Divisibility Rules

When 1781 is divided by 12, the remainder is 5. Similarly, when 1783 is divided by 12, the remainder is 7, when 1785 is divided by 12, the remainder is 9, and when 1787 is divided by 12, the remainder is 11. Replacing dividends by corresponding remainders, the question reduces to finding the remainder when ( 5*7*9*11) is divided by 12, or when 3465 is divided by 12, in which case the remainder is 9.

 

Note:

If p is divided by q, then the dividend i.e., p, can be replaced with the remainder when p is divided by q. For example, in the above question we replaced 1781, 1783, 1785, and 1787 with 5, 7, 9 & 11 respectively.

Let a = 20, b = 21, c = 22, d = 23 then a+b+c+d = 20+21+22+23 = 86

So, (20³+21³+22³+23³)/86 can be compared with  (a³+b³+c³+d³)/(a+b+c+d)

Now, (a+b+c+d) is a factor of (a³+b³+c³+d³). Hence, remainder = 0

i. 55x + 22x = (11x5x + 11x2x) = 11x(5x + 2x) -> 11x(5x + 2x)/11x = 5x + 2x -> 55x + 22x is always divisible by 11x for all values of x. Hence true.

ii. 55+22 = 77 an+bn is divisible by a+b if  n is odd -> True

iii. 55-22 = 33 an+bn is never divisible by a-b -> False

8⁶ⁿ – 7⁶ⁿ = (8²)³ⁿ – (7²)³ⁿ = 64³ⁿ – 49³ⁿ, which is divisible by 64-49 = 15. Hence option A is true.

Also, 8⁶ⁿ-7⁶ⁿ = (8⁶)ⁿ – (7⁶)ⁿ, which is divisible by 8⁶ – 7⁶

Now, 8⁶ – 7⁶  = (8³-7³)(8³+7³) = (512-343)(512+343) = 169*855 -> 8⁶ⁿ-7⁶ⁿ is divisible by both 169 and 855. Hence options B and C are also true.

Therefore, option D is true.

Since n is even and a multiple of 3

-> n is a multiple of 6

-> n = 6k

Therefore, 111….. up to  n digits = 111….. up to 6/12/18….. digits.

Note that any number of the form xyzxyz is always divisible by 1001 as well as by all factors of 1001.

Now, factors of 1001 are 1, 7, 11, 13, 77, 91, 143 and 1001.

-> xyzxyz is divisible by 1001 and by all its factors i.e. 7, 11, 13, 77, 91, 143

If k =1, n = 6 and N = 111111, which is of the form xyzxyz where x=y=z=1. Thus, 111111 is divisible by both 77 and 143.

If k=2, n =12 and N = 111111111111 which is of the form

xyzxyzxyzxyz and is also divisible by both 77 and 143.

For all higher values of k also, we get N as a number comprising all 1s such that the number of 1s is a multiple of 6

-> N will always be divisible by 1001 and all its factors.

A number is divisible by 8 if the number formed by the last three digits is divisible by 8 or when the last three digits are zeros. So, 68b has to be divisible by 8 for the given number to be divisible by 8

-> b = 0 since 680 is divisible by 8.

A number is divisible by 11 if the difference between the sum of digits in the odd and even places is zero or is a multiple of 11.

In a3720680, the difference between the sum of digits in the odd and even places = (a+7+0+8) – (3+2+6+0), which should be either 0 or multiple of 11

-> (15+a)-11 = 0 or 11k -> a+4 = 0 or 11k.

Now, a+4 can’t be 0 because a cannot be negative; also, k cannot be more than 1 as it will imply that a is a number with more than 1 digit

-> a+4 = 11

-> a = 7

A number is divisible by 11 if the difference between the sum of digits in the odd and even places is zero or is a multiple of 11. In 457x8y, the difference between the sum of digits in the odd and even places = (4+7+8) – (5+x+y) = 14 – (x+y), which is either 0 or 11k.

If 14 – (x+y) = 11k, then k cannot be more than 1 because in that case (x+y) will be negative, which is not possible

-> 14 –(x+y) = 11

-> x+y = 3 —– ❶

If 14 – (x+y) = 0, then x+y = 14 —– ❷

For y to be minimum, x has to be maximum and therefore x+y has to be maximum

-> From ❶ & ❷, x+y = 14

-> y = 14-x

Now, y is minimum when x is maximum or when x = 9

-> y_min = 14-9 = 5

(x+4)²/x = (x²+8x+16)/x = x+8+(16/x)

16/x will be an integer only if x is a factor of 16

-> x = 1, 2, 4, 8, 16

-> x can take 5 values.

For a number to be divisible by 125, the last 3 digits should form a number that is divisible by 125 -> last 3 digits should be 125 or 375.

Case I: Last three digits are 125

Here, the first three digits can be arranged in 6 ways (as the first digit can be picked out of any three, the second digit can be picked in two ways, and the third digit can be picked in only 1 way; the total number of ways will be 3*2*1 i.e., 6). This results in different numbers ending in 125. These numbers will be 347125, 374125, 437125, 473125, 734125, and 743125.

Case II: Last three digits are 375

As in the above case, we will again have 6 numbers ending in 375 i.e., 124375, 142375, 214375, 241375, 412375 and 421375

Thus, the total number of numbers = 6+6 = 12

When 9 is divided by 6, remainder is 3.

When 9+9² (=90) is divided by 6, remainder is 0.

When 9+9²+9³ (=819) is divided by 6, remainder is 3.

When 9+9²+9³+9⁴ (=7380) is divided by 6, remainder is 0.

Thus, we observe that when the number ends in an odd index of 9, the remainder is 3; and when it ends in an even index of 9, the remainder is 0.

In the given case, the number ends in 9²⁵⁴, where the index of 9 i.e., 254, is even

-> remainder is 0.