Numbers and Divisibility

 If the sum of the numbers is divisible by 3, then it is divisible by 3.

148 is divisible by 4, but not 8.

Units place is 6, i.e. even number. So, the number is always divisible by 2.

Since last 3 digits ‘824’ of 91824 are divisible by 8, so 91824 is divisible by 8.

160358 = (6 + 3 + 8) – (1 + 0 + 5) = 17 – 6 = 11 Since ‘11’ is divisible by 11, so 160358 is divisible by 11.

2801* is divisible by 3. So, 11 + * should be divisible by 3. Nearest number to 11 is 12. Hence, * = 1

Since 21 is divisible by 3, so the number will be divisible by 3.

89047* is divisible by 5 It is possible if last digit is 0 or 5.

3857* is exactly divisible by 3. So, 23 + * is exactly divisible by 3. Number of possible values of * is 2.

The number 2001 is divisible by 3 but not by 2. If 3 is added to it, then it will be divisible by 3 and 2 also. So, the nearest number which is exactly divisible by 6 is 2004.

Hence, (2) is the correct option.  

8629a is exactly divisible by 11. (10 + a) – 15 should be 0 or multiple of 11. a – 5 = 0 a = 5

The greatest number between 1000 and 1100 will be the nearest number smaller than 1100. 1100 is divisible by 4 but not divisible by 3. We should subtract 8 to get it divisible by both 4 and 3. So, 1092 is divisible by 12.

(Xn-1) will be divisible by (x + 1) only, when n is even.(817 - 1) =(92)7-1,which is divisible by (9 + 1), i. e. 10.

first term 105, last term 994common difference 7, total number of terms= [(994-105)/7]+1=128sum of all terms= [128x(994+105)]/2=70336

3000-max possible multiple of lcm(7,11,13)3000-2(1001)=998