# 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 |

**NEXT**

**If anyone faced any problem for solving Crypt problems comment below**

**Follow us on Facebook in Below FOLLOW US (Hit LIKE) link for latest Update**

## Now Everyday our website visitors can WIN Free IPL ticket according to their choice

**Category**:
Books To Follow For eLitmus Test,
Crypt Arithmatic,
Cryptarithmetic Problem,
eLitmus Syllabus And Question Paper Pattern,
Multiplication,
pH test,
Rules for solving Cryptarithmetic Problems

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

ReplyDeleteFirst 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.)

ReplyDeleteNow 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.

This comment has been removed by the author.

DeleteThis comment has been removed by the author.

DeleteU=6 s clear. How you approached for Y=2/4? Why not 8?

DeleteNice approach.

DeleteBut how to reach decision for Y as 2 or 4 only, but not as 8?

how did u take y as 2 n 4 not 6 and 8?

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

ReplyDeletewhy it cant be 6 and 8 => 8*6=48 and

ReplyDelete@admin

ReplyDeleteur approach is good i have one small doubt

in 1st step you said U=6 and Y=(2,4) but,here Y can take 8 also

what makes you think to leave that possibility(y=8)?

This comment has been removed by the author.

ReplyDeletein the beginning how did u take u as 6 and y as 2 or 4

ReplyDeleteWhy y can't b 5

ReplyDeleteHello i am completley new to these kind of problems can u help me in understanding from basics

ReplyDeleteHello i am completley new to these kind of problems can u help me in understanding from basics

ReplyDelete