43 of 100: The AHA Moment

1. H + D + A = 10 + A, so H + D = 10

2. H + A + 1 = H + 10, so A = 9 [1 in LHS carry from summing ones digits]

3. H + 1 = A, so H = 8 [1 in LHS carry from summing ones digits]

By 1st, D = 2

We have $\overline{{\color{#D61F06}H}{\color{#20A900}A}{\color{#3D99F6}D}}+\cancel{\overline{{\color{#D61F06}H}{\color{#20A900}A}}}+{\color{#D61F06}H}=100{\color{#20A900}A}+\cancel{\overline{{\color{#D61F06}H}{\color{#20A900}A}}}$ and $100 {\color{#D61F06}H}+10 {\color{#20A900}A}+{\color{#3D99F6}D}+{\color{#D61F06}H}=100 {\color{#20A900}A}$ .

Thus ${\color{#3D99F6}D}+{\color{#D61F06}H}$ is divisible by $10$ . Hence ${\color{#3D99F6}D}+{\color{#D61F06}H}=10$ , since ${\color{#3D99F6}D}+{\color{#D61F06}H}\le 17$ .

Now, we have $10 {\color{#D61F06}H}+ {\color{#20A900}A}+1=10 {\color{#20A900}A}$ . Thus ${\color{#20A900}A}+1$ is divisible by $10$ . Hence ${\color{#20A900}A}=9$ , since ${\color{#20A900}A} \le 9$ . Thus ${\color{#D61F06}H}=8$ and ${\color{#3D99F6}D}=2$ .

Since the units digit of AHA is A, this means that either D and H are both 0 or the letters sum up to 10.

Since 0 is not a choice, we go with $D+H=10$ The one carries in addition and sets up the equality $A + H + 1 = H +10$ . It could not be $A + H + 1 = H$ since A = -1, which cannot happen.

The one from getting a 10 carries another one and sets up a third equation $H+1=A$ .

From the second equation, we rearrange to get A = 9, which leads to H = 8 using the third equation. Going back to the first equation, we substitute H for 8 and get D + 8 = 10. Therefore, D = 2

Relevant wiki: Cryptogram - Problem Solving

$D+ H = 10$

$A+1 = 10 \to A=9$

$H+1=A=9 \to H=8$

$D + 8 = 10 \to D=2$

I know one can work out equations etc. But this one presents one glaring clue that solves it without that much work.

That clue is: whenever a digit in the column is passed to the sum, the rest must sum to 10 (in a column of only two or three). With that observation alone the puzzle can be solved.

The second column A + H = H forces A = 9 because A with a (forced) carry of 1 must be 10. This also forces a carry of 1 to the third column.

H and A in the third column forces H + 1 = A. Thus H = 8.

Finally D + A + H = A forces D = 2 so that 17 + 2 puts A = 9 in the right hand position of the sum, and does carry the necessary 1 to the second column.

So D must be 2.

$at first.$

$H + D + A$ = $10 + A$

$or,$ $H + D$ = $10$ .......... $[subtracting A from both sides]$

$now,$ $H + A + 1$ = $H, so, A$ = $9$

$and,$ $H + 1$ = $A, so, H$ = $8$ ..................... $[H+D=10]$

$so,$ $D$ = $2$

Add the three terms together, we have $111 H + 11 A + D = 101 A + 10 H,$ $101 H + D = 90 A.$ Since the right-hand side is a multiply of 10, so is the left-hand side. Thus $H + D = 10$ . Substitute this: $100 H + 10 = 90 A.$ $10 H + 1 = 9 A = 10 A - A.$ One more than a multiple of ten equals $A$ less than a multiple of 10. This implies $A = 9$ . From there, we easily find $H = 8, D = 2$ .

111H + 11A +D = 101A + 10H D = 90A-101H The only values for A and H where 0<D<10 are A=9 and H=8 Therefore D must equal 810-808 = 2 when you put this in the equation it indeed is true

Here's what went through my head sorry if it is hard to follow. I started with the second column and told myself A has to be either 0 or 10 more than H and if A was 0 more than H then A and H would be the same number and in the directIons it says they have to be different. So that means A must be 10 but if it was 10 that would make it so the first row of numbers would be four digits and H would be one and A would be 0 which is a contradictory. So there must be a carryover from the last column. And D+H+A at most could be 7+8+9=24 as they cannot be the same. So the second row could be eight since the carryover is 2 so we get 2+8+H=1H for a carryover of 1. Which would make H+1=A if A is one more than H and the only way we can make that work if we use the logic of columns 1 and 2 so that H=7 and A=8. And 7+8 = 15 so we are going to have to do D+15=24, or 24-15=D, which is 9. But that doesn't work because D and H must equal 10 to make the answer added to A be 1A, and the carryover is 1 instead of 2 so the A and H of column 2, and thus all of them, must be upped by 1, so instead of A=8 and B=7, it will be A=9 and B=8. That doesn't change the correctness of column 3 or 2. 8+9=17 so we must add the last digit, D, as 2 to make this right. Sorry this is a little all over.

I do not want to submit a solution, because although I did get it right I did it in exactly the same way that Akshay Gupta did it in his previous solution. So wouldn't it be a good idea if you could 'like' the solution in the way you do on Facebook. Maybe there's already a way, and maybe some-one will be kind enough to tell me this. Regards, David

Because no one approached it this (quick) way: apply modular arithmetic modulo 11, noting that $10^i\bmod 11 = (-1)^i$ . We find $(-H+A-D)\ +\ (H-A)\ +\ (-H) = -A+H-H \pmod{11}$ which simplifies to $-H-D = -2A+H \pmod{11}$ From the leftmost column and that all digits are distinct we know $A = H+1$ . Plugging this in the equation above yields $D = 2 \pmod{11}$ ; since, $D$ is a decimal digit, we have $\fbox{\$D = 2\$}$ .

1. Since D + A + H = A, then D + H = 10
2. A + H + 1 = H (carry over the 10) , so A =9
3. If another 1 carries over, H + 1 would equal A, so H=8.
4. Now going back to the beginning, D + A + H = A + 10, and we know that A = 9, H = 8, D + 9 + 8 = 19: D = 2

Since D+A+H has a sum with A as it's one's digit, D+H must equal 10 or D=10-H

A + H must have carried a 1 over to the hundred's column because H in the hundreds column doesn't equal A without a carry. 1 must have been carried due to the fact that a carry more than one isn't possible. Suppose the digits D,A,H or 9,8,and 7 (any order). That means that there would be a 2 carried over to the tens column. If A and H are 9 and 8 (any order), then 9 + 8 + 2 would equal 19. Therefore, it is only possible for the carry from the tens column to the hundreds column must be 1. Therefore A=H+1 or H=A-1.

Now, guess and check.

If A=9 then, H=A-1=8 D=10-H=10-8=2

Check if it works.

892+89+8=898

IT WORKS!!!

Therefore D=2.

1 . H+A+D=10+A, so H+D=10 2 . HAD+HA+H=AHA equals 100H+10A+D+10H+A+H=100A+10H+A equals HAD+H=A00

so... 1 .1+A=10 so A equals 9 2 .H+1=9 so H equals 9 3 .H+D=10 or 9+D=10 so D equals 2

*therefore D = 2 *

$\color{#D61F06}H \color{#333333}+ \color{#3D99F6}D \color{#333333}= 10 \\ \color{#20A900}A \color{#333333}+ 1 = 10 \\ \color{#D61F06}H \color{#333333}+ 1 = \color{#20A900}A \\ \color{#20A900}A \color{#333333}+ \color{#3D99F6}D \color{#333333}= 11 \\ 9 + \color{#3D99F6}D \color{#333333}= 11 \\ \large \color{#3D99F6}D \color{#333333}= \color{#CEBB00}\boxed{2}$

LaTeX: $\Large \text {Let's make the operation as if they were digits : }$

LaTeX: $\normalsize HAD + HA + H = AHA \iff 100H + 10A + D + 10H + A + H = 100A + 10H + A$

LaTeX: $\normalsize HAD + HA + H = AHA \iff -101\color{#D61F06}{H} \color{#333333}{+ 90} \color{#20A900}{A} \color{#333333}{=} \color{#3D99F6}{D}$

LaTeX: $\large \text {we get a linear diophantine equation where A and H are unknown, and D varies from 1 to 5 }$

LaTeX: $\large \boxed {D = 1} \; -101\color{#D61F06}{H} \color{#333333}{+ 90} \color{#20A900}{A} \color{#333333}{= 1}$

LaTex: $\large \text {Let's apply the euclidean algorithm : }$

LaTex: $\large L_1 : 101 = 90 \times 1 + 11$

LaTex: $\large L_2 : 90 = 11 \times 8 + 2$

LaTex: $\large L_3 : 11 = 2 \times 5 + \boxed {1} \; \text {The} \; gcd(101,90) = 1$

LaTex: $\large L_4 : 2 = 1 \times 2 + 0$

LaTex: $\text {101 and 90 are prime between them} \; \iff \exists \; (H;A) \in \mathbb Z^2 \; \vert \; xH + yA = 1$

LaTex: $\text {Let's modify the euclidean algorithm with the last remainder different from 0 : }$

LaTex: $\large L_3 : 11 - 2 \times 5 = \boxed {1}$

LaTex: $\large L_3 : 11 - (90 - 11 \times 8) \times 5 = \boxed {1} \; \Leftarrow [ L_2 : 90 - 11 \times 8 = 2]$

LaTex: $\large L_3 : 11 - 90\times 5 +40 \times 11 = \boxed {1}$

LaTex: $\large L_3 : - 5\times 90 +41 \times 11 = \boxed {1}$

LaTex: $\large L_3 : - 5\times 90 +41 \times (101 - 90 \times 1) = \boxed {1} \; \Leftarrow [L_1 : 101 - 90 \times 1 = 11]$

LaTex: $\large L_3 : \color{#D61F06}{41}\color{#333333}{ \times 101 }\color{#20A900}{- 46}\color{#333333}{ \times 90 =} \boxed {\color{#3D99F6}{1}}$

Latex: $\large \text {the value D=1 doesn't suit for} \; H\notin \{\pm 1, 2, 3, 4, 5, 6, 7, 8, 9\} \; \text{and} \;A\notin \{\pm 1, 2, 3, 4, 5, 6, 7, 8, 9\}$

LaTeX: $\large \boxed {D = 2} \; -101\color{#D61F06}{H} \color{#333333} \;{+ 90} \color{#20A900}{A} \color{#333333}{= 2}$

LaTeX: $\large \text {In our euclidean algorithm, we have a remainder equals to 2 in} \; L_2 : 90 = 11 \times 8 + 2$

LaTex: $\large L_2 : 90 - 11 \times 8 = 2$

LaTex: $\large L_2 : 90 - (101 - 90 \times 1) \times 8 = 2 \; \Leftarrow [L_1 : 101 - 90 \times 1 = 11]$

LaTex: $\large L_2 : \color{#D61F06}{-8} \color{#333333}{ \times 101 }\; \color{#20A900}{+ 9} \color{#333333}{ \times 90 = }\color{#3D99F6}{2 }$

LaTex: $\large D = 2 \; \text {is the right number :} \; H = \color{#D61F06}{-8} \; \color{#333333}{and} \; A = \color{#20A900}{+ 9} \color{#333333}$

LaTex: $\large H\in \{\pm 1, 2, 3, 4, 5, 6, 7, 8, 9\} \; \text{and} \;A\in \{\pm 1, 2, 3, 4, 5, 6, 7, 8, 9\}$

LaTeX: $\large \boxed {D = 3} \; -101\color{#D61F06}{H} \color{#333333} \;{+ 90} \color{#20A900}{A} \color{#333333}{= 3}$

LaTeX: $\normalsize \text {There is no remainder equals to 3, then we have to use the equation for D = 1, to multiply it by 3:}$

LaTex: $\large L_3 : \color{#D61F06}{41\times 3}\color{#333333}{ \times 101 }\color{#20A900}{- 46 \times 3}\color{#333333}{ \times 90 =} \boxed {\color{#3D99F6}{1\times 3}}$

LaTex: $\large L_3 : \color{#D61F06}{123}\color{#333333}{ \times 101 }\color{#20A900}{- 138}\color{#333333}{ \times 90 =} \boxed {\color{#3D99F6}{3}}$

LaTex: $\large H\notin \{\pm 1, 2, 3, 4, 5, 6, 7, 8, 9\} \; \text{and} \;A\notin \{\pm 1, 2, 3, 4, 5, 6, 7, 8, 9\}$

LaTex: $\large \text {We see that for D >2, the values H and A are not part of} \; \{\pm 1, 2, 3, 4, 5, 6, 7, 8, 9\}$

LaTex: $\large \text {If} \; D = 2k \; \text{and} \; (k \in \mathbb Z\setminus \{-1, 0, +1\}) \; \Rightarrow L_3 : \color{#D61F06}{-8\times k}\color{#333333}{ \times 101 }\color{#20A900}{+9 \times k}\color{#333333}{ \times 90 =} \boxed {\color{#3D99F6}{2k}}$

LaTex: $\large \text {If} \; D = n, n>2 \; \text{and} \; n \neq 2k \; \; (n;k) \in \mathbb Z^{*2} \Rightarrow L_3 : \color{#D61F06}{41\times n}\color{#333333}{ \times 101 }\color{#20A900}{- 46 \times n}\color{#333333}{ \times 90 =} \boxed {\color{#3D99F6}{n}}$

LaTex: $\Large \text {Thus, the only solution is : } \; \color{#D61F06}{\boxed {\color{#3D99F6}{D = 2,} \; \color{#D61F06}{H = 8,}\ \; \color{#20A900}{B = 9}}}$

D+H=10, given the first column

H+A+1=10+H, given the first and second columns

A+1=10 and A=9, solving for the above equation

H+1=A, given the last column

A=9, so H=8

D+H=10=D+8, substituting for the first equation

D=2

:)

892+89+8=898

D + H = 10 since the digit of 'A' stays

A + H + 1 = H + 10 since one carries over and the digit of 'H' stays

H + 1 = A the last sum since one is the most that can carry over given the conditions

sub (H+1) for A in equation 2, we get 2H + 2 = H + 10 Therefore H = 8, and D = 2 by equation 1

D+A+H=10+A
A+H+1=H so A+1=10 therefore A=9
H+1=A so A(9)-1=H=8
D+A+H=10+A goes to D+8=10 D=10-8=2

D + H = 10 1 + A = 10 A = 9 1 + H = A H = 8 D = 2

100H+10A+D+H = 100A.
=> 90A = 101H + D = 90H+(11H+D).
=> A = H + (11H + D)/90.
=> 11H + D = 90.
=> 11H = 88 + (2 - D).
=> H = 8 + (2 - D)/11.
=> D = 2

By knowing that $H=A-1$ and subtract away 111H of the question we get

$AD+A=101$ which stands for $A=9$ and so $D=2$ and $H=8$

In the units column, we have D+A+H=A, meaning that D+H must be equal to either 0 or 10(since 20 is too big for the biggest single-digit numbers to sum to it - 9+9=18<20). Since for D+H to equal to 0, both D and H must be 0, but this is not an option, meaning D+H=10. In the tens column, we have A+H=H, and we carried one over from D+H in the units column, so A has to be 9, carrying 1 over to the hundreds column. In the hundreds column, we have H+1(because we carried 1 over)=A, so H=A-1=8. From our first equation(D+H=10), we get D=10-H=10-8=2.

Thus D=2.

d= 90 a -101 h
2= 810 - 808

The key starting point for this equation was the middle column, $A+H=H$ . At this stage, it is most likely that A and H have a difference of 1 as $A + H + 10$ could equal H. If you read the last column, $H = A$ , A must be larger than H as A. List out all the possibilities of A + H that equals H and replace the letters in the left and middle column with the possibilities. Now work it out, starting from the middle and just pretend you have a 10 when solving the middle column (tens column) as you'll see why later.

If it makes sense, move onto the very right column and replace A and H. By now, you should have $A = 9$ and $H = 8$ . Now all you need to do is replace the numbers from 1 - 5 into D and you will soon get an answer which is D = 2.