WEBVTT
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Each letter in this cryptarithmetic puzzle represents a different digit, and none of the numbers use leading zeros.
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Find the value of each letter in the puzzle in order to make a correct sum.
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We’re gonna solve this puzzle using logic and elimination.
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First, I’m just gonna move this over to make some space.
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And then we’re going to list out each of the letters C, R, O, S, A, D, N, G, E.
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We list out all of the digits that each of these letters could represent.
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Almost immediately, we can look at this D and determine what number it must represent.
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To find the value of D, we’ll need to add C and R.
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But even if C and R were the largest of the digits, for example, nine and eight, they would add up to 17.
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So no matter what the digit D represents will have to be equal to one.
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It cannot be greater than one.
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We also know there are no leading zeros, which means that D cannot be equal to zero.
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D must be equal to one.
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And that means D is not equal to any of these other numbers.
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It also means that none of the other letters can be equal to one.
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That’s a great start.
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But unfortunately, that’s really the only one that’s immediately obvious to us.
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But I do recognize that S plus S equals R.
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And we can draw some conclusions from this.
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If we know that something plus itself equals R, we can eliminate some of the values.
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Because R is one digit, S must be less than five.
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Five plus five would be equal to 10.
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And 10 has more than one digit.
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We don’t know what S is, but we do know some things that S could not be.
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It can’t be larger than five.
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But what about zero?
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Could S be equal to zero?
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No, that’s not possible.
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Zero plus zero would equal zero.
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Then the statement would have to say S plus S equals S.
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We have three options for our S value.
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And because we have three options for our S value, that also means we have three options for our R value.
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Our first option, two plus two equals four.
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That means R could be equal to four.
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Three plus three equals six.
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R could be equal to six.
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And four plus four equals eight.
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These are the only three options for R.
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And we can eliminate all the other digits.
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Again, we don’t know what R is, but we do know some things that R isn’t.
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Moving on to S plus D equals E, S plus D equals E and we already know that D equals one.
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Here again, we don’t know what S is, but we do know its three options.
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One option is two plus one equals E.
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E could be equal to three, or three plus one equals E.
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In that case, E would be four.
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The final option for E is four plus one.
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E could be equal to five.
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E cannot be equal to zero, two, six, seven, eight, or nine.
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We said a few things about C plus R in the beginning, but we’ll look closer at C plus R now.
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C plus R equals something with a one in the tens place.
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If we look to R, we’ve eliminated all but three possible options.
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R would be equal to four, six, or eight.
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If R equals four, what could C be equal to?
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Let’s try six.
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C equals six.
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Six plus four equals 10.
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Okay, in this scenario, C equals six and A equals zero.
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But if A equals zero, then we have a problem, because anything plus zero equals itself.
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What would happen would be the letter O plus zero would have to be equal to the letter O and not the letter G.
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What does this tell us?
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First of all, it tells us that A cannot be equal to zero.
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It also tells us that C cannot be equal to six.
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Okay, but that’s not the only case when R is four, is it?
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How about seven?
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Could C be equal to seven?
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Well, seven plus four equals 11.
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And because we know that D represents the one, seven plus four would equal DD, which means C cannot be equal to seven.
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Is there anything else we can add to four?
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We could try eight.
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Eight plus four equals 12.
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And so far, we have no reason to eliminate these two as options.
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So we can circle C is maybe equal to eight and A is maybe equal to two.
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We have one more case if R equals four.
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That’s the case where C equals nine.
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Nine plus four equals 13, and we’ll circle nine for C as an option and three for A as an option.
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So what we’ve just said is that if R equals four, C could be eight or nine and A could be two or three.
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Now we want to look at the cases where R equals six.
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Starting with four, four plus six equals 10.
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But we’ve already seen that A cannot be equal to zero.
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And that means C could not be equal to four.
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This is not a valid option.
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We could try five.
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Five plus six equals 11.
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We’ve already seen that A cannot be equal to one.
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And if A can’t be equal to one, C cannot be equal to five.
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We’ve already eliminated C equals seven.
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C equals eight still seems like it could work.
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Eight plus six equals 14.
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If C equals eight and R equals six, then A would be equal to four.
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Those are all the cases for R equals six.
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Now we want to examine the case if R equals eight.
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If C equals two, two plus eight equals 10.
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We know that A is not zero.
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Therefore, C is not equal to two.
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Three plus eight equals 11, and A cannot be equal to one.
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Therefore, C cannot be equal to three.
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We’ve already eliminated four, five, six, and seven as choices for C.
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Our next option would be eight.
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Eight plus eight equals 16, but did you spot the issue here?
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If R equals eight, then C could not be equal to eight, which means that’s not an option for us.
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The last case would be C equal to nine.
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Nine plus eight equals 17.
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If C equals nine, R equals eight, then A would be equal to seven.
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And now what?
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If we look closely, we’ll see that we’ve actually missed one of our pairs.
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We’ve seen the case where C equals eight and R equals six.
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What about the case where C equals nine?
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If C equals nine, then A would be equal to 15.
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And so now we have a fifth option for what A could be.
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At this point, it can feel like we’re getting more and more options instead of less and less.
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But we go ahead and cross out what we know A cannot be.
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On my list, the letter C, we see that we can have zero, eight, or nine.
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Let’s consider the case if C was equal to zero.
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If C was equal to zero and R can only be four, six, or eight, we would never end up with a one as the D.
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And that helps us eliminate the zero as an option for C.
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At this point, I want to take all of the small addition problems and write them down horizontally.
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And then I’m going to try all the cases where C equals eight and C equals nine.
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Here we go.
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C equals eight and R equals four.
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There we have 12.
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If R equals four, we put the four again.
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We don’t know O; we don’t know N; we don’t know O; but A would be equal to two, and we still wouldn’t know G.
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In this case, S has to be equal to two.
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Two plus one equals three.
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E would be equal to three.
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And two plus two equals four.
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This set didn’t help us eliminate anything, so we’ll just leave it there.
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And then we’ll move on for C being nine and R being four.
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Nine plus four equals 13.
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Four plus O equals N; O plus A equals G, but we know that A equals three.
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In this case, S is equal to two and D equals one.
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Two plus one equals three.
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But look at what’s happened.
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In this case, both A and E are showing up as three.
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And we know that there cannot be duplicate digits.
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A cannot be equal to three.
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This time, we’ll choose C equals eight and R equals six.
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Eight plus six equals 14.
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Six plus O equals N.
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We can’t solve that.
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O plus four equals G.
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S plus D, three plus one, equals four.
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And we’ve run into an issue because now we have A and E showing up as four.
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We can eliminate A equal to four.
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The next round, C equals nine and R equals six.
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Nine plus six equals 15.
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Six plus O equals N.
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We can’t say anything about that.
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O plus five equals G.
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Three plus one equals four, and three plus three equals six.
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What I wanna do now is write out which digits we’ve used in both of these cases.
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As I’m writing all of this down, I notice, and maybe you have too, a mistake in this set.
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Look at how many times the two is used.
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That’s because we’ve used the value two for both A and S in this instance.
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A cannot be equal to two, and this option is not valid.
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Now that we’re down to this pink set, I want to write out all the digits we’ve already used.
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We’ve used the values one, three, four, five, six, and nine, which means we still need the values two, seven, and eight.
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We have three missing letters: O, E, and G.
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Is there a way to use two, seven, and eight and make all of these statements true?
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If I plug in two for O, then I’ll end up with six plus two equals E.
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The two seems to work.
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Six plus two equals eight.
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And the last question, “Is two plus five equal to seven?”
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It is.
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It now seems like we’ve filled all of these spaces.
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Let’s go through and make sure we haven’t made any errors.
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We’ve said that C equals nine, and that means no other value can be equal to nine.
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We’ve made R equal to six.
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So far so good.
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We’ve already identified D equals one.
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Now we have A equals five.
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Next, O equals two, G equals seven, N equals eight, S equals three, and E equals four.
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C equals nine; R equals six; O equals two; S equals three; A equals five; D equals one; N equals eight; G equals seven; E equals four.
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If you’ve made it to the end, you’re probably wondering, “Is this the only way to solve this?
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Is this the best way to solve this?”
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This is the way that you solve a problem like this by hand.
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Today, most often people use computer programs to lessen the time it takes to solve something like this.
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Computers are able to do the process we just completed, but faster and more efficiently.
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However, if you’re ever stranded on a desert island and you need to solve a cryptarithmetic puzzle, this would be the way to do it.