Cryptarithmetic Problem-1

Multiplication

			W	H	Y
			N	U	T
		O	O	N	P
	O	Y	P	Y
O	U	H	A
O	N	E	P	O	P

When you see second multiple (W H Y ) * U = (O Y P Y)

So, The number for U = 6 and Y is (2 , 4) Lets take the value of Y is 4 and replace it in figure.

			W	H	4
			N	6	T
		O	O	N	P
	O	4	P	4
O	6	H	A
O	N	E	P	O	P

Clearly if you see the yellow portion O + 6 = N and there is no carry to the next digit so we can confirm that ‘N’ is single digit number. If ‘N’ is single digit number ‘O’ should be less than (O < 3)

Let’s take O = 1 and redraw the table…

			W	H	4
			N	6	T
		1	1	N	P
	1	4	P	4
1	6	H	A
1	N	E	P	1	P

If you observe 1 + 6 = N & N + 4 = 1 so, we can clearly the value of N =7

So replace N = 7 and redraw the table.

			W	H	4
			7	6	T
		1	1	7	P
	1	4	P	4
1	6	H	A
1	7	E	P	1	P

Take the third multiple (W H 4) * 7 = 1 6 H A from this we get A = 8, and if you replace the H with 3 and w = 2 then the equation satisfies.

Like (2 3 4) * 7 = 1 6 3 8 so redraw the table with particular values.

			2	3	4
			7	6	T
		1	1	7	P
	1	4	P	4
1	6	3	8
1	7	E	P	1	P

The remaining numbers are (0,5,9) It is simple to analyse we can’t get 5 in place of p because when you multiply with 4 we cannot get the last digit as 5.

So T = 5, P = 0, E = 9

After redraw the table.

			2	3	4
			7	6	5
		1	1	7	0
	1	4	0	4
1	6	3	8
1	7	9	0	1	0

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Somdeb Burman

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shubham swarup

July 25, 2015 at 5:37 AM

In first step we can also take U=3 & Y=5 .So there are many possibilities how can you guess the right one..

Somdev Burman

July 25, 2015 at 11:00 PM

First of all Y never be 5 (cause 5* odd= 5 and 5*even= 0.So if Y=5 then Y*odd=Y and Y*even=0 so only two letter with signify 0 and 5 where as Y=5 then ) will be some other letter.Now Y*T=P and Y*N=A, so either P=0 or A=0 but both together can't be 0.So Y= 5 never be possible.)
Now U = 6 because 6*even = that number. So 6*Y=Y (Here Y maybe 2 or 4)

If any further problems are there pls comment here.

July 25, 2015 at 11:03 PM

Think in proper way.before starting direct problems just think what will be the possibilies and combinations.

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Cryptarithmetic Problem-1

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