# Challenge Questions – 2

Home Forums VMTTC Online Discussion Forum Challenge Questions – 2

• Author
Posts
• #28969
Pasyanthi
Keymaster
• #30481
KennethWilliams
Keymaster

Challenge Question 9

Can bar numbers be useful in Multiplication?

For example. find <span style=”color: #0000ff;”>23 x 29</span> using Vertically and Crosswise.

• #31131
Satya
Participant

29 can be written as 30(1) i.e. 30-1.

23 x 29 –> (23 x 30) – (23 x 1)

–> 690 – 23 = 667

Yes, bar numbers are useful  in multiplication.

• #30500
Savita P Kulkarni
Participant

Hi Mr. Ken, if we know the time zone details of all the classmates, it will be easier to post/respond  on this challenge questions post.

Now it is 8th Feb, 10pm here in California,USA. In India it is Feb 9th, 11:30am.

Thanks,

• #30501
KennethWilliams
Keymaster

Hi Savita,

If you are aksing how to know when the 24 hour wait is over for answering the challenge questions the best way is to click on ‘subscribe’ at the top right of this page. Then you will be informed when I post a question and there will no doubt be a time given on that message – so wait 24 hours from then.

• #30504
KennethWilliams
Keymaster

Challenge Question 10

How can we use the method of Lesson 14 to factorise 8051 ?

• #30507
Anupama Cherukuri
Participant

We can do 23 × 29 = 23 × 3(1)   = 6/7/(3) so 600+70-3 = 667

Another ex. 27× 29 = 3(3) × 3 (1) = 9/(12)/ 3 so 900-120+3 = 783.

It is easy no need to deal with big numbers.

• #30508
Anupama Cherukuri
Participant

We can do 23 × 29 = 23 × 3(1)   = 6/7/(3) so 600+70-3 = 667

Another ex. 27× 29 = 3(3) × 3 (1) = 9/(12)/ 3 so 900-120+3 = 783.

It is easy no need to deal with big numbers.

• #30509
Anupama Cherukuri
Participant
1. We can do 23 × 29 = 23 × 3(1)   = 6/7/(3) so 600+70-3 = 667

Another ex. 27× 29 = 3(3) × 3 (1) = 9/(12)/ 3 so 900-120+3 = 783.

It is easy no need to deal with big numbers.

• #31019
KennethWilliams
Keymaster

Thanks Anupama.

That’s right! The bar numbers are really useful.

We’ll see more of them as we go on.

• #30520
Shikha Dhingra
Participant

Challenge Question #10         Factorize 8051

Solution:    factors of   51 are       17  and 3

?  ?     -17           Taking base as 100 , the number at ? ?    should be   100 – 17 =83

?  ?      -3             taking base as 100 , the number at ? ?   should be  100 – 3 = 97

______________                            Therefore the factor of  8051 are 83  and 97

8 0     5 1

• #31020
KennethWilliams
Keymaster

Well done Shikha.

This can make factorising big numbers very easy.

• #31134
Satya
Participant

Though I got the answer in a different way, this seems the quick, easy and straight forward approach. Amazed to see how these techniques can be expanded to solve complex problems easily…looks like I need to practice more.

Thank you Shikha.

• #31022
KennethWilliams
Keymaster

Challenge Question 11

Lesson 17 showed a neat way of finding products like 74 x 76 where the first figures are the same and the last figures add up to 10.

What about products like 47 x 67 where the last figures are the same and the first figures add up to 10?

(And how would you find 94 x 67, and 32 x 36 by this method?)

• #31128
Satya
Participant

47 x 67 –> to calculate the last figures no change in the formula (when the tens place is the same digit and one place digits add to 10). so last 2 digits of the answer is 49.

First two digits can be calculated by multiplying the first 2 digits of both numbers and adding it with the units place number. So (4×6) + 7 = 31

47×67 = 3149

Similarly 85×25  = (16+5) 25 = 2125

54×54 = (25+4) 16 = 2916

94×67 can be written as 2x 47×67 –> 2x 3149 [from the above] –> 6298

32×36 can be written as (32×72) / 2 –> 2304 / 2 = 1152

• #31023
KennethWilliams
Keymaster

Challenge Question 12

A method was shown in Lesson 19 for converting temperatures from Fahrenheit to centigrade (subtract 32, add a tenth and halve), and, as mentioned in the lesson, this is approximate.

How can this be modified (by changing one word) so that the conversion is exact?

• #31126

Challenge Question 12:

Converting  Fahrenheit to Centigrade ( Subtract 32, add a tenth and halve) to give approximate conversion.

For exact conversion we have to use add a ninth instead of tenth.

<b>Subtract 32, add a <i><u>ninth</u></i> and halve to get exact conversion.</b>

Regards.

• #31129
Satya
Participant

To convert from Fahrenheit to centigrade, I think we need to change the second step. Instead of adding tenths we have to add ninth’s.

63 F –> according to the steps in the video

63-32 = 31 –> 31+3(tenth of 31) = 34 –> 34/2 = 17 but actual answer is 17.2.

Instead if we add 3.4 which is the ninth of 31, we will get 17.2.

63-32 = 31 –> 31+3.4 = 34.4 –> 34.4 / 2 = 17.2 centigrade

Similarly 147.2 F –> 115.2 + 12.8(ninth of 115.2) = 128 / 2 = 64 C

141.8F –> 109.8 + 12.2 –> 122 –> 61 C

So the steps to convert F to C could be 1. subtract 32, add ninth, halve.

Please correct if you see any flaw here. Thank you

• #31122

Challenge Question 11:

47 x 67 = 3149

We can solve this problem by using first by first and last by last sutra.

We multiply digits in tens place that add up to 10 (4×6 ) and add units place number (7) to the final answer   24+7 = 31

we simply multiply digits in units place to get the second part of the answer. 7×7 = 49

47 x 67 = 3149

94 x 67 can be solved using proportionately sutra.

94 = 47×2

67 x 47 = 3149 and we double the answer to solve 94 x 67 = 6298

<b>32 x 36 can be solved using proportionately sutra again.</b>

We double 36 to get 72.

32×72 = 2304 and we halve the result to get final answer 1152.

Regards.

• #31127
Preeti Pathak
Participant

47 * 67

For the first part of the answer,

=Product of the digits in tens place  that add up to 10 +  digit in ones place

=(4*6) + 7=24+7= 31_____(1)

For the second part,

simply multiply the digits in one’s place, so 7 * 7=49________(2)

combining (1) & (2) gives

47 * 67 = 3139

To find 94 * 67

we can use the sutra “proportionately” here.

94= 2 * 47.

47 * 67= 3139 and double it ,so 94 * 67= 2 *3139=6278

To find 32 * 36

36 * 2=72

using the “proportionately” sutra, 7 and 3 add up to 10,

32 * 72 = 2304 and halve it ,so 32 *36=1152

Thanks

• #31130
Preeti Pathak
Participant

Sorry, there was one calculation mistake in earlier post..

47 *67=3149 ( not 3139)

then, using the “Proportionately ” Sutra, to find 94 *  67,

= 2 * (47 * 67)

= 2 * (3149) gives 6298

Thanks

• #31135
Shikha Dhingra
Participant

Challenge Question #11

47 X 67   As in lesson 17 the digits at tens place are same and  the sum of digits at ones place is 10  we try to convert digits according to that.

47  X (100-33 )

47 X(100 – 33 + 10 -10)                    adding +10 and -10  so that (-33 -10) becomes -43

47 X(100 -43 +10)

4700 -(47 X43) + (47 X10)                now  in   47 X43   the digits at tens place are same and sum of digits at ones place is 10

4700 – 2021 +470  =5170 -2021 =3149

94 X 67              2 X 47 X 67

2 X (3149) =      6298              multiplication of 47 and 67 is same as above

32 X36        multiply by 2

2 X 32 X 36= 64 X36

= (64 -30 +30 ) X36           adding +30 and -30  to make ( 64-30 = 34)

=  (34  +30 ) X36

= 34 X36 + ( 30 X 36)            now in 34 and 36 the tens digits are same and sum of ones digits is 10

= 1224 + 1080 = 2304

now divide by 2 as we have multiply by 2  in the beginning

therefore 2304 ÷ 2 = 1152

• #31183
KennethWilliams
Keymaster

Challenge Question 13

In Challenge Question 11 we looked at products like 47 x 67 (where the first figures add up to 10 and the last are the same), and the method was 4×6+7 / 7×7 = 3149.

How can this method be used to find 57 x 37?

[[perhaps this question should be barred]]

• #31334
Satya
Participant

57 x 37 can be written as 6(3) x 4(3) – using bar numbers

We can now follow the same rule that we use to calculate 47 x 67 .

So first half –> (6*4) + (-3) as 3 is a bar number we have to actually subtract it.

24-3 = 21.

now second half –> (-3) * (-3) = 9. we have to write 09.

so 57×37 = 6(3) x 4(3) = 2109

• #31184
KennethWilliams
Keymaster

Challenge Question 14

How can bar numbers be used to simplify division by 9 (lesson 18)?

For example, 3172 divided by 9.

• #31216

Challenge Question 14:

Using bar numbers we can eliminate carry forward and simplify calculations.

Using bar numbers:

9 ) 3 2 (3)  2

___________
3 5 2 r 4

Without using bar numbers:

9 ) 3 1 7 2

_________

3 4 1 r 13

1

_________

352 r 4

Regards.

• #31335
Satya
Participant

3172 can be written as 318(8) or 32(28) using bar numbers.

3172 divided by 9 gives 352 R 4

<hr />

Now 32(28) divided by 9:

= 3 2 -2 -8

= 3 5 (5-2=3)  R (3-8 = -5 )

= 3 5 3 R -5

= 3 5 2 R 4 (after removing bar number)

<hr />

Now 3 1 8 (8) divided by 9:

= 3 1 8 -8

= 3 5 3 R (3-8 = -5)

= 3 5 3 R (-5)

= 3 5 2 R 4 after removing the bar number.

<hr />

• #31213
Shikha Dhingra
Participant

Challenge Question#13

57 X 37    = 6(3)  X  4(3)

= 6 X 4 -3 / (-3) X(-3)

= 24-3 / 09

= 21 / 09

57 X 37 = 2109

• #31613
KennethWilliams
Keymaster

Challenge Question 15

Two 3-figure numbers are multiplied to get a 6-figure number. But some of the digits get smudged, so we can only see:

321 x ??? = ???123

Can you find the missing digits?

• #32988
Satya
Participant

Challenge question 15:

321 x ? ? ? = ? ? ? 123

I used vertically and crosswise multiplication method to find the missing digits.

From the above sentence it is clear that the ending digit of the second multiplier is 3

3 2 1

x y 3

———————-

a b c 1 2 3

(2*3) + (1*y) should be equal to a 2 digit number that ends in 2 i.e. 12 or 22 or 32 etc. y here has to be single digit so y has to be 6.

3 2 1

x 6 3

—————

a b c 0(+1) 2 3

Now (3*3) + (2*6) + (1*x) = multiple of 10 –> 21 + x = multiple of 10. Since x has to be single digit, it should be 9

3 2 1

9 6 3

————-

a b c 1 2 3

=  3 0 9 1 2 3

• This reply was modified 10 months, 3 weeks ago by Satya.
• #31614
KennethWilliams
Keymaster

Challenge Question 16

Squaring fractions

How can the general squaring method shown in Lesson 22 be applied to squaring fractions.

For example (3 5/6)^2 (that’s three and five sixths squared)?

• #31729

Challenge Question 16 : Squaring fractions.

(3 5/6)^2 = D(3) D(3 5/6) D(5/6)

= 9  (2 x3x 5/6 ) 25/36

= 9 (30/6) 25 /36

= 9 + 5 25/36

= 14 25/36

Conventional way :

(3 5/6 ) ^2 =  ( 23 / 6)^2

=  529 /36

= 14 25/36

Regards.

• #32990
Satya
Participant

Challenge Question 16: Square of 3 5/6

3 5/6 can be written as (3 + 5/6)

3+5/6 square = 3 square + 2 * 3 * 5/6 + 5/6 square

= 9 + 5 + 25/36

= 14 + 25/36 which is 14 25/36

• #31668
Anupama Cherukuri
Participant

Challenge Question – 15

I used vertical and crosswise method to solve

321 * 963  = 309123

Challenge Question – 16

Square of 3 5/6  = (3*3)    +   (3 * 5/6  +  3* 5/6 )  + (5/6 * 5/6)

9          +         5                            +   25/36

14  25/36

• #32700
KennethWilliams
Keymaster

Thanks for these answers – all correct

Challenge Question 17

There are certain whole numbers which divide into at least one of the elements of any perfect triple.

For example, 2 is such a number as all triples have at least one even element.

[We do not count triples that contain a zero, like 1,0,1]

Can you find them?

• #33007
Satya
Participant

Please let me know if I understand the question correctly. I think in each ‘perfect triple’ combo group there will be one even number – divisible by 2, one number divisible by 3 and another/same number that is divisible by 5 too.

• #33011
KennethWilliams
Keymaster

Well done Satya. That’s right, in fact 1, 2, 3, 4, 5 will always divide into at least one of the elements of a perfect triple.

• #33026
Anupama Cherukuri
Participant

so is (6,8,10) is perfect triple.. bcoz 6 is divisible by 2, 3 ; 8 is divisible by  2 ,4 ; 10 is divisible by 2,5

• #33027
Anupama Cherukuri
Participant

is (12,5, 13), (9,12,15), (8,15,17)  – they are perfect triples bcoz atleast 2 numbers are divisible with 2,3,5 ?

• #33052
KennethWilliams
Keymaster

Not quite Anupama.

6,8,10 is a perfect triple because 6^2 + 8^2 = 10^2.

The 2, 3, 5 property is only a property and does now prove ‘tripleness’ or non-tripleness.

• #32701
KennethWilliams
Keymaster

Challenge Question 18

Missing Digits – Left to Right

Use vertically and crosswise to find the missing digits:

123 x ??? = 321??

• #32839

Hi,

123 x ??? = 321??

if I pair 1st two numbers

12 3 x ?? ? = 321 ??

if I divide 321 with 12 I get 26 remainder 9

12 3 x 26 ? = 312 and 9 is a carry from crosswise multiplication.

26 x 3 = 78 +12 x 1= 90 so  last digit should be 1 in the multiplier.

123 x 261 = 32103

Regards.

• #33006
Satya
Participant

Challenge Question 18: 123 * ??? = 321??

To use Vertically and crosswise I chose

12 3

x  y

———–

3 2 1 a b

12 * x has to be equal to 321 including any carryon’s from the right side. Closest number to 321 in 12th table is 312 [12*26 = 312]. So I thought x can be 26, to start with.

12  3

26  y

———–

3 2 1 a b

321 – 312 = 9. So (26 * 3) + (12 * y) = 9[a] –> 78 + 12y = 9[a]. 9[a]-78 has to be a multiple of 12 and the only value that satisfies it is when a = 0. So

78 + 12y = 90 thus y = 1.

Certainly b = 3

12  3

26  1

———–

3 2 1 0 3

123 * 261 = 32103

• #33012
KennethWilliams
Keymaster

Good answers – well done Mallika and Satya.

• #33015
KennethWilliams
Keymaster

Challenge Question 19

How can the digit sum check be used to verify that 3 numbers form a perfect triple?

For example, I added two triples, got 77,36,85 and now wish to check this is a triple.
How can digit sums be used to verify that 77,36,85 is a triple?

• #33022
Preeti Pathak
Participant

Hi,

challenge question-19

we need to verify if 77 , 36 and 85 is a triple by checking their digit sums.

77 ^2 + 36 ^2 = 85^2

Their corresponding digit sums will be,

L.H.S= 5 ^2 + 9 ^2

=digit sum(25 + 18)

=7+9= 7

R.H.S=4 ^2

=digit sum(16)=1+6=7

L.H.S=R.H.S

since their digit sums are same, we can say that 77,36 and 85 is a triplet.

Thanks

• #33053
KennethWilliams
Keymaster

That’s it Preeti, nicely explained.

• #33028
Satya
Participant

This is what I noticed while checking if a given 3 numbers set form a triple, using digit sum –

step 1: find the digit sum of the 3 numbers
step 2: square each number
step 3: find the digit sum after squaring
step 4: one of the numbers is 9 while the other two are equal.

e.g: 77, 36, 85
step 1 – 5, 9, 4
step 2 – 25, 81, 16
step 3 – 7, 9, 7

e.g2: 39, 80, 89
step 1 – 3, 8, 8
step 2 – 9, 64, 64
step 3 – 9, 1, 1

65, 72, 97
2, 9, 7
4, 81, 49
4, 9, 4

3,4,5
9,16,25
9,7,7

• #33054
KennethWilliams
Keymaster

Good observations Satya.

So you could say triples these reduce to 9,7,7 and 9,4,4 and 9,1,1.

Are there triples that reduce to 9,2,2 or 9,3,3 etc?

And does that mean that all triples can fall into 9 classes?

• #33060
Satya
Participant

Hi Ken,

With your response I explored some more and I am unable to find any triples that reduce to 9,2,2 or 9,3,3 etc. I am not sure if there are any. Would you please throw some light on triples in connection to this sequence.

Thank you

Satya

• #33067
KennethWilliams
Keymaster

Hi Satya,

I have to say I have not noted this pattern before. It seems 1, 4, 7 are the only possible digits.

How that might be used or extended are open questions. Please keep looking – I will too.

In fact you have rather anticipated the next Challenge Question which I will post in a moment.

• #33016
KennethWilliams
Keymaster

Challenge Question 20

Finding the square root of 3 could be a bit tricky as the first figure is 1 and the divisor is therefore 2, i.e. quite small, leading to many subsequent reductions.

How can the square root of 3 be found more easily?

• #33074
Anupama Cherukuri
Participant

can we try like this

square root of 3 = closest perfect square root is  4

square root 0f 4 = 2 – 1/8 =  1.875  not perfect but approx.

I tried for different examples

square root of 30 = square root of  square root of 25  + 5/10 = 5.5     or

square root of 30 = square root of 36 – 6/12 = 5.5

• #33083
KennethWilliams
Keymaster

Yes Anupama. Provided it is a 2-figure whole number.

It is equivalent to what we are doing – dividing the remainder by twice the first digit.

• #33062
Preeti Pathak
Participant

Hi ,

one way is to multiply and divide by square root of 25.

square root of 3=square root (3  * 25)/ square root(25).

• #33072
KennethWilliams
Keymaster

That is a great answer Preeti.
square root of 3 = a fifth of the square root of 75.

Or, similarly: square root of 3 = a half of the square root of 12.

Another method is to note that since the square root of 300 is about 17 we can use that, i.e. have a divisor of 34.

• #33073
Preeti Pathak
Participant

Yes Ken. So, we can do this by at least 3 ways .Thank you.

• #33071
KennethWilliams
Keymaster

Challenge Question 21

Noting that one element of a perfect triple is always divisible by 3 (see Challenge Question 19) how can the digit sum check for triples be simplified?

• #33105
Preeti Pathak
Participant

Hi,

challenge question #21

since in a triple, one  element is always divisible by 3.That means , if we consider one of the elements, for example ,to be either

3 / 6 /9 / 12 / 15 / 18 / 21 /24/27/30  etc. (multiples of 3).Their corresponding squares will be,

9/36/81/144/324/441/729/900.And their corresponding digit sums will be

9/9/9/9/9/9/9 .And 9 ‘s can be casted out.

so the digit sum of squares of one element(other than the one divisible by 3) will be equal to the digit sum of square of the 3rd element(of course, the remaining element).

we can verify this by applying the digit sum check.

For example

1.  TRIPLET  3, 4 5———————-digit sum check—————->3 IS DIVIBLE BY 3 ,SO RULED OUT.
2.                   L.H.S=  DS(4^2 )=DS(16)=7,R.H.S=DS(5^2)= DS(5^2)=DS(25)=7 ,L.H.S=R.H.S

2.TRIPLET 24, 7 25——————–digit sum check——————->24 IS DIVIBLE BY 3 ,SO RULED OUT.

L.H.S=DS(7^2)=DS(49)=4,R.H.S=DS(25^2)DS(625)=4,L.H.S=R.H.S

Thanks

• #33086
Shikha Dhingra
Participant

Challenge Question # 21

5 ,12,13     and      11,60,61 are triplets

Now in both triplets one of the element is divisible by 3 i.e. 12 and 60

In first case   5^2 + 12^2 = 13^2

The element which is divisible by nine always comes out to be 9 after digit sum so we can cast out 9.

L.H.S              25 + 3^2 =7+9 =16 =7 ( by digit sum)

In L.H.S. if cast out 9 then also the digit sum comes out to be 7

R.H.S.              13^2= 4^2 = 16 = 7

In triplet 11,60,61

60 is divisible by 3 , the square of 60 is comes out to be 9 after digit sum ( 60^2= 6^2= 36= 9)

So we can only take the digit sum of 11 in L.H.S.

11^2 =1+1 ^2= 2^2=4

R.H.S.   61^2 = 7^2=49=4+9=13=1+3=4

L.H.S. =R.H.S.

So in triplet we can skip the element which is divisible by 3 and only take the digit sum of other element on L.H.S.

• #33091
KennethWilliams
Keymaster

Thanks Shikha,

You have it, and summed it up perfectly with

‘So in triplet we can skip the element which is divisible by 3 and only take the digit sum of other elements’.

• #33100