Friday 16 August 2013

Multiplication Cryptarithmetic Puzzle - 4


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Elitmus pH Test Pattern & Syllabus

Tips For Elitmus Test

Tips To Solve Cryptarithmetic Puzzles

The puzzle that we will be solving today looks something like this,


W H Y * N U T
--------------------
        O O N P
      O Y P Y +
  O U H A + +
--------------------
 O N E P O P


Follow the steps to solve the above puzzle:

Step 1: The term (W H Y)* U = O Y P Y, we can guess that U=6 and Y=2,4,8. Whenever there are more than one occurence of the same character in these kind of problems, Y is usually 4(No all the times though). So let us take y=4 as proceed to see if the assumption is correct.

 W H 4 * N 6 T
--------------------
        O O N P
      O 4 P 4 +
  O 6 H A + +
--------------------
 O N E P O P

Step 2: Looking at the sum term, O+6=N, and also since there is no carry from this sum to the next term, we conclude that N<=9, and as a result O could be 1/2/3. Considering O=1 as proceeding we get,

 W H 4 * N 6 T
--------------------
        1 1 N P
      1 4 P 4 +
  1 6 H A + +
--------------------
 1 N E P 1 P


Step 3: Now looking at the term N+4=1, we can easily guess that N=7, since there is no carry being carried to this term from P(which is the previous term). Rewriting with N=7 would yield,

 W H 4 * 7 6 T
--------------------
         1 1 7 P
      1 4 P 4 +
  1 6 H A + +
--------------------
 1 7 E P 1 P


Step 4: Looking at the product term (W H 4)*7, we can guess that, A=8. And since the last term in all the sum term of individual products (i.e. "1 1 7 P", "1 4 P 4", "1 6 H A") is always 1, the minimum value W can take is 2. So considering W=2, we get

  2 H 4 * 7 6 T
--------------------
         1 1 7 P
      1 4 P 4 +
  1 6 H 8 + +
--------------------
 1 7 E P 1 P

Step 5: Looking at the product term, (2 H 4)* T = 1 1 7 P, we can guess that T=5, since (2 H 4)*T = 1 1 7 P. And, further it is safe to assume P=0, since (2 H 4)*T(=5). Rewriting we get,

  2 H 4 * 7 6 5
--------------------
         1 1 7 0
      1 4 0 4 +
  1 6 H 8 + +
--------------------
 1 7 E 0 1 0

Step 6: Now, (2 H 4)*5=1 1 7 0, we find H to be 3, and hence E turns out to be 9 (1+4+3+carry_of_1=E). This gives us the solution of,

2 3 4 * 7 6 5
--------------------
         1 1 7 0
      1 4 0 4 +
  1 6 3 8 + +
--------------------
 1 7 9 0 1 0

which is the required result.


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14 comments:

  1. Solve these cryptarithmetic problems-1.OTO*HAS
    --------------------

    TICO
    CHCK
    CLAI
    --------------
    COOKOO
    ------------------


    2.PXB*WYA
    OAZO
    ONXW
    OXNP
    ---------------
    OANZNO
    --------------------
    3. Solve these questions of time,speed and distance
    a)Without stoppage a train travels at an average speed of 75 km/h and with stoppage it covers the same distance at an average speed of 60km/h.How many minutes per hour does the train stop ?
    b)Ravi,who lives in a country side,caught a train for home earlier than usual yesterday.His wife normally drives to the station to meet him.But yesterday he sat on foot from the station to meet his wife on the way.He reached home 12 minutes earlier than he would have done had he waited at the station for his wife.The car travels at a uniform speed,which is 5 times Ravi's speed on foot.Ravi reached home exactly at 6 O'clock.At what time would he have reached home if his wife ,forwarned of his plan,had met him at the station.

    ReplyDelete
  2. The cryptarithmetic problems doesn't seem to have a solution. i'm pretty sure of that. Please make sure you have typed in the correct alphabets.

    Coming to the time distance problems, the answers would be:

    a. 12 min/hr

    b. 5:36 o'clock

    ReplyDelete
  3. please explain solution of time distance problems

    ReplyDelete
  4. Ah, sure.

    1) train's speed on it's own = 75 kmph
    with stoppages it loses 15 kmph, which it would have covered in 12 minutes. (i.e. since it covers 75kms in 1 hour, it will cover 15kms in 12 mins)
    so it stops for 12 minutes.



    2. Given that Ravi reaches home 12 mins earlier than he wud have if he waited for his wife at the railway station. So this implies that his wife drives for a duration which is 12mins less than the usual days. This 12 mins should be divided into the to and fro journey i.e. she saved 6 mins each way.
    Now since Ravi's speed is 5 times less than the car's, he walked for a duration of 30mins to cover the same distance where he meets his wife on the way.
    now, what this implies is that Ravi spends (30-6 = 24)mins extra as he walks to the point of meeting.
    So, if he had forewarned his wife, he would have been at his home at (6:00 - 0:24) which is 5:36


    Hope it is clear!

    ReplyDelete
  5. why (30-6) at the end ... plz explain

    ReplyDelete
  6. Okay... So here I try again,

    Distance between stations and home is d and he meet at d1 distance that means any how his wife has to travel d -d1 distance , she get relaxation to not to cover that d1 distance which he covered , if husband had to wait at station wife had to cover 2d1 in to and fro that means 2d1 is equals that time which he saved that is 12 min so d1 is 6 min.

    Had he forewarned his wife, she would have come and waited for him, so he would have travelled D1 by car in 6mins instead of 30mins walk.
    so he would have saved 30-6 = 24 mins.

    ReplyDelete
  7. it was really best explanationa ... thanx a lot :)

    ReplyDelete
  8. ROZ
    x MUG
    -----------------
    TAUA
    TOYZ
    TUGA
    -----------------
    RATTTA

    PLEASE SOLVE THIS PROBLEM. WAS ASKED IN ELITMUS 1ST SEPT 2013 TEST.
    THANKS

    ReplyDelete
  9. The solution would be
    345*876
    -----------
    2070
    2415
    2760
    -----------
    302220

    ReplyDelete
  10. Narendra Kumar Jha9 September 2013 at 09:40

    thanks. after trying out for million tyms, i really thought the question was wrong.. bt u made it.. :)

    ReplyDelete
  11. Admin thanks a lot from today onwards i will visit regularly and one request from admin and all friends please post the correct answer of each and every question because when i stude from m4math there is lot of confusion because different people give different-2 answer so please post correct answer with suitable explanation. Thanks a lot great work :)

    ReplyDelete
  12. plz share also these cryptarithmetic puzzles also.I have tried too much.

    ReplyDelete
  13. the answer of OTC*HAS is 432*567
    u have posted the wrong question @namrata

    ReplyDelete
  14. the second cryptarithmetic question is also wrong so plz chck it out first and then post the right questions it vl b beneficial to all of us and u 2.

    PXB*A=OAZO yu shd. strt wit tis by using the values 1 to 9 for O. A*B=O.. so, lets take O=1 Possible values are (3,7 and 7,3) and for O=2 Possible values are (3,4),(4,3),(6,7), (7,6), (4,8),(8,4),(9,8),(8,9) and for 3 only possible value is (7,9) and for 4 7,2 2,7 8,3 3,8 6,9 9,6 and dnt take 5 coz Z+W=N So z must nt be 0, and so on.. yu hv to chck.. none of the combination iS possible. so ques. is wrng.

    ReplyDelete