SSC CGL Previous Year Questions: Number System and HCF & LCM- 1 - SSC CGL MCQ

The smallest three digit number divisible by 6 and 7 is = 42 × 3 = 126

⇒ A = 126

And to find the largest four digit number digit number divisible by 6 and 7, we have to divide 9999 by 42

⇒ 9999 = 42 × 238 + 3

⇒ The largest four digit number divisible by 42 is 9999 – 3 = 9996

⇒ B = 9996

So, the required difference = 9996 – 126

∴ The difference between B and A is 9870

Given data is

41, 43, 46, 50, 85, 61, 76, 55, 68, 95

Now, the given data is in ascending order
41, 43, 46, 50, 55, 61, 68, 76, 85, 95

The median of this data is the middle most number of this data

So, the required median is the average of 55 and 61

⇒ (55 + 61)/2

∴ The required median 58

8439 x 53 is divisible by 99 i.e. given number is divisible by 11
∴ (3 + x + 3 + 8) - (5 + 9 + 4) = 0
x = 4

Any number which divisible by 44, must be divisible by 11 also is
And for any number divisible by 11 the difference of sum of its digits at odd and even places be divisible by 11.
For the given number 15x ¹y²
(x + y + 1) – (5 + 1 + 2) = 0
x + y = 7

L. C. M. of 11, 13 and 7
= 11 × 13 × 7 = 1001.
Now, from given option '259259' is divisible by '1001'.
Hence, '259259' is divisible by 11, 13 and 7.

Let the two numbers are 4x and 7x.
H. C. F. of 4x and 7x = x.
Now, x = 26.
So, two numbers are 4 × 26 and 7 × 26.
Sum of two numbers = 4 × 26 + 7 × 26 = 11 × 26 = 286.

Any number that is divisible by 72 must be divisible by 3, 4, 8, and 9.
Now, a number is divisible by 4 when a number formed by its last two digits of that number is divisible by 4. 'y6' is divisible by 4, for y = 1, 3, 5, 7, 9
Again a, number is divisible by 8, when a number formed by its last 3 digits is divisible by 8.
'9y6' is divisible by '8' for y = 3, 7
Now, for divisibility by 9, sum of its digits should be divisible by 9.
for y = 3, '7 + 4 + x + 2 + 9 + 3 + 6' = 31 + x
so, for x = 5, 36 is divisible by 9.
Now, is '7452936' which is divisible by '24' also, so, it is divisible '72'.
Now, (2x + 3y) = 2 × 5 + 3 × 3 = 19.

L.C.M. of 12, 16 and 54.
12 = 2 × 2 × 3.
16 = 2 × 2 × 2 × 2.
54 = 2 × 3 × 3 × 3
L.C.M. = 2 × 2 ×2 × 2 × 3 × 3 × 3 = 432
Remainder = 7.
So, required number = 432 + 7 = 439.
But this is not divisible by 13.
so, next number is 432 × 2 + 7 = 871.
Number 871 is divisible by 13.
Hence, required number is 871.
Sum of its digits = 8 + 7 + 1 = 16.

A number is divisible by 11, if difference of the sum of the digits at the even places and sum of digits at odd places is either 0 (zero) or a multiple of 11.
Now,
(3 + N) – 4 = 0
3 + N – 4 = 0
N – 1 = 0
∴ N = 1

Here,

So,  is the smallest number.

According to question

Now, (27)² = 729
∴ 729 – 709 = 20
∴ 20 must be added to 709 to make it a perfect square.

The smallest number of 5 digits = 10000
Now, 10000/88 = 113, and remainder is 56
∴ Required number = 10000 + (88 – 56) = (10000 + 32) = 10032.

Let digit at ten's place be x and digit at unit's place be y.
∴ The number = 10x + y
When digit are interchanged, the new number
= 10y + x
According to question,
Product of digits = 27 i.e., xy = 27 ...(i)
Also,
10x+ y + 54 = 10y + x
9x – 9y = – 54
x – y = – 6
∴ x = y – 6 ...(ii)
From (i) and (ii),
y (y – 6) = 27
y² – 6y – 27 = 0
y² – 9y + 3y – 27 = 0
(y – 9) (y + 3) = 0
∴ y = 9 or y = – 3
∴ x = 3
When x = 3, and y = 9
∴ Required number = 10x + y
= 10 × 3 + 9
⇒ 30 + 9 = 39.

LCM of 57 and 93,

⇒ 3 × 19 × 31 = 1767.
So, Required answer is 1767.

According to option,

Hence, 2500 is a perfect square of 50.

LCM of 5 and 7 = 35
So, the numbers divisible by both 5 and 7 are multilpe of 35. Between 300 and 650. We have 10 multiple of 35. They are : 315, 350, 385, 420, 455, 490, 525, 560, 595, 630.