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Given multiplication is $\begin{array}{cccccc} & & & P & E & N \\ & & & I & N & K \\\hline & & L & K & P & R \\ & I & R & T & N & \\ E & A & K & N & & \\\hline L & E & T & T & E & R \end{array}$

how to find the values of the letters?

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    If you get stuck: http://www.math.ubc.ca/~israel/applet/metic/metic.html2011-04-20

3 Answers 3

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Start from N*N ends in N, so N must be 5 or 6 (can't be 0 or 1 from the second line in the sum). Then I*N ends in N (and I !=1) says N is 5 and I is odd. Then K is even and R is 0. Continue

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    It has in all cryptarithms I have ever solved and matches the way I was taught to do multiplication.2011-04-20
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Rules:

1) If $ A+B=A$ then $B= 0$ or $9$.

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6 1 5 * 354

 2 4 6 0 

3 0 7 5

1 8 4 5

2 1 7 7 1 0

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    observe column KTN T ==> K+N=10 ; it cannot be 5+5.. if K cannot greater than 5 since adding 6 or greater will result in value greater than 10 and condition K+T+N= T fails. therefore K must be less than ; also observe 1st partial product P * K = K (+L ->carry) implies K is even and P is 6. so K can be now either 2 or 4. since 0 6 are already consumed. Now observe N(5)+P= E +(10). We know P=6; therefore E= 1; Now you know P E N K R values, do multiplication and you will get values in first partial product. which gives L=2 and from column L+R+K=T ==> T=L+5;2013-01-29