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HI folks!!!!

Welcome.

CAT 16 is done and dusted.

Let's gear up for CAT 17.

Often the saying goes true "" The early bird always catches the worms !!""

**ALL THE PUNE FOLKS ARE WELCOME HERE, FOR THE CLASSROOM PROGRAMMES**

**Divisibilty Methods** For checking divisibility by any 'prime/odd' number except for factors of 5', you have the concept of base number.

Number Add Base Number Subtract Base Number

---------------------------------

3: 1 2

7: 5 2

9: 1 8

11: 10 1

13: 4 9

17: 12 5

19: 2 17

21: 19 2

23: 7 ?

27: ? 8

29: 3 ?

...

...

And so on....(i'll describe a method to get these below)

Now for checking divisibility either add the last digit*add base number to the number "formed by removing last digit" or you can subract last digit*subtract base number from the "formed by removing last digit"

for e.g. check for 51/17 either 5- 1*5 =0 or 5 + 1*12 =17 hence divisible.

to check 312/13 we can 31+2*4 =39 hence dvisible or we can 31-2*9 = 13 ence divisible.

to check 61731/19 = 6173 + 1*2 = 6175 = 617 + 5*2 = 627 = 62 + 7*2 = 76 hence divisible.

to check 357976/29 = 35797 + 6*3 = 35815 = 3581 +5*3 = 3596 = 359 + 6*3 = 377 = 37 + 7*3 = 58 hence divisible..

to check 382294/11 = 38229-4*1 =38225 = 3822-5 = 3817 = 381 - 7 = 374 = 37 -4 =33 Hence divisible..

The Subract base Number for a number can be obtained as the {(samllest multiple of number which ends in one)-1}/10

i.e. for 3 or 7 it is (21-1)/10 =2

for 13 it is 91-1/10 = 9.

The AddbaseNumber for a number can be obtained as the {(samllest multiple of number which ends in nine)+1}/10

i.e. for 13 it is (39+1)/10 =4.

for 7 it is 49+1/10 = 5

Proof:

For SubtactBaseNumber say the number abcde...

I want to check divisibility by 17 where subtractbasenumber is 5

I can always write abcde... as 10X+Y (where Y is last digit and X is number formed by removing last digit)

Now X-Y*5 = (10X -50Y)/10 = (10X + Y -51Y)/10 = (OriginalNumber - 51 Y ) / 10

The number '51 Y' is a multiple of 17 so if "OriginalNumber" is divisible by 17 then "OriginalNumber - 51*Y" got to be.. i.e. "10X - 50Y"

as 10 and 17 are co-prime if "10X- 50Y" is divisible the "X-5Y" got to be.....

same theory hold's for addbasenumbers too....

This also defines why it is so easy to check divisibility by 3 or 9 just keep on adding the digits...

And you can check divisibility by 11 just by keeping on subrating digits form previous number.. (which is same as taking sum of even/odd location separately..) **Divisibility by 7**

**Only for those interested in Number theory (Not a Cat short-cut)**

say the number is :

38,391,787

Separate into pairs of digits

38 39 17 87

Consider the difference between each pair of digits and the nearest multiple of seven, beginning for the first pair at right, lower (upper) for the first, upper (lower) for the second and so on, alternating for each new pair.

**4** -----**4** (21-17)

38 39 17 87

**---4** ------**3** (87-84)

The resulting digits, read from right are **3444** (which is also a number multiple of 7).

Proceed in the same way with 3,444

**1**

34 44

**----2**

The final pair **21** is a multiple of seven, so is the original number 38,391,787.

**ANOTHER EXAMPLE**

Look how fast this method is.

Consider the 15-digit number 531,898,839,909,822

**2 ----2--- 3 ----0**

5 31 89 88 39 90 98 22

**---3 ---4 ----6 ----1**

Now we have **10,634,232**

**4 -----0**

10 63 42 32

**----0 ----4**

And now **4,004**

**2**

40 04

**---4**

Which gives **42**, a multiple of 7.

We only need three steps for a 15-digit number. This is called TOJA's method of divisibility. Incidentally this also works for 11 and 13. Just a little manupulation is required, (in case you get a remainder of more than 9)

Let A = 5,962

**7**

59 62 77 which is a multiple of 1

**--7**

EXAMPLE 2

Let A = 5, 971,845

**6---- 4**

5 97 18 45

**--9 ----1**

**8**

**14 96 -> 88 ->divisible**

**----8**

EXAMPLE 3

Let A = 80,714,546

**8 ----10**

80 71 45 46

----**5**-------**2**

The resulting numbers ( 2 10 5 8 ) don’t form a decimal number, so proceed in this way: Put the exceeding number 1 from 10, below the 2 and sum.

**2 0 5 8 -> 3 0 5 8**

**1**

**3**

30 58 33

**---3**

https://www.facebook.com/groups/632112500304610/?ref=bookmarks

Solve

Solve

A dishonest shopkeeper marks his goods 15% above the cost price and offers them a discount of 15% but he actually uses a false weight and gives only 900 grams in place of 1 kg. Find his overall gain or loss?

Please answer with the approach.

Sum of two numbers and their LCM is 41. How many pairs of numbers satisfy this condition? Plz help with detail solution, answer is not given!! Thanks

**## HOW MANY IN FOR SESSION ON BASICS OF NUMBERS ####**

**________/\____ attendance needed , then we will chalk out a time!!**

**PG sessions / events**

Plz Solve this, thanks

**### BASICS #####**

**NOTES _____/\_____ :beers**

** To Find out Pythagorean triplets
**

**There is a formula
**

**(N^2-1)/2,( N^2+1/2) of any number N.
**

**For Example if the number is 3, the pythagorean triplets are (3)^2-1/2 = 4
and (3)^2+1/2 = 5.
**

**For any other number , i.e. !5, the triplets are (15^)2-1/2 = 112 and (15)^2+1/2 = 113.**

## FIRST BLOOD ##

** A three digit number abc is a perfect square and its number of factor is also a perfect square where a b c are distinct . **

**If a + b+ c is also a perfect square ..what is the factor of the 4 digit number bbcc? **

🍻

**## bAAAASICSSS##**

**FIND THE TTOTAL NO OF TRAILING ZEROES IN THE EXPANSION OF**

** 125 ^ 67 * 48^ 67**

🍻

**## NOTES ###**

**NUMBER OF TRAILING ZEROES IN ANY NUMBER N**

**BREAK N IN THE FORM OF **

**N= 5^K * 2^M * ( REMAINING)**

**NOW OUT OF K, M WHICHEVER IS LESS , THAT MANY NUMBER OF**

**ZEROES DOES THE NUMBER HAVE ****🍻**** ****🍻**** **

**" TRAILING " TERM IS IMPORTANT ****___/\___**

**## basics ##**

**Find the unit digit in the expansion of **

**(100 -1)( 100 -2) (100-3) ....(100 - 199)**

🍻

di-lr thread

https://www.pagalguy.com/discussions/official-dilr-thread-for-cat-2017-5049610114433024

verbal thread

https://www.pagalguy.com/discussions/official-verbal-thread-for-cat-2017-5220159373443072

If the product of all the factors of a number is equal to the square of the number and the sum of all the factors of the number other than the number itself is 21, then find the number of possible values for the number.

A)0 B)1 C)2 D)More than 2

A number when divided by a divisor leaves 5 as remainder and when divided by thrice the divisor leaves 25 as remainder.How many divisors satisfy these conditions and what are they?

Hey guys, I have made a simple website here - http://practicetables.000webhostapp.com/ to practice tables. Can you please take a look at it and give inputs, if you have any? Of course, you can also use it for your practice :P

If (T)i = i where i is a natural number then for how many values of i where 1<= i<=100 is (T)i + (T)i+1 a perfect square?

A)5 B)4 C)6 D)More than 6

Any shortcut pls