SSC CGL Previous Year Questions: Number System and HCF & LCM- 3 - SSC CGL MCQ

∵ 13b7 is exactly divisible by 11.
∴ b = 9 ∴ a = 1
∴ a + b = 9 + 1 = 10

35 – 18 = 17
45 – 28 = 17
55 – 38 = 17
i.e., difference between the divisor and correseponding remainder is same.
LCM of 35, 45 and 55 = 3465
∴ Required number
= 3465 – 17 = 3448

∴ Required number
= 75070 + (65 – 60) = 75075

Let the 4 terms in A.P are a – 3d, a – d, a + d, a + 3d
According to question
a – d + a + d = 110 ... (1)
(a – 3d) (a + 3d) = 2125 ... (2)
From equation (1)
a – d + a + d = 110
2a = 110 ⇒ a = 55
From equation (2)
(a – 3d) (a + 3d) = 2125
⇒ a² – 9d² = 2125
⇒ (55)2 – 9d² = 2125
⇒ 3025 – 9d² = 2125
⇒ 900 = 9d² ⇒ d² = 100 ⇒ d = 10
∴ a = 55, d = + 10
series would be :
25, 45, 65, 85
IIIrd term would be 65.

Required number = H.C.F of (140 – 4), (176 – 6) and (264 – 9) = H.C.F. of 136, 170 and 255.

∴ Required number = 17
"Alternate Method:
Here divisible terms are 140 – 4 = 136, 176 – 6 = 170
and 264 – 9 = 255
Now, difference between these numbers
170 – 136 = 34
225 – 170 = 85
H.C.F of difference = 17
Hence required number = 17."

Required time = LCM of 200, 300, 360 and 450 seconds = 1800 seconds.

If ( n ) is divisible by 5 and gives a remainder of 2, we can express this as:

[ n = 5k + 2 ]

for some integer ( k ).

To find the remainder when ( n^2 ) is divided by 5, we first calculate ( n^2 ):

[
n^2 = (5k + 2)^2 = 25k^2 + 20k + 4
]

Now, when we divide ( n^2 ) by 5, we focus on the remainder of the last term, since ( 25k^2) and ( 20k ) are both divisible by 5:

[
n^2 = 4 (mod 5)
]

Thus, the remainder when ( n^2 ) is divided by 5 is 4.

Let the numbers be 10x and 10y where x and y are prime to each other.
∴ LCM = 10 xy
⇒ 10xy = 120 ⇒ xy = 12
Posssible pairs = (3, 4) or (1, 12)
∴ Sum of the numbers = 30 + 40 = 70
"Alternate Method:
If two different numbers are in form of ax and bx
H.C.F of these numbers is x and L.C.M of these numbers is abx
Now a and b are co-prime terms in L.C.M.
10 × a × b = 120
a × b = 12 → 1 × 12
2 × 6 this is not a pair of co prime terms.
3 × 4"

5⁷¹ + 5⁷² + 5⁷³
= 5⁷¹ (1 + 5 + 5²) = 5⁷⁰ × 5 × 31
= 5⁷¹ × 155 which is exactly divisible by 155.

a * b = a^b
∴ 5 * 3 = 5³ = 5 × 5 × 5 = 125

4¹ = 4; 4² = 16; 4³ = 64; 4⁴ = 256; 4⁵ = 1024
Remainder on dividing 372 by 4 = 0
Remainder on dividing 373 by 4 = 1
∴ Required unit digit
= Unit’s digit of the sum = 6 + 4 = 0

Remainder when (x – 1)ⁿ is divided by x is (–1)ⁿ
∴ (17)²⁰⁰ = (18 – 1)²⁰⁰
∴ Remainder = (–1)²⁰⁰ = 1
"Alternate Method:
aⁿ – bⁿ is completely divisible by a + b. If n is an even number in the case of 17²⁰⁰ ÷ 18 (17²⁰⁰ - 1²⁰⁰) is completely divisible by 17 + 1 = 18
Here, 1 is remainder.
Or in other words if an is divided by a + 1 and n is even number then it always left 1 as remainder."

The largest 4-digit number = 9999

∴ Required number = 345 – 339 = 6

LCM of 24, 36 and 54 seconds
= 216 seconds = 3 minutes 36 seconds
∴ Required time = 10 : 15 : 00 + 3 minutes 36 seconds = 10 : 18 : 36 a.m.

Here, the first divisor i.e. 49 is multiple of second divisor i.e. 7.
∴ Required remainder = Remainder obtained on dividing 32 by 7 = 4

HCF must be a factor of LCM from option 35 is not factor of 120.

Last digit of (1001)²⁰⁰⁸ + 1002 = 1 + 2 = 3

3¹ = 3; 3² = 9; 3³ = 27; 3⁵ = 81; 3⁵ = 243
i.e. unit’s digit is repeated after index 4.
Remainder after dividing 21 by 4 = 1
∴ Unit’s digit in the expansion of (3)²¹ = 3
∴ Remainder after dividing by 5 = 3

11² = 121, 12² = 144, 13² = 169, 14² = 196
15² = 225, 16² = 256, 17² = 289
Square no above 120 = 121 of 11
Square no below 300 = 289 of 17
Total 11, 12, 13, 14, 15, 16, 17, i.e. 7 no.
"Alternate Method:
First square number above 120 is 121
11² > 120 and 18² > 300
Hence, required number of squares between 120 to 300
= 18 – 11 = 7"

n³ – n = (n² – 1)
⇒ n (n +1) (n – 1)
For n = 2, n³ – n = 6
2³ – 2 = 6
i.e. n³ – n is always divisible by 6.

Let the numbers be 3x and 3y.
∴ 3x + 3y = 36
⇒ x + y = 12 ...(i)
and 3xy = 105 ...(ii)
Dividing equation (i) by (ii), we have

First number × second number
= HCF × LCM
⇒ 84 × second number = 12 × 336
∴ Second number

p × q = HCF × LCM

xyxy = xy × 100 + xy
= xy (100 + 1) = 101 × xy
Hence, the number is exactly divisible by 101.

If the first divisor be a multiple of the second divisor, then required remainder = remainder obtained by dividing the first remainder (36) by the second divisor
(17) = 2
∵ 17 is a factor of 136
∴ Remainder when 36 is divided by 17 = 2

2 + 4 = 6
6 + 5 = 11
11 + 6 = 17
17 + 7 = 24
24 + 8 = 32

Let the numbers be 3x and 4x.
∴ Their LCM = 12x
∴ 12x = 84

∴ Larger number
= 4x = 4 × 7 = 28

If the first divisor is a multiple of second divisor.
Then, remainder by the second divisor.
∴ Remainder = 21 ÷ 19 = 2

0 + 3 = 3
3 + 5 = 8
8 + 7 = 15
15 + 9 = 24
24 + 11 = 35
35 + 13 = 48
48 + 15 = 63
63 + 17 = 80

∴ LCM = 2 × 2 × 5 × 7 × 8 = 1120
∴ Required number = 5834 – 1120 = 4714