Mind Your Decisions. I’m Presh Talwalkar. Solve for the value of each color
in the following arithmetic problem. Each color represents a different digit from 0 to 9. I thanks Lucas for suggesting this problem. It originally appeared as a Belgian Olympiad problem with the letters M, A, and T. Pause the video if you’d like to give this problem a try and when you’re ready, keep watching
to learn how to solve this problem. We’ll get started by focusing on the letter M. Suppose M equals 5. In that case, the sum will be at least 1500. And this is going to be too large of the value because we need the sum to be 1416. Therefore M equals 5 produces a sum that’s too large and it’s not possible. The same logic is true for M
greater than or equal to 5. We can eliminate all of these possibilities. What if M equals 3? In that case, the largest value of the sum will be 1197. And that’s too small because we need
the sum to be 1416. Therefore M equals 3 cannot be a possibility and M less than or equal to 3 can also be eliminated. What remains is M is equal to 4. This will simplify our calculation. If we substitute M is equal to 4 we know that just from the terms involving M we get a sum of 1204. We subtract those out we get the resulting sum of 212 and we have a new problem. We now need to solve for A and T. Let’s focus on T. In the unit’s column, T plus T can either equal 2 or it can equal 12. Consider the first case. If T plus T equals 2 then T is equal to 1. Which then means the entire problem has to be 10 A plus 10 A plus 1 plus 11 is equal to 212. This means 20 A equals 200 and A is equal to 10. But that’s not possible because A has to be
a single-digit from 0 to 9. Therefore we must have T plus T is equal to 12. This means T is equal to 6. And we can then go ahead and solve that A must be equal to 7. Therefore this is the answer to the problem. M is equal to 4, A is equal to 7, and T is equal to 6. And we can, in fact, verify this does work out. Thanks for making MindYourDecisions
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