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February 8, 2021 at 12:00 pm #28969PasyanthiKeymaster

February 8, 2021 at 7:49 pm #30481KennethWilliamsKeymaster
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.

February 9, 2021 at 4:08 pm #30500Savita P KulkarniParticipant
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,

February 9, 2021 at 6:24 pm #30501KennethWilliamsKeymaster
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.


February 9, 2021 at 6:42 pm #30504KennethWilliamsKeymaster
Challenge Question 10
How can we use the method of Lesson 14 to factorise 8051 ?

February 10, 2021 at 12:56 am #30507Anupama CherukuriParticipant
We can do 23 × 29 = 23 × 3(1) = 6/7/(3) so 600+703 = 667
Another ex. 27× 29 = 3(3) × 3 (1) = 9/(12)/ 3 so 900120+3 = 783.
It is easy no need to deal with big numbers.

February 10, 2021 at 12:57 am #30508Anupama CherukuriParticipant
We can do 23 × 29 = 23 × 3(1) = 6/7/(3) so 600+703 = 667
Another ex. 27× 29 = 3(3) × 3 (1) = 9/(12)/ 3 so 900120+3 = 783.
It is easy no need to deal with big numbers.

February 10, 2021 at 1:14 am #30509Anupama CherukuriParticipant
 We can do 23 × 29 = 23 × 3(1) = 6/7/(3) so 600+703 = 667
Another ex. 27× 29 = 3(3) × 3 (1) = 9/(12)/ 3 so 900120+3 = 783.
It is easy no need to deal with big numbers.

February 14, 2021 at 8:34 pm #31019KennethWilliamsKeymaster
Thanks Anupama.
That’s right! The bar numbers are really useful.
We’ll see more of them as we go on.

February 10, 2021 at 11:23 pm #30520Shikha DhingraParticipant
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

February 14, 2021 at 8:37 pm #31020KennethWilliamsKeymaster
Well done Shikha.
This can make factorising big numbers very easy.

February 16, 2021 at 4:03 am #31134SatyaParticipant
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.


February 14, 2021 at 8:47 pm #31022KennethWilliamsKeymaster
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?)

February 16, 2021 at 12:26 am #31128SatyaParticipant
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


February 14, 2021 at 8:52 pm #31023KennethWilliamsKeymaster
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?

February 15, 2021 at 9:41 pm #31126Basa Mallika GogulamudiParticipant
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.

February 16, 2021 at 12:45 am #31129SatyaParticipant
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
6332 = 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.
6332 = 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


February 15, 2021 at 9:12 pm #31122Basa Mallika GogulamudiParticipant
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.

February 15, 2021 at 10:12 pm #31127Preeti PathakParticipant
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

February 16, 2021 at 1:13 am #31130Preeti PathakParticipant
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

February 16, 2021 at 4:40 am #31135Shikha DhingraParticipant
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 (10033 )
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 ( 6430 = 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

February 17, 2021 at 6:41 pm #31183KennethWilliamsKeymaster
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]]

February 21, 2021 at 2:39 am #31334SatyaParticipant
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.
243 = 21.
now second half –> (3) * (3) = 9. we have to write 09.
so 57×37 = 6(3) x 4(3) = 2109


February 17, 2021 at 6:42 pm #31184KennethWilliamsKeymaster
Challenge Question 14
How can bar numbers be used to simplify division by 9 (lesson 18)?
For example, 3172 divided by 9.

February 18, 2021 at 11:05 pm #31216Basa Mallika GogulamudiParticipant
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 4Without using bar numbers:
9 ) 3 1 7 2
_________
3 4 1 r 13
1
_________
352 r 4
Regards.

February 21, 2021 at 2:51 am #31335SatyaParticipant
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 (52=3) R (38 = 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 (38 = 5)
= 3 5 3 R (5)
= 3 5 2 R 4 after removing the bar number.
<hr />


February 18, 2021 at 7:16 pm #31213Shikha DhingraParticipant
Challenge Question#13
57 X 37 = 6(3) X 4(3)
= 6 X 4 3 / (3) X(3)
= 243 / 09
= 21 / 09
57 X 37 = 2109

February 22, 2021 at 2:30 am #31613KennethWilliamsKeymaster
Thanks for these excellent answers
Challenge Question 15
Two 3figure numbers are multiplied to get a 6figure number. But some of the digits get smudged, so we can only see:
321 x ??? = ???123
Can you find the missing digits?
March 6, 2021 at 4:52 am #32988SatyaParticipant
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.


February 22, 2021 at 2:31 am #31614KennethWilliamsKeymaster
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)?

February 24, 2021 at 12:59 am #31729Basa Mallika GogulamudiParticipant
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.

March 6, 2021 at 5:42 am #32990SatyaParticipant
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


February 23, 2021 at 8:52 am #31668Anupama CherukuriParticipant
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

March 2, 2021 at 1:52 am #32700KennethWilliamsKeymaster
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?

March 6, 2021 at 1:19 pm #33007SatyaParticipant
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.

March 6, 2021 at 6:38 pm #33011KennethWilliamsKeymaster
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.

March 7, 2021 at 3:09 am #33026Anupama CherukuriParticipant
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

March 7, 2021 at 3:35 am #33027Anupama CherukuriParticipant
is (12,5, 13), (9,12,15), (8,15,17) – they are perfect triples bcoz atleast 2 numbers are divisible with 2,3,5 ?

March 7, 2021 at 7:16 pm #33052KennethWilliamsKeymaster
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 nontripleness.


March 2, 2021 at 1:53 am #32701KennethWilliamsKeymaster
Challenge Question 18
Missing Digits – Left to Right
Use vertically and crosswise to find the missing digits:
123 x ??? = 321??

March 3, 2021 at 7:57 pm #32839Basa Mallika GogulamudiParticipant
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.

March 6, 2021 at 12:59 pm #33006SatyaParticipant
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

March 6, 2021 at 6:39 pm #33012KennethWilliamsKeymaster
Good answers – well done Mallika and Satya.


March 6, 2021 at 6:44 pm #33015KennethWilliamsKeymaster
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?
March 6, 2021 at 9:28 pm #33022Preeti PathakParticipant
Hi,
challenge question19
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

March 7, 2021 at 7:19 pm #33053KennethWilliamsKeymaster
That’s it Preeti, nicely explained.

March 7, 2021 at 8:02 am #33028SatyaParticipant
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, 7e.g2: 39, 80, 89
step 1 – 3, 8, 8
step 2 – 9, 64, 64
step 3 – 9, 1, 165, 72, 97
2, 9, 7
4, 81, 49
4, 9, 43,4,5
9,16,25
9,7,7 
March 7, 2021 at 7:25 pm #33054KennethWilliamsKeymaster
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?

March 8, 2021 at 1:20 pm #33060SatyaParticipant
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

March 8, 2021 at 7:25 pm #33067KennethWilliamsKeymaster
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.


March 6, 2021 at 6:45 pm #33016KennethWilliamsKeymaster
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?

March 9, 2021 at 1:01 am #33074Anupama CherukuriParticipant
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

March 9, 2021 at 8:17 pm #33083KennethWilliamsKeymaster
Yes Anupama. Provided it is a 2figure whole number.
It is equivalent to what we are doing – dividing the remainder by twice the first digit.


March 8, 2021 at 2:18 pm #33062Preeti PathakParticipant
Hi ,
one way is to multiply and divide by square root of 25.
square root of 3=square root (3 * 25)/ square root(25).

March 8, 2021 at 7:40 pm #33072KennethWilliamsKeymaster
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.

March 8, 2021 at 8:20 pm #33073Preeti PathakParticipant
Yes Ken. So, we can do this by at least 3 ways .Thank you.


March 8, 2021 at 7:34 pm #33071KennethWilliamsKeymaster
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?

March 11, 2021 at 2:58 pm #33105Preeti PathakParticipant
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
 TRIPLET 3, 4 5———————digit sum check—————>3 IS DIVIBLE BY 3 ,SO RULED OUT.
 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


March 10, 2021 at 12:40 am #33086Shikha DhingraParticipant
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.

March 10, 2021 at 7:33 pm #33091KennethWilliamsKeymaster
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’.


March 11, 2021 at 1:50 pm #33100Seth KinkadeParticipant
Test message
 This reply was modified 10 months, 2 weeks ago by Seth Kinkade.


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