Some good questions on Numbers

1) If the product of four consecutive natural number increased by a natural number p, is a perfect square, then find the value of p?

Solution:

I always like to take values/variables which simplifies calculations.

So let us take the numbers as

(a - 1.5), (a - 0.5), (a + 0.5), (a + 1.5)

(Don't think the use of decimals is going to slow you down, we are at most going to add some decimals, not multiplying them)

Now, Multiplying the four numbers we get,

(a^2 - 0.25)(a^2 - 2.25) = a^4 - 2.5 * a^2 + 2.25 * 0.25

Trying to make a perfect square let's consider the middle term 2.5 * a^2 which should be of the form 2*x*y where x is a^2.

Hence, y must be 2.5/2=1.25

Now, we need to make 1.25^2 by adding something to the last term, in order to have a perfect square

The last term is 2.25 * 0.25=(1.25+1) * (1.25-1)=1.25^2 - 1

Hence, to make this product 1.25^2, we must add 1

p = 1

2) The sum of two numbers is 144 and their LCM is 420. What are the numbers?

Solution

There's a famous process in numbers, which I'll try to simplify in the following manner:

Let us start with an example, 48 = 12 * 4, 36 = 12 * 3; notice here that 12 being their HCF, the 4 and 3 that are left out are co-prime to each other.

Also, let us add these two numbers: we must get 12 * (3+4), which also has the HCF 12. Further, the extra 3+4 should be co-prime to both 3&4.

So, the sum and LCM must also have the HCF as 12

Now, let us take our example. The sum is 144 and the LCM is 420. HCF = HCF (144,420) = 12

If we divide the LCM by the HCF, we should get the product of the co-prime parts 420/12=35 Which can be either 35 * 1 or 5 * 7

Recall that the numbers will be HCF * (one co-prime part) So, they can be either 12 * 1 & 12 * 35

Or, in 12 * 5 & 12 * 7

In the second case, we have the sum as 144; so the numbers should be 60 & 84

3) Find the remainder when 25^82 is divided by 27

Solution

25 divided by 27, we can say the remainder is (-2) So, 25^82 is equivalent to (-2)^82 = 2^82

Now, Euler of 27 is 18... So 2^18 when divided by 27 leaves a remainder of 1 And as 2^18 = (2^9)²... Thus 2^9 when divided by 27 must leave a remainder of +1 or -1

If u check 2^9 when divided by 27 leaves a remainder of -1 Thus 2^82 = 2 * (2^9)^9 = 2 * (-1)^9 = -2... Hence 25

4) The micromanometer in a certain factory can measure the pressure inside the gas chamber from 1 unit to 999999 units. Lately this instrument has not been working properly. The problem with the instrument is that it always skips the digit 5 and moves directly from 4 to 6. What is the actual pressure inside the gas chamber if the micromanometer displays 003016?

Solution

One digit missing implies 9 digits .. implies base 9

The only difference is the correspondence of this number system and the base 9 system is that the digits are different

So in this number system, the digits are 0,1,2,3,4,6,7,8,9 But in base 9, the digits are 0,1,2,3,4,5,6,7,8

So using the above correspondence: 3016 in this number system will be 3015 in base 9

3015 in base 9 implies 3 * 9^3 + 1 * 9 + 6 = 2201

5) N = 7777....7777(7 repeats itself 429 times). Find the remainder when N is divided by 1144

Solution

1144 = 11 * 8 * 13 Any digit repeated even number of times is a multiple of 11

Any 6 digits number of the form abcabc = abc*1001, will be a multiple of 1001 and hence a multiple of 13 as well as 11...

If the last 3 digits of the number is divisible by 8, then the number is divisible by 8, as 1000 is a multiple of 8 If we separate the last 3 digits of the above number, we are left with 426 sevens followed by 3 zeroes ...

426 is a multiple of 6, hence we can group the number into multple groups of 777777, which is of the form abcabc, hence multiple of 13 and 11

As, the last 3 digits are 000, the number should also be a,multiple of 8

So, the remainder will be given by the 777 that we have separated, as the remaining part is divisible by 11, 13 and 8 => divisible by 1144 777 divided by 1144, remainder is 777