Test: SSC CGL Previous Year Question: Number System and HCF & LCM (2023-2024) - 2 - SSC CGL MCQ

S₁ = 2, 9, 16, 23 , 30, 37, 44, 51 , ……., 632
S₂ = 7, 11, 15, 19, 23, 27, 31, 35, 39, 43, 47, 51 ,
Common terms between given series = 23, 51,......
Common difference(d) = 51 - 23 = 28
ATQ,
23 + (n - 1)28 ≤ 632
(n - 1)28 ≤ 609 ⇒ n - 1 ≤ 21.75
n ≤ 22.75 ⇒ so, n = 22
Now, S = (22/2) [ 2 × 23 + (22 − 1) 28]
S = 11 [ 46 + 21 × 28 ]
S = 11 [46 + 588] = 11 × 634 = 6974

0.0625 = 1/16,
So, 
Again, x + y + z = 2(3/12) = 27/12
x + z = .

Sequence of quotient from bottom to top:
17 × 2 + 0 = 34
34 × 11 + 17 = 391
391 × 1 + 34 = 425
So, the no’s are 391 and 425
Required sum = 391 + 425 = 816

12 = 2 × 2 × 3 ;16 = 2 × 2 × 2 × 2
18 = 2 × 3 × 3; 20 = 2 × 2 × 5
25 = 5 × 5
LCM of 12,16,18,20 and 25 = 2 × 2 × 2 × 2 × 3 × 3 × 5 × 5 = 3600
⇒ x must be = 3600 k + 4
Where 3600k + 4 is multiple of 7
The condition gets satisfied when k = 5
Required number = 3600(5) + 4 = 18004
⇒ digit at the thousands' place in x  = 8

Let the quotient be x.
So, n = 8x + 3
6n - 1 = 6(8x + 3) - 1 = 48x + 17
48 is multiple of 8 so 48 will be exactly
divisible by 8. But when we divide 17 by 8
the remainder is 1.

Short-trick:
choose the smallest value of n for which
remainder is 3 when the number is
divided by 8. ⇒ Let n = 11
6n - 1 = 6(11) - 1 = 65
Remainder when 65 is divided by 8 = 1

Since 5y5884805x6 is divisible by 72, it must be divisible by 9 and 8 (coprime factors of 72). So the
sum of digits of this number must be divisible by 9 and last three digits by 8.
5 + y + 5 + 8 + 8 + 4 + 8 + 0 + 5 + x + 6
⇒ 49 + x + y,
Possible values of x + y = 5, 14
For x + y = 5
Possible values of x, y
= (1, 4), (2, 3), (3, 2), (4, 1)
For x + y = 14
Possible values of x,y
= (5, 9), (6, 8), (7, 7), (8, 6), (9, 5)
Among these values last three digits of the number are divisible by 8 only when x = 3 or 7
But for x = 3, y = 2…...(x ≠ y)
Clearly the desired values of x and y are 7 and 7 respectively.

7823326867 X is divisible by 18. So , the given no. must be divisible by 9 and 2. For divisibility of 9
= 7 + 8 + 2 + 3 + 3 + 2 + 6 + 8 + 6 + 7 + X
= 52 + X
52 + X , divisible by 9 only if the value of
X = 2
The given no = 78233268672 …(also divisible by 2)

Let the numbers are a and b, then
According to the question,
a + b = 98 --------- Eq (1)
And a - b = 28 --------- Eq (2)
on adding both the equations, we get
2a = 98 + 28 ⇒ a = 126/2 = 63
Second number (b) = 98 − 63 = 35.

Factor of 484 = 1 × 2 × 2 × 11 × 11
Required sum = 1 × (2⁰ + 2²) x (11⁰ + 11²) = 1 x 5 x 121 = 610

If a number is in the form of aⁿ + bⁿ, where n is odd, then the number is divisible by (a + b).

it is completely divisible by (x + 1). 
Hence, remainder = 0.

According to the question, abba is divisible by 4
Then , last two digit (ba) be divisible by 4 {where a < b}
So, possible values of ba = (32, 52 , 64 , 72 , 76 , 84 , 92 , 96 ) = 8

Sum of digits of the given number,
5826 = 21 , 5964 = 24 , 6039 = 18,
6336 = 18, 6489 = 27, 6564 = 21,
6867 = 27 and 6960 = 21 .
Clearly, the numbers whose sum of digits are 18 and 27 are also divisible by 9.
So, 4 numbers are divisible by 3 but not by 9.

11368 = (1 + 3 + 8) - (1 + 6) = 5,
12638 = (1 + 6 + 8) - (2 + 3) = 10
11863 = (1 + 8 + 3) - (1 + 6) = 5
11638 = (1 + 6 + 8) - (1 + 3) = 11
Clearly, 11638 is divisible by 11.

LCM of (3, 7, 11) = 231
Given number = 750PQ
Let 750PQ = 75099.
Rem.(75099/231) = reminder (24)
Required no. = 75099 − 24 = 75075
So , P = 7 and Q = 5
Now, P + 2Q = 7 + 2 × 5 = 17

If, Rem. (N/7) = 4 Then, Remainder (N²/7) = Rem.(4²/7) = 2
Short Trick :-
Assume a number which when divisible by 7 leaves remainder 4 i.e. 11
Now square the number i.e. 121 leaves rem. 2 , when divided by 7.

On dividing a certain number by 363, we get 17 as remainder,
Then the number is 363x + 17
⇒ (363x + 17) = (11 × 33x + 17)
Multiple of 11 will always be divisible by 11.
Remainder on dividing (363x + 17) by 11 = 6

L.C.M. (7, 9, 11) = 693
Now, largest five digit number is 99999
When we divide 99999 by 693 , the remainder will be 207.
No. completely divisible by 693 is 99999 - 207 = 99792
Now, the largest number which when divided by 7 , 9 and 11 leaves remainder 3 in each case :- 99792 + 3 = 99795

Factor of 49 = 7 × 7
Factor of 147 = 3 × 7 × 7
Factor of 322 = 2 × 7 × 23
Clearly, 7 is the largest which divides 49, 147 and 322, leaving 0 as remainder in all cases.

LCM of (3, 4, 5, 7) = 420
When 35460 is divided by 420 leaves, remainder 180.
So, the no which should be added = 420 - 180 = 240.

Let the four digit pin be abcd.
a , b = 2a, c = 4a, d = 8a
Putting a = 1. We get the four digit pin as 1248 which is divisible by 2, 3 and 13.

It is given that,

So, the total possible values of N
= (343 - 216) - 1 = 126

I. Let the two digit no be 10x + y
ATQ, (10x + y) - (10y + x) = 36
9x - 9y = 36 ⇒ x - y = 4
II. The value of no can be 84 as 84 - 48 = 36
III. the number formed in the given case may or may not be composite no. i.e. 84 which is a composite no and 73 which is a prime no.
Clearly, we can see that option (d) is the correct one.

As we know, that sum of two odd numbers gives an even no and two even no. also gives an even no.
Putting x = 2 which is the only even prime no, we have y + z = 70 - 2 = 68
The possible value of y and z is (7, 61) and (31, 37)
So, the value of z = 61 or 37
Checking the options, we get z = 37.
Hence, the correct option is (d).

Given series is the combination of two series :

 


So, the required sum = 84/13 + 132/25

For greatest number take a = 9 and b = 9
239689, now for making it divisible by 3 but not by 9.
we will change the value of b , 2 + 3 + 9 + 6 + 8 + 9 = 37
37 - 1 = 36 (divisible by both 9 and 3)
37 - 4 = 33 (divisible by 3 but not by 9)
So b = 9 - 4 = 5
Required number = 239685

22 = 11 × 2
For any number to be divisible by 5, the unit digit of that number must be either 5 or 0.
So, b ≠ 0 (as the given no should not be divisible by 5). Therefore, option (a) and (b) gets eliminated.
For a given expression to be divisible by 2, the unit digit must be even no. So, option (d) gets eliminated.
Now, For 234a5b to be divisible by 11, (5 + 4 + 2) - (3 + a + b) = 8 - (a + b) = 0
⇒ a + b = 8
Putting b = 2 and 8,we get a = 6 and 0 respectively.
As we need the greatest number, the correct answer is 234652.

Let the numbers be “a” and “b”
a + b = 65, √ab = 26 ⇒ ab = 676

(433)²⁴ - (377)³² + (166)⁵⁴
= 3⁴ - 7² + 6² = 1 - 9 + 6 = - 2
Unit digit for - 2 = 10 - 2 = 8

LCM (36, 72, 80, 88) = 7920
Here the difference between each divisor and each remainder given in question = 20
So, for finding the required least number, we need to deduct the difference as obtained above from the LCM of the divisors.
Required least number = 7920 – 20 = 7900
So, sum of digits = 7 + 9 + 0 + 0 = 16

As prime numbers have 2 factors
Only the squares of prime numbers will have three factors.
Let r = 9 and s = 4
And q = 49 and p = 25

Putting s = 4 in all options we find only option (a) satisfies this value.