Challenge Questions

Home Forums VMTTC Online Discussion Forum Challenge Questions

Viewing 28 reply threads
  • Author
    Posts
    • #28407
      Pasyanthi
      Keymaster

      These questions are not part of the course assessment, but are open to everyone on the course. They are only for those who are interested: you can offer your thoughts on all, some or none of them.

      The idea is to look at a Vedic method/sutra from another perspective, in a way that extends the range of application of VM. I hope you will get the idea from the first questions. There are currently two questions here.

      Please wait for 24 hours before posting a reply so that everyone has the chance to have a go before answers start appearing

       

      Challenge Question 1

      We will write bar numbers here using brackets.

      So 46 will be 5(4)

      and 467 will be 5(33).

       

      Convert the following to standard form (i.e. where all the digits are positive)

       

      1. 1(1)

      2. 2(22)

      3. 3(303)

      4. 6(7)8(9)

      5. 77(03)1

      6. 8(01)9(01)

       

       

       

      Challenge Question 2

      The product of 2 numbers is seen but one of the digits is smudged. So you can only see:

       

      4 ? 2 x 7 4 = 3 3 4 4 8

       

      where the question mark is the smudged digit.

       

      Can you use digit sums to find the smudged digit (explain how you got the answer)?

      • This topic was modified 9 months ago by Pasyanthi.
      • This topic was modified 8 months, 3 weeks ago by Pasyanthi.
      • This topic was modified 1 month ago by Pasyanthi.
    • #29123
      Anupama Cherukuri
      Participant

      for Challenge Question 2 :

      In the Multiplication, the DM of the product of their DM will be equal to the DM of the Product of the original numbers.

      4?2 X 74 = 33448

      (4+? + 2 ) X DS of 74 is 2 = DS of 33448 is 4

      (6 + ?)  times 2 = 4

      so the Digit sum of the 6+? must be 11 in order to get the Digit Sum of 2 so the ? must be 5.

      452 X 74 = 33448

      DM check  = 2 X 2 =4

    • #29195
      Satya
      Participant

      How I solved the Challenge question2 : 4?2 X 74 = 33458

      Going by the digit sums above equation can be written as –> ? X 2 = 4

      so digit sum of the first number has to be 2 for the above equation to be true. Digit sum of the first number with already given numbers there is 6. If a 5 can be added, digit sum of it would become 11 thats equal to 2. So the smuged number in the first figure is 5.

      452 X 74 = 33458

      Still trying to understand Challenge question1.

    • #29196
      Satya
      Participant

      sorry typo – its 33448 not 33458

    • #29228

      Challenge 1.

      We can convert bar numbers to normal numbers by applying “one less” and “all from 9 last from 10”.

      1. 1(1) = 09 ( one less than 1(in 10s place) is 0 and applying  all from 9 last from 10 to bar 1 will result in 9  )

      2. 2(22) = 178 (one less than 2(in 100s place) is 1 and apply all from 9 last from 10 to bar 22  will result in 78)

      3. 3(303) = 2697

      4.  6(7)8(9) = 5371

      5. 77(03)1 = 76971

      6. 8(01)9(01) = 799899

      Thanks.

    • #29229
      Anupama Cherukuri
      Participant

      Challenge 1 :

      1.9

      2.178

      3.2697

      4.5371

      5.76971

      6.799899

      Thank you so much for the very interesting challenge questions.

       

    • #29231
      Satya
      Participant

      yeah, understood after Lesson 10.

      How I solved challenge question 1:

      1. 1(1) = 10-1 = 9

      2. 2(22) = 200-22 = 178

      3. 3(303) = 3000-303 = 2697

      4. 6(7)8(9) = 60-7 80-9 = 5371

      5. 77(03) 1 = (7700-03) 1 = 76971

      6. 8(01)9(01) = 800-01 900-01 = 799899

    • #29249
      KennethWilliams
      Keymaster

      Thanks for these answers – all correct.

      Challenge Question 3

      How could you use the Vedic formula The First by the First and the Last by the Last to find 4187 divided by 79, given that the answer is exact?

      [Please wait 24 hours before answering]

      Challenge Question 4

      In how many ways can the number 2019 be represented as a 4-figure number, and what are those representations?

      [Please wait 24 hours before answering]

      • #29284
        Satya
        Participant

        Solution to challenge question 3: 4187 / 79

        Last digit of the dividend[4187] is 7 and the last digit of the divisor[79] is  9. Going by the Last by the Last, the number that has a 7 in units place in 9th table is when it gets multiplied with 3/13/23 etc – 9×3 = 27. So the last digit(unit place) of the quotient is 3.

        Now the first 2 digits of the dividend are 41 and the first digit of the divisor is 7. Going by ‘the first by the first’, the closest value to 41 in 7 table is 7×6 = 42 but since 41 is less than 42, we will consider 7×5 = 35. So the first digit(tens place) of the answer is 5.

        so 4187 / 79 = 53

        Verify: 79×53 = 4187

      • #30261
        Preeti Pathak
        Participant

        Few number of ways in which 2019 can be is  represented as follows(in terms of bar numbers):

        202(1)

        21(81)

        3(981)

        Thanks

         

    • #29357

      Challenge Question 4

      In how many ways can the number 2019 be represented as a 4-figure number, what are those representations?

      Scenario 1 : When repetition of digits are allowed, we can write 2019 4-figure number in 3*4*4*4 = 192 ways.

      Thousand’s place will have only 3 possible ways (2,9,1) because,  if we use 0 it will become 3-figure number.

      In Hundred’s, ten’s and unit’s place we can use any digits as repetition is allowed.

      T h(9,1,2) H(0,1,2,9) T (0,1,2,9) U(0,1,2,9)

      Example : 9999,1209, 2221, 1192 etc

      Scenario 2: when repetition is not allowed, we can write 2019 4-figure number in 3*3*2*1 = 18 different ways.

      Thousand’s place value will have 3 possible options  – 2,1,9   Example: we take 9

      Hundred’s place value will have 3 possible options – any of three numbers except the one used in thousand’s place. Example:  we can take any one digit other than 9 so let us take 1;91

      Ten’s place value will have 2 possible options –  any one of the two numbers that are not used in thousand’s and hundred’s place. Example : we can take 2; 912

      Unit’s place value will have 1 possible option – use digit that not used in above three. Example 0. ; 9120

      Th (9,1,2) H(one of any 3 digits other than the one used in Th’s place) T(one of two digits other than digits used in Th’s and H’s place) U(Whatever digits is left).

      2019 , 2091, 2901, 2910, 2109 ,2190

      9012, 9021, 9102, 9120, 9210, 9201

      1029, 1092 , 1902, 1920, 1209, 1290

      Thanks.

      • #29362
        KennethWilliams
        Keymaster

        Thanks for this Mallika. But these are not alternative representations of 2019 as they are different numbers.

        The alternative representation must have the same value. There are actually 5 of them.

    • #29363

      🙂 Sorry Ken. I got carried away by permutations and combinations. 🙂

      I am glad I signed up for this course – this course is helping challenge my thought process. 🙂

      Here are those 5:

      202(1)

      3(99)9

      3(981)

      1(8)02(1)

      1(799)9

      Are these same 5 you had in mind? Or are there other combinations?

      I validated these with the digit sum. Is there a way to know how many combinations possible by looking at the number?

      Regards.

      • #29366
        KennethWilliams
        Keymaster

        Yes, you’ve got 3 of them now because only 4-figure answers were requested.

        2 more to go.

    • #29367

      21(81)

      21(9)9

      😊

    • #29799
      Shikha Dhingra
      Participant

      I am late to answer these challenge questions.

      challenge 1:

      1. 1(1) = 10-1 =9

      2. 2(22) = 200 – 22=178

      3. 3(303) = 3000 – 300=2700 – 3 = 2697

      4. 6(7)8(9) = 6000-700=5300 ;80-9=71; the number becomes 5371

      5. 77(03)1 = 70000 + (7000 – 30 ) +1 = 70000 +6960+ 1 = 76961

      6. 8(01)9(01) = 8(01)/9(01) = (800-1)/(900-1) = 799899

      • #29800
        KennethWilliams
        Keymaster

        Thanks Shikha. Correct except no 5 is 76971.

    • #29805
      Savita P Kulkarni
      Participant

      I am getting the challenge 1  no. 5 is – 76961, I know it is not correct but nor able to solve.

      • #29814

        Hi Sativa,

        77(03)1 –

        7 in ten thousands place is written as it is.

        Second step, subtract 03 from 700. 700-03=697

        1 in units place is written as it is.

        76971

        Thanks.

    • #29817
      KennethWilliams
      Keymaster

      Challenge Question 5 – Two Smudged Digits

      The product of two numbers is seen but two of the digits are smudged. So you can only see:

      3 a b   x   6 1   =   1 9 9 4 7

      where a and b are the smudged digits.

      Can you find the smudged digits (please explain your answer)?

      • #29852
        Anupama Cherukuri
        Participant

        3+ a+ b X 61 = 19947

        3 +a+b  X 7 = 3

        21+ a+b = 3

        So 21 Digit sum is already 3 so a+b must be 9

        So options are 2,7 0r 1,8 0r 3, 6 0r 4, 5 , I tried with those options and got the answer of 327

        327 X 61 = 19947

      • #29855
        Satya
        Participant

        Hi, I was waiting for 24hrs to complete before posting the answer…Since Anupama has already posted her answer (as she got confused with the time zone) here is how I solved Challenge Question 5:

        3 a b X 61 = 19947

        Going by ‘Last by the Last’, I know that value of b has to be 7 as the unit place of answer is 7.

        3 a 7 * 61 = 19947 –> now apply digit sums here

        1+a * 7 = 3 [after casting out 9s while adding digit sums]

        a has to be 2 for the above expression to be true –> 3 * 7 = 21 whose digit sum is 3.

        So a = 2, b = 7 making the above 327 * 61 = 19947

    • #29818
      KennethWilliams
      Keymaster

      Challenge Question 6 – multiplication by 111

      We know how to easily multiply by 11.

      Can this be extended to multiplication by 111?

      For example 234 x 111.

      • #29853
        Anupama Cherukuri
        Participant

        234 X 111 =

        first digit 2, second digit 2+3 , third digit 5 (2+3) + 4, fourth one  3+4 and last digit 4  so answer is

        25974

        I checked with so many examples even with carries

        327 X 111 = 36297

        443 X 111 = 49173

      • #29860
        Satya
        Participant

        Challenge Question 6: Multiplication by 111

        I think it works this way –

        a b c * 111 =

        a (a+b) (a+b+c) (b+c) c.

        So 234 * 111 = 2 (2+3) (2+3+4) (3+4) 4 = 25974

        Similarly 4 digit number * 111 –>

        a b c d * 111 =  a (a+b) (a+b+c) (b+c+d) (c+d) d

        So 1234 * 111 = 1 (1+2) (1+2+3) (2+3+4) (3+4) 4 = 136974

        2345 * 111 = 260295 (there are carry on’s).

        2 digit number * 111 –>

        a b * 111 = a (a+b) (a+b) b

        applying this to 51 * 111 = 5 6 6 1

        72 * 111 = 7 9 9 2

        99 * 111 = 9 (18) (18) 9 = 10 9 8 9

        Please correct if my observation has any limitations. Thank you!

      • #30290
        Preeti Pathak
        Participant

        Hello,

        Anwer to challenge question 6    234*111=???

        unit digit  : 4(same digit)

        digit in ten’s  place : 3+4=7

        digit in hudredth’s place: 2+3+4=9

        digit in Thousanth’s place: 2+3=5

        digit in ten-thousandth’s place : 2

        so answer will be 25973.If there is any carry over,we have to add them .

        we can do it from left to right also.Don”t forget to add the carry overs.

        so in general, if pqr is a 3 digit number.

        So we need to find pqr*111=?

        ans will be:

        p (p+q) (p+q+r) (q+r) r (taking into consideration the carry overs).

        similarly, 854*111 =94794(in this case ,we have carry overs,we need to take those into account).

        Thanks

    • #29820
      Savita P Kulkarni
      Participant

      Thanks Mallika,

      I will split it after the bar  77(03)/1= 7700-3/1=7697/1=76971

    • #29854
      Anupama Cherukuri
      Participant

      Sorry If I posted the answers before 24hrs. I got confused with the Time Zone now  I can’t delete the post.

    • #29861
      Anupama Cherukuri
      Participant

      Challenge Question 5 : applying vertical and cross wise sutra:

      3  a  b  X 61 – here 0 is in hundred place – so 061

      The last digit in answer 19947 is 7 so b must be 7  – 3 a 7 so far

      3x 0      (3×6)+(aX0)         (3X1) + (7X0)+(aX6)       (aX1)+(7X6)         (1X7)

      0                 18                         3 +0+6a                                42+a                    7

      18                        3+(6X 2)+ 4 (carry over)  to get 19 here ‘a’ must be 2           42 + 2             7

      18                                15                              44                 7

      = 19947

      so the answer is 327

       

       

       

       

    • #29868
      KennethWilliams
      Keymaster

      Thanks for all these great and creative answers.
      Be sure to read other people’s answers.

      Please post your answer even if someone else has already posted it.

    • #29881
      Savita P Kulkarni
      Participant

      Challenge 5 –

      3ab*61 = 19947

      Digit Sum = 3–*7=3

      I tried 12 which gives DS of L as 4 and we want it to be 3.

      Next – 10 will give us left side DS as 1

      11 will make the DS 8

      13 gives 4

      23 gives 2

      24 gives 9

      45 gives DS as 345 *7= 21=3

      So the answer is 345*7 = 3(DS)

    • #29882
      Savita P Kulkarni
      Participant

      Challenge 5 –

      But I realized that the multiplication is not 19947.

      so to make sure that even if the digit sum is 3 the multiplication has to match.

      Is DS method not effective here?

       

    • #29883
      Savita P Kulkarni
      Participant

      i have to try with vertical  multiplication, but it is 3 figure by 2 figure multiplication.

    • #29884
      Savita P Kulkarni
      Participant

      I am watching video 13 and at practice 4 , so i am hoping I will be able to solve this 3 digit by 2 digit multiplication missing nos.

    • #29885
      Savita P Kulkarni
      Participant

      Lesson 13 did not have 3 figure by 2 figure multiplication. So i looked at Anupama’s post. It does follow the vertical multiplication and crosse wise product pattern with 1 extra step.

      • #29887
        KennethWilliams
        Keymaster

        Hi Savita,

        The way to tackle this question is to first use the Last By Last Sutra. By looking at the last digits of the three numbers you can see that b must be 7, it cannot be any other number.

        Then the digit sums will give the value of a.

        Multiplication of 3-digit by 2-digit numbers is dealt with in lesson 21.

    • #29888
      KennethWilliams
      Keymaster

      Thanks for these answers to Questions 5 and 6 – excellent.

      Challenge Question 7

      Can bar numbers be useful in fractions work?
      For example to find 4 9/11 + 3 7/8 (that’s four and nine elevenths plus three and seven eighths)?
      Note: the numbers are close to 5 and 4 respectively.
      [Please wait 24 hours before answering]

      • #30281

        Challenge Question 7:

        4 9/11 + 3 7/8

        we add wholes and convert 9,8,7 to bar numbers.

        7  1(1) /11 + 1(3)/1(2)

        We multiply crosswise and add to get numerator.

        7  1(3)2 + 1(23) / 1(12)

        We multiply denominators to get denominators

        7   2(51) / 1(12)

        7  149/88

        8  61/88

        We can use bar numbers in fractions. But I found it little complicated to use them. Looking forward to see different approaches

        Regards,

        Mallika

      • #30284
        Anupama Cherukuri
        Participant

        I tried the same way too. Did try with different numbers and thought this approach is very complicated. I think there must be another way to make it simple even with Bar numbers. Looking forward to different approaches.

      • #30292
        KennethWilliams
        Keymaster

        Yes it is complicated that way. But you don’t have to introduce bar numbers everywhere – be selective.

        In fact just introducing one bar number simplifies the addition considerably.

      • #30396

        Got it. Thanks Ken.

    • #29889
      KennethWilliams
      Keymaster

      Challenge Question 8 – Missing Digits

      Two 2-figure numbers are multiplied to get a 4-figure number.
      But some of the digits get smudged, so we can only see:
      21 x ?? = ??12.
      Can you find the missing digits?

      • #30285
        Satya
        Participant

        Challenge Question 8:  21 * ? ? = ? ? 12

        Going by the Last by the last digit, we know that that the value of second ? would be 2. So it becomes

        2 1

        a 2 *

        —-

        b c 1 2

        —-

        Going by ‘Vertically and cross wise’ we know that 2*2 + a*1 = c 1

        i.e. 4 + a = c 1 –> number that gives 1 in units place of the answer when 4 gets added to it is 7

        which makes the above  4 + 7 = 1 1 means a = 7 and there is a carry on of 1 that needs to be added to the hundred’s place value so

        2 1

        7 2 *

        —-

        b c(+1) 1 2

        —–

        7*2 is 14 and there is a carry on of 1 making it 15.  So b is 1 and c is 5.

        2 1 * 7 2 = 1 5 1 2

         

        • This reply was modified 8 months, 2 weeks ago by Satya.
    • #30265
      Shikha Dhingra
      Participant

      Challenge question 8         21 x ?? = ?? 12

      21 x 72 = 1512

      explanation:

      As 2 is at ones place in answer therefore one place digit in second number is 2

      21 x ? 2 = ?? 12

      according to crosswise and vertical method

      21

      ?2              the digit at ? should be 7   so that (2 x 2 ) + (7 x 1 ) = 1 at tens place and 7 x2 =14  and carry 1  which is 14 + 1=15

      _____

      1 5 1 2

    • #30291
      Preeti Pathak
      Participant

      Hello,

      Answer of challenge question 5

      3 ab * 61=19947

      First of all, see  the digit in the units place of the answer.It is 7.

      Applying “Last by Last”sutra,we get b=7.

      Now,

      3 a*7* 61=19947

      21a*61= 19947

      Applying digit sums,

      digit sum of 19947=3(casting out 9’s)

      Numbers                            Digit sums

      21a                                        3+a

      61 *                                        7 *

      ——–                                ———

      19947                                   3

      In the digit sums column,

      (3+a) * 7 = 3

      21 * 7a =3

      again, applying digit sums,

      3 * 7a=3

      for this to be true, 7a=9,implies a=2

      so the smudged digits are b=7 and a=2

      327 * 61=19947

      Thanks

       

       

       

    • #30471
      KennethWilliams
      Keymaster

      Thanks for these excellent answers.

      For adding/subtracting fractions using bar numbers the best way is to put one of the fractions into alternative form.

      So, just as 19 can be written as 2(1) as 19 is close to 20, we note that 3 7/8 is close to 4, so we can write it as 4 (1)/8 – that is 4 minus an eighth.

      Then add: 4 9/11 + 4 (1)/8 =…

      Please note we will move onto the Challenge Questions – 2 thread for future questions.

      • #30496
        Anupama Cherukuri
        Participant

        Thank you. Not thought that way at all.

        we can use Bar numbers but still feeling it won’t simplify the procedure much.

         

      • #30503
        KennethWilliams
        Keymaster

        Do try it Anupama. It is certainly much easier with bar numbers.

Viewing 28 reply threads
  • You must be logged in to reply to this topic.
error:
Scroll to Top
Visit Us On FacebookVisit Us On Youtube