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Replace the same characters by the same numerals so that the mathematical operations are correct.

(J+O+I+N+T)*(J+O+I+N+T)*(J+O+I+N+T) = JOINT

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No. 20 - solution

(1+9+6+8+3)*(1+9+6+8+3)*(1+9+6+8+3) = 19683

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• 5 months later... If different letters could be the same number, 17576 also works. Interestingly, the cube root of 19683 and 17576 are 27 and 26, respectively.

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• 6 months later... very clever ##### Share on other sites
• 1 month later... There might be a technicality against this but 00000, 00001, 00512, 04913, and 05832 should also fit the criteria.

##### Share on other sites i think all the other ones that work wouldnt if it said no 2 letters are the same. im not sure if this is implied.

the reason they are the cube roots as some1 said is because this can be re-writen as (j+o+i+n+t)3 = JOINT

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• 5 weeks later... There might be a technicality against this but 00000, 00001, 00512, 04913, and 05832 should also fit the criteria.

No technicality, the numbers only add to three digits not five, you would have to enter the zeroes yourself which would defeat the point, the numbers have to fit you can not put them where you wish

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• 5 months later... No. 20 - Back to the Cryptograms and Algebra Puzzles

Replace the same characters by the same numerals so that the mathematical operations are correct.

(J+O+I+N+T)*(J+O+I+N+T)*(J+O+I+N+T) = JOINT

No. 20 - solution

(1+9+6+8+3)*(1+9+6+8+3)*(1+9+6+8+3) = 19683

This is a good one!

Can you expand the solution more?

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• 1 year later... This is a good one!

Can you expand the solution more?

PS: My english aint all that

Hey Swima, well my answer to that was more logical than other things.

(j+o+i+n+t)3= JOINT

lets say : j+o+i+n+t = Y

Means that if we add those numbers potency 3 = 5 digits

Using calculator you realise that the number 22 is the first potency 3 has 5 digits.

22*22*22=10648

the last number is 46

46*46*46= 97336

so we know that the total of the numbers is between 22 and 46.

we will check the minimum / maximum sum of five numbers ( j+o+i+n+t)

if we do 1+2+3+4+5 = 15 ( not possible as between 22 and 46)

2+3+4+5+6= 20 not possible as between 22 and 46.

3+4+5+6+7= 25 possible, now we know its between 25 and 46

The maximum will be : 9+8+7+6+5= 35 this max so the sum is between 25 and 35.

Using calculator you just have to do 25*25*25= 15'625 not good one as number five is shown twice

26*26*26= 15 756 not good one as number five as well is shown more than one time

27*27*27= 19 683 this could be used as total of 1+9+6+8+3= 27

28*28*28= 21 952 not good one as number two is shown twice

29*29*29= 24 389 this could be used however their sum isnt 29

32*32*32= 32 768 this could be used however their sum isnt 32

Hop dat helpd lil bit, as said its more logical than mathematical.

Peace m8

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