Robert Bosch Divisibility Quiz-1

Let the numbers be x and y.
Then, xy = 130 and x2 + y2 = 316
(x + y)2 = x2 + y2 + 2xy = 316 + (2 x 130) = 576
x + y = \sqrt{576} = 24

Let the middle digit be x.
Then, 2x = 10 or x = 5. So, the number is either 253 or 352.
Since the number increases on reversing the digits, so the hundred's digits is smaller than the unit's digit.
Hence, required number = 253.

Let the numbers be x and y.
Then, x + y = 24 and x - y = 10.
4xy = (x + y)2 - (x- y)2
= (24)2 - (10)2
= (576 - 100)
= 476
xy = 119.

Let the numbers be x and x + 2.
Then, (x + 2)2 - x2 = 104
4x + 4 = 104
4x = 100
x = 25.
The required sum = x + (x + 2) = 2x + 2 = 52

We are given a 5 digit number
Let first digit be 'X'
then 5th digit is '3X'
let 2nd digit be 'Y'
then 3rd digit is 'Y - 3'
and 4th digit is 'Y + 4'
then the no is '(X)(Y)(Y - 3)(Y + 4)(3X)'
from the above we can say 3X <= 9
so X<=3 and any of the digit in the number is <= 9
Also Y–3 ≥ 0 and Y+4 ≤ 9
⇒ Y ≥ 3 and Y ≤ 5 i.e. Y = 3,4 or 5
for x = 1, all the conditions won’t be satisfied.
For x = 2,
Let Y = 5
The number is 25296

Let’s find a number that is completely divisible by 4, 6, 8, 12, and 16.
To find such number, we need to find the LCM of all these numbers
LCM of (4,6) , 8 , 12 , 16
= 12 , 8 , 12 , 16
= LCM of (12 , 8) , LCM of (12 , 16)
= 24 , 48
= LCM of (24 , 48)
= 48
Hence 48 + 2 i.e. 50 will leave a remainder of 2 if divided by any of the following numbers: 4, 6, 8, 12 and 16.

Since we need to use all 4 digits of the given number, we need to see that the sum of digits 6,4,8,1 = 19
And divisibility rule of 9 says that
For a number to be divisible by 9, the sum of its digits should be divisible by 9.
Hence, we can say that no numbers can be formed using all 4 digits and still divisible by 9.

Let us have N be a number which is always divisible by 22, 35, 48 and 61.
Hence
N = LCM(26, 39,52, 65)
N = LCM[LCM(26,39) , LCM(52,65)]
N = LCM[78,260]
N = 780
Since the difference between each divisor and its respective remainder is 4.Then N - 4 will suit all the conditions given.
Hence, N - 4 = 780 - 4 = 776

Let those 2 numbers be p and q .
We are given that HCF (p, q) = 13 and
LCM (p, q) = 273.
We know that product of numbers = HCF × LCM
Hence, p × q = 13 × 273
p × q = 13 × 13 × 3 × 7
P cannot be 13 × 3 or 13 × 13 or 7 × 3 or 13 or 3 or 7 as these numbers do not come between 60 and 140.
Hence we can say that p = 13 × 7 = 91

We need to find the approximate value of the square root of 4321 using a long division method or trial and error.
We find that 652 = 4225 < 4321 and 662 = 4356 > 4321
So the least number to be added = 4356 – 4321 = 35.