[varunkumar]

A no. when divided by 899 gives a remainder 63.What remainder will be obtained by dividing the same no. by 29.

[made_for_iims]

Originally Posted by varunkumar

A no. when divided by 899 gives a remainder 63.What remainder will be obtained by dividing the same no. by 29.

let the number be 899x+ 63

now get the remainder

899x % 29 + 63 % 29

0 + 5%29

=> remainder is 5

[themallucat]

answer is 5. 899 is a multiple of 29..so divide the reminder (63)by 29 and find the reminder.it gives the answer..so 63/29= reminder 5

[ajc]

Ans is 5..Give something for the brain to chew abt

[cat_catcher]

Originally Posted by varunkumar

A no. when divided by 899 gives a remainder 63.What remainder will be obtained by dividing the same no. by 29.

let the no. be :899x+63. where x is the quotient.
now check it for 29: 899 on dividing by 29 gives remainder
0 whereas 63 gives the remainder 5. hence the remainder is 5.:grab:

[cat_catcher]

Originally Posted by ajc

Ans is 5..Give something for the brain to chew abt

:grab: 23^23^4 when divided by 7 will yield remainder=............:grab:

[manishpurohit29]

What is the least number that should be multiplied to 100! to make it perfectly divisible by 3**50?


Manish

[cat_catcher]

Originally Posted by manishpurohit29

What is the least number that should be multiplied to 100! to make it perfectly divisible by 350?


Manish

i thk that 1 is the least number that shuld be multiplied to 100! to make it perfectly divisible by 350. as 100! itself contains factors which makes up 350 as a number.

[estranged_gnrs]

Originally Posted by cat_catcher

:grab: 23^23^4 when divided by 7 will yield remainder=............:grab:

<snip> rong soln.

P.S. I will be really glad if we can shift this discussion to already existing quant threads.

[manishpurohit29]

Originally Posted by cat_catcher

:grab: 23^23^4 when divided by 7 will yield remainder=......2......:grab:

(23/7)*(23/7)*(4/7)
= REMAINDER(2*2*4)
= 16/7 = Remainder 2