Quant by Arun Sharma

MBA CAT 2022
1094
3^15676*3^2

(81)^3919*3^2

now multiply 3(tens digit) with power's last digit (6)

so got 21

so final and=21*9=89


Hi Jigar n atheist 😃 thanks

i Understood wat was explained by atheist:grin: but din't get the above part(in bold) can u pls elaborate...
1095
Hi Jigar n atheist 😃 thanks

i Understood wat was explained by atheist:grin: but din't get the above part(in bold) can u pls elaborate...



I have attached 1 figure plz read it u will find solution
1096

3^15678

81^15676 * 9

81*9 = 29 ==> Isn't this we are supposed to do from the above pic ??

1097
3^15678

81^15676 * 9

81*9 = 29 ==> Isn't this we are supposed to do from the above pic ??


why u written 81^15676 its 81^3919
1098

I have attached 1 figure plz read it u will find solution



thanks Jigar...thanks for sharing...
1099

Originally Posted by chetna
hi ppl

solve some questions for me too:)

1)Some chocolates were distributes equally few children and three chocolates were left.Had there been four times the no. of children,then ten choclate would have been left.find the number of children.



chetna
see the shortcut....
(C-3)/n and (C-10)/4n are perfectly divisible ... =>wid a difference of 7 the division can be performed by 4 times the number of children ...it can only be possible , when the number is a multiple of 7 as well as multiple of 4 .which means C-10= 28. Hence C=38 nd n=7


Hi... i found the above problem in number syatem thread
link:
http://www.pagalguy.com/discussions/number-system-25000852

why has he only taken c-10=28,he could have taken c-3=28 then the value of c will be 31..pls go thru this and help.... ..or have i missed something.

1100
Originally Posted by chetna
hi ppl

solve some questions for me too:)

1)Some chocolates were distributes equally few children and three chocolates were left.Had there been four times the no. of children,then ten choclate would have been left.find the number of children.



chetna
see the shortcut....
(C-3)/n and (C-10)/4n are perfectly divisible ... =>wid a difference of 7 the division can be performed by 4 times the number of children ...it can only be possible , when the number is a multiple of 7 as well as multiple of 4 .which means C-10= 28. Hence C=38 nd n=7


Hi... i found the above problem in number syatem thread
link:
http://www.pagalguy.com/discussions/number-system-25000852

why has he only taken c-10=28,he could have taken c-3=28 then the value of c will be 31..pls go thru this and help.... ..or have i missed something.



1)let total number of chocolate be N,number of children x and perhead they get p chocolate
then N=px+3
now if the number of children becomes 4x then say per head chocolate is q
then N=4qx+10
hence
px+3=4qx+10
i.e
(p-4q)x=7
here either p-4q=7 and x=1
or p-4q=1 x=7
now if we take x=1,p=7+4q
for q=1 p=11
hence initially N=14
now if we take x=7,p=1+4q
for q=1 p=5
hence initially N=38

as we have been told about few children number of children=7
1101

as we have been told about few children number of children=7


Hi Jigar thanks...i had overlooked the part in bold:)

1102

Hi here is one more problem...

How many keystrokes are needed to type numbers from 1 to 1000?

1103
Hi here is one more problem...

How many keystrokes are needed to type numbers from 1 to 1000?


for

1-9 =9
10-99=90*2
100-999=3*900
1000=4

ans is 9+180+2700+4=2893
1104

please solve this question

what is the remainder
(50^56^52)/7 ????

Thanks

1105
please solve this question

what is the remainder
(50^56^52)/7 ????

Thanks

50 = 49 + 1
so, 50^56^52 = (49 + 1)^56^52
think Binomial Theorem here
remainder = 1^56^52 = 1
1106
Hi,
can someone let me know the easy way of solving this kind of problems..
1)Find the last two digits of 3 to the power of 15678.

I tried to solve this one by dividing it by 100,but it is a tedious approach :lookround:and i got stuck ...Pls ans this question.



May b I am late in posting the way of solving such questions but here it goes!!!!!!!!



I am dividing this method into four parts and we will discuss each part one by one:


a. Last two digits of numbers which end in one
b. Last two digits of numbers which end in 3, 7 and 9
c. Last two digits of numbers which end in 2
d. Last two digits of numbers which end in 4, 6 and 8

Before we start, let me mention binomial theorem in brief as we will need it for our calculations.



Last two digits of numbers ending in 1

Let's start with an example.

What are the last two digits of 31786?

Solution: 31786 = (30 + 1)786 = 786C0 1786 + 786C1 1785 (30) + 786C2 1784 302 + ..., Note that all the terms after the second term will end in two or more zeroes. The first two terms are 786C0 1786 and 786C1 1785 (30). Now, the second term will end with one zero and the tens digit of the second term will be the product of 786 and 3 i.e. 8. Therefore, the last two digits of the second term will be 80. The last digit of the first term is 1. So the last two digits of 31786 are 81.

Now, here is the shortcut:


Here are some more examples:

Find the last two digits of 412789
In no time at all you can calculate the answer to be 61 (4 9 = 36). Therefore, 6 will be the tens digit and one will be the units digit)

Find the last two digits of 7156747
Last two digits will be 91 (7 7 gives 9 and 1 as units digit)

Now try to get the answer to this question within 10 s:

Find the last two digits of 51456 61567
The last two digits of 51456 will be 01 and the last two digits of 61567 will be 21. Therefore, the last two digits of 51456 61567 will be the last two digits of 01 21 = 21

Last two digits of numbers ending in 3, 7 or 9

Find the last two digits of 19266.

19266 = (192)133. Now, 192 ends in 61 (192 = 361) therefore, we need to find the last two digits of (61)133.
Once the number is ending in 1 we can straight away get the last two digits with the help of the previous method. The last two digits are 81 (6 3 = 18, so the tens digit will be 8 and last digit will be 1)

Find the last two digits of 33288.

33288 = (334)72. Now 334 ends in 21 (334 = 332 332 = 1089 1089 = xxxxx21) therefore, we need to find the last two digits of 2172. By the previous method, the last two digits of 2172 = 41 (tens digit = 2 2 = 4, unit digit = 1)

So here's the rule for finding the last two digits of numbers ending in 3, 7 and 9:


Now try the method with a number ending in 7:

Find the last two digits of 87474.

87474 = 87472 872 = (874)118 872 = (69 69)118 69 (The last two digits of 872 are 69) = 61118 69 = 81 69 = 89

If you understood the method then try your hands on these questions:

Find the last two digits of:

1. 27456
2. 7983
3. 583512

Last two digits of numbers ending in 2, 4, 6 or 8

There is only one even two-digit number which always ends in itself (last two digits) - 76 i.e. 76 raised to any power gives the last two digits as 76. Therefore, our purpose is to get 76 as last two digits for even numbers. We know that 242 ends in 76 and 210 ends in 24. Also, 24 raised to an even power always ends with 76 and 24 raised to an odd power always ends with 24. Therefore, 2434 will end in 76 and 2453 will end in 24.

Let's apply this funda:

Find the last two digits of 2543.

2543 = (210)54 23 = (24)54 (24 raised to an even power) 23 = 76 8 = 08

(NOTE: Here if you need to multiply 76 with 2n, then you can straightaway write the last two digits of 2n because when 76 is multiplied with 2n the last two digits remain the same as the last two digits of 2n. Therefore, the last two digits of 76 27 will be the last two digits of 27 = 2

Same method we can use for any number which is of the form 2n. Here is an example:

Find the last two digits of 64236.
64236 = (26)236 = 21416 = (210)141 26 = 24141 (24 raised to odd power) 64 = 24 64 = 36

Now those numbers which are not in the form of 2n can be broken down into the form 2n odd number. We can find the last two digits of both the parts separately.

Here are some examples:

Find the last two digits of 62586.
62586 = (2 31)586 = 2586 3586 = (210)58 26 31586 = 76 64 81 = 84

Find the last two digits of 54380.
54380 = (2 33)380 = 2380 31140 = (210)38 (34)285 = 76 81285 = 76 01 = 76.

Find the last two digits of 56283.
56283 = (23 7)283 = 2849 7283 = (210)84 29 (74)70 73 = 76 12 (01)70 43 = 16

Find the last two digits of 78379.
78379 = (2 39)379 = 2379 39379 = (210)37 29 (392)189 39 = 24 12 81 39 = 92
Now try to find the last two digits of
1. 34576
2. 28287


DISCLAIMER : THIS IS NOT MY INVENTION I FOUND IT IN A WEBSITE SO THOUGHT IT INTERESTING AND COPIED AND PASTED IT HERE.

HOPE U FIND IT USEFUL.
1107
please solve this question

what is the remainder
(50^56^52)/7 ????

Thanks


50^56^52

let,56^52 be k

50^k mod 7

(49+1)^k mod 7

remainder is 1.

hope that helps.
1108
Hi here is one more problem...

How many keystrokes are needed to type numbers from 1 to 1000?



For 1 to 9 =9
For 10 to 99=90*2=180
For 100 to 999=3*900= 2700
For 1000=4

Total =2893

Cheers
Sukrit
1109
For 1 to 9 =9
For 10 to 99=90*2=180
For 100 to 999=3*900= 2700
For 1000=4

Total =2893

Cheers
Sukrit



I believe it form's a pettern ,

Look for 1 to 9 we can calculate it as 9 times or 9*1 , (Since till 9 we hav single digit numbers)
From 10 to 99 it is nothing but 90*2 , (Since from 10 to 99 we hav only 2 digit numbers)
Similarly from 100 to 999 we hav 900*3 (Since from 100 to 999 we hav only 3 digit numbers)
For 1000 we can again count , we hav to type the digits only 4 times.
I think it this way if u find anything wrong with my logic plz let me know!

1110

guys i m founding myself in a mess.
i have bought 2nd edition 2007 reprint arun sharma from my friend.
but some of my friends are saying that this book is of old edition.
it has red cover page.
i have searched on tmh site, they are showing book of new edition with black cover page.
friends can u confirm that arun sharma new edition has come in the market or not .............pls:banghead:

1111

plz send me the ebook on quantitaive , DI by arun sharma

mail id
regards
studymba

1112
plz send me the ebook on quantitaive , DI by arun sharma

mail id
regards
studymba



Plz don't ask links for e-books here, it is against forum rules and regarding the edition go thru any edition it hardly matters.

Problems r same altogether in all the editions with little difference.
1113
At what integral value of x will the function (x2 + 3x + 1)/(x2 - 3x + 1) attain its maximum value??

a) 3
b) 4
c) -3
d) none of this
LOD 3, chapter: Function