This is very hard!

Logic Level 4

T H I S + V E R Y H A R D \begin{array}{ccccc} & T& H &I & S \\ + &V & E & R & Y \\ \hline & H & A & R & D\\ \hline \end{array}

In the cryptogram above, each letter represents a distinct single digit non-negative integer. What is the largest value of H A R D ? \overline{HARD}?


The answer is 9754.

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2 solutions

Charlz Charlizard
Mar 17, 2018

The things to note here are,I has two possible values, either 0 or 9, but in order to have either of H,E,A equal to 9, best shot is to assign I=0.Moving further to the value of H as it will going to define highest, we can assume its value equals to 9 and E=8, and as there is no carry to the hundred's digit count, we can add H+E to A=7 with a carry of 1 to the thousand's digit count.Now here comes the complex part, to decide the value of T and V, we have two options either (T,V)=(6,2) or (5,3).If we take (T,V)=(5,3) the choice left for the value of R is 1, as no other criterion fits, and so will be the value of (S,Y)=(4,2).The final answer comes up for HARD=9716, but we don't know whether it is maximum value or not, so have to check the other possibility which is (T,V)=(6,2).Now here we have 2 values of R, either 1 or 5, that can fit without creating contradictions.So in order to maximize the last result, let's assume R=5 and so the value of (S,Y)=(3,1), which gives the final value of HARD=9754 which is higher than the previous one.

Now here, no other choices have been left, and we have assumed values in order to maximize the last result, so we can be sure that this is the only maximum value.

Also the values in the brackets are not ordered pairs, as we have given no conditions to determine the exact value for them.

Hey hey how about 6105??

David Lin - 1 year, 1 month ago

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Hey David, Aren't we talking about finding the largest value of HARD, here?

Charlz Charlizard - 1 year, 1 month ago

2698+3407=6105.....

David Lin - 1 year, 1 month ago
Nicola Mignoni
Jan 14, 2018

Even if it's not the most elegant way, this is a simple MATLAB code I used to brute-force the cryptogram. Total time elapsed during the execution was circa 4.6 minutes.

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