Number System - Questions & Discussions

MBA CAT 2022
542
divishth Says
Answer is 370

how the answer is 370 ?
543

70770777077770.................up to 119 digits when divided by 440 leaves a remainder of (a)107 (b)77 (c)370 (d)70 ?plz explain the process to get remainder.thanks in advance.

70770777077770.................up to 119 digits wen divided by 8 gives remainder=2
by 5 gives 0 and with 11 gives 7
thus 370 satisfies all conditions

544
destiny1400 Says
70770777077770.................up to 119 digits when divided by 440 leaves a remainder of (a)107 (b)77 (c)370 (d)70 ?plz explain the process to get remainder.thanks in advance.:-P

i m getting 370
545
70770777077770.................up to 119 digits when divided by 440 leaves a remainder of (a)107 (b)77 (c)370 (d)70 ?plz explain the process to get remainder.thanks in advance.

70770777077770.................up to 119 digits wen divided by 8 gives remainder=2
by 5 gives 0 and with 11 gives 7
thus 370 satisfies all conditions


how had u checked for divisibility by 11. by divisibility rule of 11 or divisibility rule of 7,11and 13.Is there any other way without goin through options?there was one more option as none of these then also goin through option will work?
546
destiny1400 Says
how had u checked for divisibility by 11. by divisibility rule of 11 or divisibility rule of 7,11and 13.Is there any other way without goin through options?there was one more option as none of these then also goin through option will work?

well
write it
70:770:7770...........
for every 3 such there will b equal number o 7
only in the end i.e after 12 such series
last digits
will be
77777777777770777777777777770 wen divided by 11 gives remainder as 7
547
destiny1400 Says
answer is not 70 but 370 but explain the process plzz


70770777077770......119 digits....

so the number would contain (1+2+3+...+n) + n digits...

n(n+1)/2 + n = 119

which is feasible for n=14...

so number of 7's = 105...number of zeros = 14....

now..440 = 4*11*10....apply Chinese remainder theorem

70770777077770......119 digits.../ 10 = 70770777077770..118 digits...last digit 7

70770777077770......118 digits...mod 4 = 1
70770777077770......118 digits...mod 11 = 4

so 4a + 1 = 11b + 4
so b=3,a = 9...hence remainder 4*9 + 1 = 37....multiply with 10..coz we divided the number by 10 previously...

hence 370....
548

Which of the following completely divides the expression 1005 +1015 +1025 +1035 ++1595
(a) 31 (b) 37 (c) 41 (d) 47

549
Which of the following completely divides the expression 1005 +1015 +1025 +1035 ++1595
(a) 31 (b) 37 (c) 41 (d) 47



my answer is coming none of these which is actually not in the options..:shocked::shocked::shocked::shocked::shocked::shocked::shocked::shocked:
550
Which of the following completely divides the expression 1005 +1015 +1025 +1035 ++1595
(a) 31 (b) 37 (c) 41 (d) 47

What is the answer???I m getting none of these:shocked: Is there an option 39????
551
Slayer23 Says
What is the answer???I m getting none of these:shocked: Is there an option 39????



sorry , the term is 1005^5 +1015^5 +1025^5 +1035 ^5 ++1595^5
552
viveknitw Says
sorry , the term is 1005^5 +1015^5 +1025^5 +1035 ^5 ++1595^5

I think the terms are 100^5 + 101^5 + 102^5 + ...+159^5.
Then it is divisible by 39.
553

what is the remainder when 11^11^11 is divided by 9?


a)5 ,b)4 , c)3 ,d)2 ,e)none of these? i m getting ans as 2 but it is given as 5 how ?

554
what is the remainder when 11^11^11 is divided by 9?


a)5 ,b)4 , c)3 ,d)2 ,e)none of these? i m getting ans as 2 but it is given as 5 how ?

11^11^11mod9
euler's no. of 9 = 6
11^11mod6
eulers no. of 6 = 2
hence 11^1mod6 = 5
so 11^5mod9 = 2^5=32mod9 = 5
555
what is the remainder when 11^11^11 is divided by 9?


a)5 ,b)4 , c)3 ,d)2 ,e)none of these? i m getting ans as 2 but it is given as 5 how ?



by eulers thm rem(11^6)/9 = 1
so we need 11^11 in the form of 6k+r
now also agin by rem formula rem(11^2)/6 = 1
thus rem(11^11 )/6 = 5
thus now we want rem(11^5)/9 which is 2^5/9 = 32/9 = 5
am i clear
556
Let N be the largest integer divisible by all the positive integers less than its cube root. Find the no of divisors of number N..
557
Last three digits is nothing but finding the remainder when the number is divided by 1000

3^1994 mod 1000= 3^394mod 1000= 3^4 *(3^10)^39 mod 1000

NOw 3^10mod1000 = 049

(3^10)^39 mod 1000 = 49^39 mod 1000 =7^78 mod 1000

Closest multiple of 4 near 78 is 76

7^76 mod 1000

= 2401^19 mod 1000= (2400+1)^19 mod 1000= (1 + 19C1 *2400)mod 1000 = 45601mod1000 =601

So 3^4*7^2*601 mod 1000= 81*49*601 mod 1000 =369


Another method,
= 3^1994 % 1000, since E(1000) = 400, 3^400 % 1000 = 1
= 3^394 % 1000
3^6 * 3^394 % 1000 = 1
729 * k % 1000 = 1 (product of 729 and k shall end with 001)
729 * k = x001, hence k = 369
558
imlavmishra Says
Let N be the largest integer divisible by all the positive integers less than its cube root. Find the no of divisors of number N..


N = 2^3 * 3^3 * 5^3 * 7^3 *......
no of divisors = (4*4*4*....) = 4^n, where n is the largest possible prime number.
What is the answer??
559
imlavmishra Says
Let N be the largest integer divisible by all the positive integers less than its cube root. Find the no of divisors of number N..

I can only think of N = 8.:shocked::shocked:
560

Ans is N=8

561
imlavmishra Says
Ans is N=8

N = 2^3 * 3^3 * 5^3 * 7^3 *......
no of divisors = (4*4*4*....) = 4^n, where n is the largest possible prime number.
What is the answer??


Ans is N=8