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I got this question from a math contest, and I need to know how to find the following. I am rather new to cryptarithms, and I really need to master that skill. For the contest, I needed to find out the following:

If $CARIBOU+CARIBOO=3456789$

find $C+A+R+I+B+O+U$

I originally started with writing it as an addition problem, and realized that $U+O=9$ and listed all of the sums down for each of the numbers. However, I don't know how to proceed with it further. Is there a strategy to do a question like this? Am I missing something out?

RobPratt
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ShadowKnight700
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  • For similar simple puzzles try: https://puzzling.stackexchange.com/questions/129615/pqr-qr-1012-alphametic AND https://puzzling.stackexchange.com/questions/126218/jmo-jmo-jmo-imo-alphametic – Will.Octagon.Gibson Dec 25 '24 at 05:45

5 Answers5

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Hint:

$\rm CARIBO\times 10 + U + CARIBO\times 10 + O = 3456789$, so $\rm 2\times CARIBO = 345678$

jjagmath
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since O+O=8 or O+O=18 now 2xO has to be 18 since if it is 8 then no carry over this means that B+B=7. Which is a contradiction. Hence we get O=9,B=3 or B=8 again applying same logic B+B will be 6 as if B+B is 16 then carry over is observed implying that I+I+1=6. Which is a contradiction as I cannot be a fraction. Hence I+I=6 Which means B=3 and I=8 now similarly R+R+1=5 this means R=7 or R= 2 now carryover wont occur implies R=2,A=7 AND C=1 Therefore sum is 1+7+2+8+3+9+0=30

Anuradha Banthia
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2

Imagine simply dividing $3456789$ by $2$ to see where it lands. It is roughly $1728394.5$, suggesting each summand is close to $172839x$. One of them ends in $U$, the other in $O$, and their last digits must add to $9$. This makes it natural to check $1728390$ and $1728399$, which indeed sum to $3456789$. Matching these to CARIBOU and CARIBOO reveals:

$C = 1$, $A = 7$, $R = 2$, $I = 8$, $B = 3$, $O = 9$, $U = 0$

Since all letters match distinct digits and produce the correct sum, the cryptarithm is solved. Finally, the sum $C + A + R + I + B + O + U$ is $30$.​​​​​​​​​​​​​​​​

Raynard Bond
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The cryptarithm you are trying to solve is:

  CARIBOU
+ CARIBOO
  -------
  3456789

In the units column, $\ U+O \le 9+9=18 \ $ but $\ U+O \ $ is a sum that ends in $\ 9.$
Therefore, $\ U+O=9 \ $ and there is no carry into the tens column.

Therefore, $\ 2 \times CARIBO = 345678.$

So, $\ CARIBO = \dfrac{345678}{2} = 172839.$

So, $\ C = 1, \ A = 7, \ R = 2, \ I = 8, \ B = 3, \ O = 9.$

Since $\ U+O=9 \ $ and $\ O = 9, \ $ we know $\ U=0.$

Finally, $\ C+A+R+I+B+O+U = 1+7+2+8+3+9+0 = 30.$

Will.Octagon.Gibson
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2

Any digit added to itself will produce an even number. And a value of at most 1 can carry over from 1 digit to the next. So we know any odd numbers (except the least significant digit) is the result of carry-over.

Now you can just go from left to right and directly calculate every value.

$C + C + carry = 3$

Since the result is odd, we know there was a carry ($= 1$), thus $C + C = 2$, and $C = 1$.

Then:

$A + A = 14$ (remembering the carry from the previous step)

Since the value is even, there was no carry. So $A = 7$.

And do the same for R, I, B and O.

When you get to the final digit, we already know $O = 4$, thus $U = 9-O = 5$.

Then add those all together and you have your answer.

NotThatGuy
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