AMCAT Divisibility Quiz-1

Answer : 992
Explanation :
Out of all the given numbers, 992 is the greatest 3 digit number divisible by 8.

Answer : 34
Explanation :
11 numbers are divisible between 1 and 100 by 9.
25 numbers are divisible by 4 between 1 and 100.
Therefore, total numbers = 36.
Also, there are numbers which are divisible by both 4 and 9 and are counted twice.
= 11 + 25 – 2 = 34

Answer: 100380
Explanation :
The least number of 6 digits is 100000.
On dividing 100000 by 956 we get 576 as a remainder
100000 - 576 = 99424
Required 6 digit number 99424 + 956= 100380
Hence the required number is 100380.

Answer : 24
Explanation :
Out of all the given numbers, only 24 is divisible by 4 , 6 and 8 also.

Answer: 5533935
Explanation :
A Number is divisible by 11 only if the difference between the sum of the digits in the odd places and difference between  the sum of the digits in the even places is divisible by 11.
5+3+9+5= 22
5+3+3= 11 (22-11= 11)

Answer: 96
Explanation :
By checking the divisibility for both 3 and 4 together, then the number will be directly divisible by 12. Out of all the given numbers, only 96 is divisible by 12.

Answer : 5
Explanation :
On dividing 9805 by 8, the remainder is 5. So, 5 is the smallest number which should be subtracted from 9805 to make it divisible by 8.

Answer : none of the above
Explanation :
80806 when divided by 3 leaves a remainder of 1.
80806 when divided by 6 leaves a remainder of 4.
80806 when divided by 4 leaves a remainder of 2.
Thus, 80806 is divisible by none of them.

Answer : 9
Explanation :
684 when divided by 3 leaves a remainder of 0.
684 when divided by 7 leaves a remainder of 5.
684 when divided by 5 leaves a remainder of 4.
Thus sum of remainders = 0 + 5 + 4 = 9.

Answer : n+ 60
Explanation :
If a number is divisible by 3, 5, and 12, it should be a multiple of the LCM of 3, 5, and 12.
LCM of 3, 5, and 12 = 60
Or, n = 60k
n + 60 is divisible by 60, as
n + 60 = 60 k + 60 = 60 (K + 1)