Property Owned As Tenants In Common
Law1 min ago
Codebreaker
1 Answers
Each and every letter of the alphabet is allocated a different number 1-26. Find the value of every letter given the following words, eg. IVY has got to be 1+2+3 total 6, but I could be either 1.2.or 3 and so on:-
1. Holly= (total) 40, 2. Ivy=6, 3. Xmas=31, 4. Present=81, 5. Turkey=96, 6.Jingle Bells=134, 7.December=126, 8. Boxing Day=100, 9.Snowflake=100,
10. Frozen=74, 11. Snow=22, 12. Icicle=66
Any unused letter is allocated any unused number
Help please
1. Holly= (total) 40, 2. Ivy=6, 3. Xmas=31, 4. Present=81, 5. Turkey=96, 6.Jingle Bells=134, 7.December=126, 8. Boxing Day=100, 9.Snowflake=100,
10. Frozen=74, 11. Snow=22, 12. Icicle=66
Any unused letter is allocated any unused number
Help please
Answers
Only 1 answer
Best Answer
No best answer has yet been selected by mimosa1. Once a best answer has been selected, it will be shown here.
For more on marking an answer as the "Best Answer", please visit our FAQ.
OK, so I, V and Y are (1,2,3) in some order.
SNOW is 22...none of these letters can be less than 4 (I-V-Y are 1,2,3) so they must be 4 ,5, 6 and 7 (or SNOW would be more than 22). So SNOW are (4,5,6,7) in some order.
XMAS is 31. We know that S is between 4 and 7, and XMA must each be 8 or higher. So if XMAS is not to exceed 31, XMA must be (8,9,10) and S must be 4. [S=4]
HOLLY is 40. We know that Y is between 1 and 3, and O is between 5 and 7. If HOLLY is not to exceed 40, L must be 11(the lowest available number) and H must be 12. This means that Y is 1 and O is 5. [L=11, H=12, Y=1, O=5]
PRESENT is 81. S is 4 and N must be between 6 and 7, so P, R, E, E, T must be 70 or 71. The lowest possible total is 71, if E is 13 (again, the lowest available number) and PRT are (14,15,16). N must be 6, which means W is 7. [E=13, N=6, W=7]
The lowest available 'slot' is now 17. Over to you...
SNOW is 22...none of these letters can be less than 4 (I-V-Y are 1,2,3) so they must be 4 ,5, 6 and 7 (or SNOW would be more than 22). So SNOW are (4,5,6,7) in some order.
XMAS is 31. We know that S is between 4 and 7, and XMA must each be 8 or higher. So if XMAS is not to exceed 31, XMA must be (8,9,10) and S must be 4. [S=4]
HOLLY is 40. We know that Y is between 1 and 3, and O is between 5 and 7. If HOLLY is not to exceed 40, L must be 11(the lowest available number) and H must be 12. This means that Y is 1 and O is 5. [L=11, H=12, Y=1, O=5]
PRESENT is 81. S is 4 and N must be between 6 and 7, so P, R, E, E, T must be 70 or 71. The lowest possible total is 71, if E is 13 (again, the lowest available number) and PRT are (14,15,16). N must be 6, which means W is 7. [E=13, N=6, W=7]
The lowest available 'slot' is now 17. Over to you...
Only 1 answer
