PE 1 - Numbers | Algebra - Number Theory
The least number which on division by 25 leaves the remainder 15, on division by 35 leaves the remainder 25 and on division by 45 leaves the remainder 35 is
- (a)
1565
- (b)
1575
- (c)
1585
- (d)
1595
- (e)
None of these
Answer: Option A
Explanation :
This is a direct application of LCM Model II, as the complement of the remainder is the same in all the three cases.
Complement of remainder = Difference between divisor and remainder
Case I d1 - r1 = 25 – 15 = 10 = x (say)
Case II d2 - r2 = 35 – 25 = 10
Case III d3 - r3 = 45 – 35 = 10
Hence, smallest number = LCM(d1, d2, d3) - x
In this case, LCM of 25, 35 and 45 = 1575
Required number = 1575 – 10 = 1565
Alternatively,
The answer should be such a number that when 10 is added to it, it is divisible by 25, 35 as well as 45, i.e. by 1575.
Add 10 to each option and check which option is divisible by 1575.
Hence, option (a).
Workspace:
A group of students is divided into groups of 3, 4 and 5, and each time one student is left out. The least number of students in the heap is
- (a)
31
- (b)
41
- (c)
51
- (d)
61
- (e)
71
Answer: Option D
Explanation :
This is a direct application of LCM Model I, as the remainder is the same in all the three cases.
∴ Number of students = LCM(3, 4, 5) + 1 = 60 + 1 = 61
Hence, option (d).
Workspace:
What is the highest power of 288 that divides 35!?
- (a)
2
- (b)
6
- (c)
4
- (d)
5
- (e)
None of these
Answer: Option B
Explanation :
288 = 25 × 32
The highest power of 3 in 35! is + + = 11 + 3 + 1 = 15.
∴ The highest power of 32 will be = 7.
The highest power of 2 in 35! is + + + = 17 + 8 + 4 + 2 + 1 = 32.
∴ The highest power of 25 will be = 6.
⇒ The highest power of 288 will be lower of 7 and 6 i.e., 6.
Hence, option (b).
Workspace:
369 has been divided into three parts such that half of the first part, one-third of the second part and one-fourth of the third part are equal. The largest part is
Answer: 164
Explanation :
= = = x
⇒ A = 2x; B = 3x; C = 4x
⇒ A : B : C = 2 : 3 : 4
Largest part = 369 × = 164
Hence, 164.
Workspace:
Seven bells commence tolling together and toll at intervals of 3, 6, 9, 12, 15, 18 and 21 s respectively. In 300 mins, how many times do they toll together?
- (a)
4
- (b)
11
- (c)
10
- (d)
15
- (e)
None of these
Answer: Option D
Explanation :
LCM of 3, 6, 9, 12, 15, 18 and 21 is 1260.
So, the bells will toll together after every 1260 s, i.e. 21 mins.
In 300 mins, they will toll together for (300/21) + 1 = 15 times.
Hence, option (d).
Workspace:
When ‘n’ is divided by 5 the remainder is 3. What is the remainder when n2 is divided by 5?
- (a)
2
- (b)
1
- (c)
3
- (d)
4
- (e)
0
Answer: Option D
Explanation :
n = 5k + 3
⇒ n2 = 25k2 + 30k + 9
=
= + + = 0 + 0 + 4 = 4.
Hence, option (d).
Workspace:
The expression + + + ... + for any natural number n, is
- (a)
always less than 1
- (b)
always greater than 2
- (c)
always equal to 1
- (d)
always lies between 1 and 2
- (e)
None of these
Answer: Option A
Explanation :
N = + + + ... +
N = + + + ... +
N = - + - + - + ... + -
N = - + - + - + ... + -
N = -
Since 'n' is a natural numbers N will always be less than 1.
Hence, option (a).
Workspace:
The ratio between a two-digit number and the sum of the digits of that number is 4 : 1. If the digit in the unit’s place is twice the digit in the ten’s place, what is that number?
- (a)
48
- (b)
63
- (c)
84
- (d)
Cannot be determined
Answer: Option D
Explanation :
Let tens digit be x. Then, units digit = 2x
Sum of the digits = x + 2x = 3x
Number = 10x + 2x = 12x
Now, =
This is true for all values of x.
Hence, option (d).
Workspace:
What is the greatest positive power of 5 that divides 40! exactly?
- (a)
5
- (b)
6
- (c)
7
- (d)
8
- (e)
9
Answer: Option E
Explanation :
The number of 5s in 30! is calculated as follows:
+ = 8 + 1 = 9
Hence, option (e).
Workspace:
If the sum of two natural numbers is multiplied by each number separately, the products so obtained are 2772 and 1584. What is the difference between the numbers?
- (a)
18
- (b)
22
- (c)
26
- (d)
35
- (e)
27
Answer: Option A
Explanation :
Suppose numbers are x and y, we have
x(x + y) = 2772 (i)
y(x + y) = 1584 (ii)
Adding (i) and (ii), we get
(x + y)2 = 4356
∴ x + y = 66 (iii)
Subtracting (ii) from (i), we get
x2 – y2 = 1188
⇒ (x + y) × (x – y) = 1188
⇒ 66 × (x – y) = 1188 using (iii)
∴ x – y = 18
Hence, option (a).
Workspace: