Official Quant thread for CAT 2013

MBA CAT 2024
12621
@vbhvgupta said:
find remainder when 21^3 + 23^3 + 25^3 + 27^3 is divided by 96.
zero..

consider no. as 4k+1 and 4k-1

so when (4k+1)^3+(4k-1)^3 is divided by 32 remainder is zero..

now observe that given expression is divisible by 3..(find out how)

so..the given number is divisible by 32*3=96


12622
@vbhvgupta said:
find remainder when 21^3 + 23^3 + 25^3 + 27^3 is divided by 96.
21^3+23^3+25^3+27^3 = 48*(21^2+27^2-21*27+23^2+25^2-23*25 ) = 48*even

0 remainder
12623
Is pq36 a perfect square?
12624
@Zedai said:
Is pq36 a perfect square?
May or may not be........for a perfect square ending with 6, 10th place digit should be odd
12625
@Zedai said:
Is pq36 a perfect square?
yes if p = q = 0
even so if there is odd before 3 then it is a perfect square
12626
@Zedai said:
Is pq36 a perfect square?
depends on p and q .. But yes it can be perfect square

44^2 and 56^2 ends in 36
12627
@krum said:
21^3+23^3+25^3+27^3 = 48*(21^2+27^2-21*27+23^2+25^2+23*25 ) = 48*even0 remainder
bhai, there should be "subtraction" instead of "addition".
12628
@vbhvgupta said:
Please explain......If x is a composite no is the statement given below true?If there r 3 factors not greator than (2)^1/2 , then there r 3 factors not less than (2)^1/2
@krum can u please solve this one?
12630
@Zedai said:
bhai, there should be "subtraction" instead of "addition".
A^3 + b^3 = (a+b)(a^2 + b^2 + ab)
12631
@vbhvgupta said:
A^3 + b^3 = (a+b)(a^2 + b^2 + ab)
its -ab (how can you negate terms like a^2b
12632
@Redemption_CMAT said:
How to calculate value of 2.3[56] in fractions where [ ] denotes recurring decimal.
let x=2.3[56]

multiply by 10
10x=23.[56].... (1)

multiply by 100
1000x=2356.[56]... (2)

subtract (1) from (2)

990x=2333

therefore: x=2333/990
12633
@wovfactorAPS said:
zero..consider no. as 4k+1 and 4k-1so when (4k+1)^3+(4k-1)^3 is divided by 32 remainder is zero..now observe that given expression is divisible by 3..(find out how)so..the given number is divisible by 32*3=96
so when (4k+1)^3+(4k-1)^3 is divided by 32 remainder is zero..
hw u concluded dis?? though it's visible bt still..
12634
FIND THE UNIT DIGIT OF :
888^9235! + 222^9235! + 666^2359! + 999^9999!
a> 1 b>7 c>9 d>3

12635
@dreamz007 said:
FIND THE UNIT DIGIT OF : 888^9235! + 222^9235! + 666^2359! + 999^9999! a> 1 b>7 c>9 d>3
9
12636
@dreamz007

Is it 9??

Approach:

888^9235!..Here we luk for 8 cyclcity ..its 8^1=8, 8^2=6, 8^3=6, 8^4=2, 8^5=8..9235! is an obvious factor of 4..hence the unit digit will be 6

222^9235!..For cyclcity of 2..its 2^1=2, 2^2=4, 2^=8, 2^4=6, 2^5=2...thus same for 9235! its 6..

666^2359!..unit digit is 6..

999^9999!...9^1=9, 9^2=1, 9^3=9..thus, the unit digit is 1

Sum=6+6+6+1 =19..hence unit digit is 9..
12637

It's 9

888^[4x(any number)] ends with 6.
222^[4x(any number)] ends with 6.
666^anything ends with 6
999^[(anything even)] ends with 1.
hence last digit is 6+6+6+1=9

12638
@catconquerer CANU tell 9235! is divisible by 4 ? if u can show
12639

9235!=9235x9234x9233x.......x4x3x2x1

12640
@dreamz007 said:
FIND THE UNIT DIGIT OF : 888^9235! + 222^9235! + 666^2359! + 999^9999! a> 1 b>7 c>9 d>3
c>9

6+6+6+1
12641
@dreamz007 said:
FIND THE UNIT DIGIT OF : 888^9235! + 222^9235! + 666^2359! + 999^9999! a> 1 b>7 c>9 d>3
8^4 + 2^4 + 6^4 + 9^4 = 6+6+6+1 = 9.