A=5,, L=9,, I=4,, N=6,, D=7,, R=3,, O=8,, M=1,, E=2 $$
answered 32 mins ago
hexomino
32.5k295155
add a comment |
up vote
2
down vote
As a start (partial answer):
One-letter conclusions:
$L+I+M+E<M+E+A+L$ so $I<A$
$D+R+I+P<D+R+O+P$ so $I<O$
$D+R+I+P<R+I+N+D$ so $P<N$
$M+E+A+L<L+E+A+D$ so $M<D$
$L+I+A+R<N+A+I+L$ so $R<N$
$D+I+N+E<R+I+N+D$ so $E<R$
$D+R+O+P<R+I+N+D$ so $O+P<I+N$. Logically, $I<O$ so $P<N$ (as was already known) and $O-I<N-P$ (which may prove useful)
More observationally:
L seems to have a high value (it's in the last four and only one other one besides) and M seems to have a low value (it's only in two words on the low end of the scale)
answered 2 hours ago
kanoo
76914
IA<OE, OR<AD, ED<IA ...go brain figure sth out
– Jannis
1 hour ago
@Jannis and by ED<IA<OE you prove ED<OE so D<O
– gabbo1092
1 hour ago
@brain thank you ;) that means R<A
– Jannis
1 hour ago
@Jannis where did E+D<I+A come from?
– kanoo
1 hour ago
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
One possible solution is
$$ P=0,, A=5, ,L=9, , I=4, ,N=6, , D=7, , R=3, , O=8, , M=1, , E=2 $$
Partial Reasoning
Each word takes a different value and the words are ordered left to right. This means, for example, that $$PAIN + 8 < NAIL Rightarrow P+8 < LRightarrow P=0, ,L=9$$ $$MEAL +5 < LEAD Rightarrow M + 5 < D Rightarrow D>6 Rightarrow DRIP > 9$$ In fact, since $E < R$ and $M < D-5$ the smallest value that $DRIP$ can attain is $13$. $$DROP > DRIP + 3 Rightarrow O > I + 3 Rightarrow I < 5$$ This means that $NAIL$ can be at most $28$. In fact, we also observe that since $D>6$ and $O > I+3$ this further restricts the maximum value of $NAIL$ to be $24$. Since the words all have distinct values and are ordered left to right, we must have $DRIP + 9 < NAIL$.
If we set $DRIP=13$, we quickly find that $NAIL < 23$ (a contradiction) so it must be that $DRIP=14$ which forces $NAIL=24$. From there, the value of each word is fully determined and the corresponding letters that work are $$ P=0,, A=5,, L=9,, I=4,, N=6,, D=7,, R=3,, O=8,, M=1,, E=2 $$
answered 32 mins ago
hexomino
32.5k295155
add a comment |
up vote
4
down vote
One possible solution is
$$ P=0,, A=5, ,L=9, , I=4, ,N=6, , D=7, , R=3, , O=8, , M=1, , E=2 $$
Partial Reasoning
Each word takes a different value and the words are ordered left to right. This means, for example, that $$PAIN + 8 < NAIL Rightarrow P+8 < LRightarrow P=0, ,L=9$$ $$MEAL +5 < LEAD Rightarrow M + 5 < D Rightarrow D>6 Rightarrow DRIP > 9$$ In fact, since $E < R$ and $M < D-5$ the smallest value that $DRIP$ can attain is $13$. $$DROP > DRIP + 3 Rightarrow O > I + 3 Rightarrow I < 5$$ This means that $NAIL$ can be at most $28$. In fact, we also observe that since $D>6$ and $O > I+3$ this further restricts the maximum value of $NAIL$ to be $24$. Since the words all have distinct values and are ordered left to right, we must have $DRIP + 9 < NAIL$.
If we set $DRIP=13$, we quickly find that $NAIL < 23$ (a contradiction) so it must be that $DRIP=14$ which forces $NAIL=24$. From there, the value of each word is fully determined and the corresponding letters that work are $$ P=0,, A=5,, L=9,, I=4,, N=6,, D=7,, R=3,, O=8,, M=1,, E=2 $$
answered 32 mins ago
hexomino
32.5k295155
add a comment |
up vote
4
down vote
up vote
4
down vote
One possible solution is
$$ P=0,, A=5, ,L=9, , I=4, ,N=6, , D=7, , R=3, , O=8, , M=1, , E=2 $$
Partial Reasoning
Each word takes a different value and the words are ordered left to right. This means, for example, that $$PAIN + 8 < NAIL Rightarrow P+8 < LRightarrow P=0, ,L=9$$ $$MEAL +5 < LEAD Rightarrow M + 5 < D Rightarrow D>6 Rightarrow DRIP > 9$$ In fact, since $E < R$ and $M < D-5$ the smallest value that $DRIP$ can attain is $13$. $$DROP > DRIP + 3 Rightarrow O > I + 3 Rightarrow I < 5$$ This means that $NAIL$ can be at most $28$. In fact, we also observe that since $D>6$ and $O > I+3$ this further restricts the maximum value of $NAIL$ to be $24$. Since the words all have distinct values and are ordered left to right, we must have $DRIP + 9 < NAIL$.
If we set $DRIP=13$, we quickly find that $NAIL < 23$ (a contradiction) so it must be that $DRIP=14$ which forces $NAIL=24$. From there, the value of each word is fully determined and the corresponding letters that work are $$ P=0,, A=5,, L=9,, I=4,, N=6,, D=7,, R=3,, O=8,, M=1,, E=2 $$
answered 32 mins ago
hexomino
32.5k295155
One possible solution is
$$ P=0,, A=5, ,L=9, , I=4, ,N=6, , D=7, , R=3, , O=8, , M=1, , E=2 $$
Partial Reasoning
Each word takes a different value and the words are ordered left to right. This means, for example, that $$PAIN + 8 < NAIL Rightarrow P+8 < LRightarrow P=0, ,L=9$$ $$MEAL +5 < LEAD Rightarrow M + 5 < D Rightarrow D>6 Rightarrow DRIP > 9$$ In fact, since $E < R$ and $M < D-5$ the smallest value that $DRIP$ can attain is $13$. $$DROP > DRIP + 3 Rightarrow O > I + 3 Rightarrow I < 5$$ This means that $NAIL$ can be at most $28$. In fact, we also observe that since $D>6$ and $O > I+3$ this further restricts the maximum value of $NAIL$ to be $24$. Since the words all have distinct values and are ordered left to right, we must have $DRIP + 9 < NAIL$.
If we set $DRIP=13$, we quickly find that $NAIL < 23$ (a contradiction) so it must be that $DRIP=14$ which forces $NAIL=24$. From there, the value of each word is fully determined and the corresponding letters that work are $$ P=0,, A=5,, L=9,, I=4,, N=6,, D=7,, R=3,, O=8,, M=1,, E=2 $$
answered 32 mins ago
hexomino
32.5k295155
edited 12 mins ago
answered 32 mins ago
hexomino
32.5k295155
answered 32 mins ago
hexomino
32.5k295155
answered 32 mins ago
hexomino
32.5k295155
32.5k295155
add a comment |
add a comment |
up vote
2
down vote
As a start (partial answer):
One-letter conclusions:
$L+I+M+E<M+E+A+L$ so $I<A$
$D+R+I+P<D+R+O+P$ so $I<O$
$D+R+I+P<R+I+N+D$ so $P<N$
$M+E+A+L<L+E+A+D$ so $M<D$
$L+I+A+R<N+A+I+L$ so $R<N$
$D+I+N+E<R+I+N+D$ so $E<R$
$D+R+O+P<R+I+N+D$ so $O+P<I+N$. Logically, $I<O$ so $P<N$ (as was already known) and $O-I<N-P$ (which may prove useful)
More observationally:
L seems to have a high value (it's in the last four and only one other one besides) and M seems to have a low value (it's only in two words on the low end of the scale)
answered 2 hours ago
kanoo
76914
IA<OE, OR<AD, ED<IA ...go brain figure sth out
– Jannis
1 hour ago
@Jannis and by ED<IA<OE you prove ED<OE so D<O
– gabbo1092
1 hour ago
@brain thank you ;) that means R<A
– Jannis
1 hour ago
@Jannis where did E+D<I+A come from?
– kanoo
1 hour ago
add a comment |
up vote
2
down vote
As a start (partial answer):
One-letter conclusions:
$L+I+M+E<M+E+A+L$ so $I<A$
$D+R+I+P<D+R+O+P$ so $I<O$
$D+R+I+P<R+I+N+D$ so $P<N$
$M+E+A+L<L+E+A+D$ so $M<D$
$L+I+A+R<N+A+I+L$ so $R<N$
$D+I+N+E<R+I+N+D$ so $E<R$
$D+R+O+P<R+I+N+D$ so $O+P<I+N$. Logically, $I<O$ so $P<N$ (as was already known) and $O-I<N-P$ (which may prove useful)
More observationally:
L seems to have a high value (it's in the last four and only one other one besides) and M seems to have a low value (it's only in two words on the low end of the scale)
answered 2 hours ago
kanoo
76914
IA<OE, OR<AD, ED<IA ...go brain figure sth out
– Jannis
1 hour ago
@Jannis and by ED<IA<OE you prove ED<OE so D<O
– gabbo1092
1 hour ago
@brain thank you ;) that means R<A
– Jannis
1 hour ago
@Jannis where did E+D<I+A come from?
– kanoo
1 hour ago
add a comment |
up vote
2
down vote
up vote
2
down vote
As a start (partial answer):
One-letter conclusions:
$L+I+M+E<M+E+A+L$ so $I<A$
$D+R+I+P<D+R+O+P$ so $I<O$
$D+R+I+P<R+I+N+D$ so $P<N$
$M+E+A+L<L+E+A+D$ so $M<D$
$L+I+A+R<N+A+I+L$ so $R<N$
$D+I+N+E<R+I+N+D$ so $E<R$
$D+R+O+P<R+I+N+D$ so $O+P<I+N$. Logically, $I<O$ so $P<N$ (as was already known) and $O-I<N-P$ (which may prove useful)
More observationally:
L seems to have a high value (it's in the last four and only one other one besides) and M seems to have a low value (it's only in two words on the low end of the scale)
answered 2 hours ago
kanoo
76914
As a start (partial answer):
One-letter conclusions:
$L+I+M+E<M+E+A+L$ so $I<A$
$D+R+I+P<D+R+O+P$ so $I<O$
$D+R+I+P<R+I+N+D$ so $P<N$
$M+E+A+L<L+E+A+D$ so $M<D$
$L+I+A+R<N+A+I+L$ so $R<N$
$D+I+N+E<R+I+N+D$ so $E<R$
$D+R+O+P<R+I+N+D$ so $O+P<I+N$. Logically, $I<O$ so $P<N$ (as was already known) and $O-I<N-P$ (which may prove useful)
More observationally:
L seems to have a high value (it's in the last four and only one other one besides) and M seems to have a low value (it's only in two words on the low end of the scale)
answered 2 hours ago
kanoo
76914
edited 1 hour ago
answered 2 hours ago
kanoo
76914
answered 2 hours ago
kanoo
76914
answered 2 hours ago
kanoo
76914
76914
IA<OE, OR<AD, ED<IA ...go brain figure sth out
– Jannis
1 hour ago
@Jannis and by ED<IA<OE you prove ED<OE so D<O
– gabbo1092
1 hour ago
@brain thank you ;) that means R<A
– Jannis
1 hour ago
@Jannis where did E+D<I+A come from?
– kanoo
1 hour ago
add a comment |
IA<OE, OR<AD, ED<IA ...go brain figure sth out
– Jannis
1 hour ago
@Jannis and by ED<IA<OE you prove ED<OE so D<O
– gabbo1092
1 hour ago
@brain thank you ;) that means R<A
– Jannis
1 hour ago
@Jannis where did E+D<I+A come from?
– kanoo
1 hour ago
IA<OE, OR<AD, ED<IA ...go brain figure sth out
– Jannis
1 hour ago
IA<OE, OR<AD, ED<IA ...go brain figure sth out
– Jannis
1 hour ago
@Jannis and by ED<IA<OE you prove ED<OE so D<O
– gabbo1092
1 hour ago
@Jannis and by ED<IA<OE you prove ED<OE so D<O
– gabbo1092
1 hour ago
@brain thank you ;) that means R<A
– Jannis
1 hour ago
@brain thank you ;) that means R<A
– Jannis
1 hour ago
@Jannis where did E+D<I+A come from?
– kanoo
1 hour ago
@Jannis where did E+D<I+A come from?
– kanoo
1 hour ago
add a comment |
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