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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