Divisibility Quiz – 1

Let 2^{32} = x. Then,\left ( 2^{32}+1\right )= (x + 1).

Let (x + 1) be completely divisible by the natural number N. Then,

\left ( 2^{96}+1\right ) =\left ( \left ( 2^{32}\right )^{3}+1\right )=\left ( x^{3}+1 \right )=\left ( x+1 \right )\left ( x^{2}-x+1 \right ) which is completely divisible by N, since (x + 1) is divisible by N.

23) 1056 (45
92
---
136
115
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21
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Required number = (23 - 21)
= 2.

132 = 4 x 3 x 11
So, if the number divisible by all the three number 4, 3 and 11, then the number is divisible by 132 also.
264 -->11,3,4 (/)
396 -->11,3,4 (/)
462 -->11,3 (X)
792 -->11,3,4 (/)
968 -->11,4 (X)
2178 -->11,3 (X)
5184 -->3,4 (X)
6336 -->11,3,4 (/)
Therefore the following numbers are divisible by 132 : 264, 396, 792 and 6336.
Required number of number = 4.

Largest 4-digit number = 9999

88) 9999 (113
88
----
119
88
----
319
264
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55
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Required number = (9999 - 55)
= 9944.

4 | x y = (5 x 1 + 4) = 9
--------
5 | y -1 x = (4 x y + 1) = (4 x 9 + 1) = 37
--------
| 1 -4

Now, 37 when divided successively by 5 and 4, we get

5 | 37
---------
4 | 7 - 2
---------
| 1 - 3

Respective remainders are 2 and 3.

Let the two consecutive even integers be 2n and (2n + 2). Then,

\left ( 2n+2 \right )^{2}=\left ( 2n+2+2n \right )\left ( 2n+2-2n \right )
= 2(4n + 2)
= 4(2n + 1), which is divisible by 4.

99 = 11 x 9, where 11 and 9 are co-prime.

By divisibility rule, we find that 114345 is divisible by 11 as well as 9. So, it is divisible by 99

Divisibility rule of 9=if the sum of digits of the number is divisible by 9

Divisibility rule of 11= the number of digits is even, add the first and subtract the last digit from the rest. The result must be divisible by 11. If the number of digits is odd, subtract the first and last digit from the rest. The result must be divisible by 11.

Sum of marbles = 23
Let the total sum of marbles be N
Remainder when number of marbles is divided by 9
N/9 = 5 {Remainder on division by 9 for any number is equal to the remainder of dividing the sum of the digits of the number by 9}
=> Remainder of N/3 = 2 {A number of the form 9k + 5 divided by 3 leaves a remainder 2}
N = 11k + 7
N = 3m + 2
11k + 7 => Possible numbers are 7, 18, 29, 40, 51
3m + 2 => Possible numbers are 2, 5, 8, 11, 14, 17, 20, 23, 26, 29

The number that is of the form 11k + 7 and 3m + 2 should be of the form 33b + 29. How did we arrive at this result?

The first natural number that satisfies both properties is 29.
Now, starting with 29, every 11th number is of the form 11k + 7, and every 3rd number is of the form 3m + 2.
So, starting from 29, every 33rd number should be on both lists (33 is the LCM of 11 and 3).
Or, any number of the form 33b + 29 will be both of the form 11K + 7 and 3m + 2, where b, k, m are natural numbers.
The remainder when the said number is divided by 33 is 29.
The question is "What is the remainder when N is divided by 33?"
Hence the answer is "29".

This again is a question that we need to solve by trial and error. Clearly, N is an odd number. So, the remainder when we divide N by 24 has to be odd.

If the remainder when we divide N by 24 = 1, then N2 also has a remainder of 1. we can also see that if the remainder when we divide N by 24 is -1, then N2 a remainder of 1.

When remainder when we divide N by 24 is ±3, thenN^{2} has a remainder of 9.
When remainder when we divide N by 24 is ±5, thenN^{2} has a remainder of 1.
When remainder when we divide N by 24 is ±7, then N^{2} has a remainder of 1.
When remainder when we divide N by 24 is ±9, then N^{2} has a remainder of 9.
When remainder when we divide N by 24 is ±11, thenN^{2}has a remainder of 1.

So, the remainder when we divide N by 24 could be ±1, ±5, ±7 or ±11.

Or, the possible remainders when we divide N by 24 are 1, 5, 7, 11, 13, 17, 19, 23.

Or, the possible remainders when we divide N by 12 are 1, 5, 7, 11.
The question is "What are the possible remainders we can get if we divide N by 12?"
Hence the answer is "1, 5, 7, 11".

When we add two numbers that are not coprime, the sum of these two numbers cannot be prime. Awesome intuitive stuff, but very often forgotten. q can be 1.
If q =2, number would be of the form 28n + 2 which is a multiple of 2.

Similarly, when q = 4, number would be of the form 28n + 4 which is again a multiple of 2. Any number of the form 28n + an even number will be a multiple of 2.

When q = 7, number would be of the form 28n + 7 which is a multiple of 7.

So, the only remainders possible are remainders that share no factors with 28. Or numbers that are co-prime to 28.

There is a formula for this and a shorter way of finding the number of numbers co-prime to a given natural number. A more detailed discussion on this is provided here and here.

1, 3, 5, 9, 11, 13, 15, 17, 19, 23, 25 and 27. q can take 12 different values.
The question is "How many values can q take?"
Hence the answer is "12"