PE 2 - Numbers | Algebra - Number Theory
How many even numbers from 10 to 10,000 are not perfect squares?
Answer: 97
Explanation :
The lowest such possible perfect square is 16 (42).
The highest such possible perfect square is 10000 itself (1002).
∴ Number of perfect squares from 10 to 10,000 is (100 – 4 + 1) = 97.
Hence, 97.
Workspace:
If the last 2 digits of a four-digit number are interchanged, the new number obtained is greater than the original number by 72. What is the difference between the last two digits of the number?
- A.
9
- B.
1
- C.
6
- D.
8
- E.
Cannot be determined
Answer: Option D
Explanation :
Consider a 4-digit number – a b c d.
The number formed by reversing the last two digits is – a b d c.
Given,
1000a + 100b + 10c + d - (1000a + 100b + 10d + c) = 72
9c - 9d = 72
c -d = 8.
Alternately,
We could have safely ignored the first 2 digits, as they would have not mattered after subtraction, this gives us {for last 2 digits}.
(10c + d) – (10d + c ) = 72
⇒ 9(c – d) = 72
⇒ c – d = 8
Hence, option (d).
Workspace:
LCM of two numbers x and y is 143. Find the value of |x – y|, given that y > x, x > 1.
- A.
30
- B.
15
- C.
41
- D.
165
- E.
Cannot be determined
Answer: Option E
Explanation :
Factorizing 143 we get, 143 = 11 × 13
The pairs of x & y, which satisfy LCM(x, y) = 143 are
1. 1, 143
2. 11, 143
3. 13, 143
4. 143,143
5. 11, 13
Out of these cases, (1) & (4) can be eliminated as they do not follow the criteria y > x; x > 1.
Still there are three different cases possible.
Hence, option (e).
Workspace:
XX and YY are 2-digits numbers such that their sum is ABA. Find the value of A + B + A?
- A.
4
- B.
3
- C.
2
- D.
1
- E.
5
Answer: Option A
Explanation :
Here, XX and YY are multiples of 11, hence their sum ABA should also be a multiple of 11.
Also, sum of 2 2-digit numbers cannot exceed 198.
∴ A = 1 and ABA = 1B1.
Now, 1B1 should be a multiple of 11. This is only possible when B = 2.
Hence, A = 1 and B = 2
⇒ A + B + A = 1 + 2 + 1 = 4
Hence, option (a).
Workspace:
Three consecutive whole numbers are such that product of smallest and the largest number is one less than the square of the middle number. Find the middle number.
- A.
6
- B.
18
- C.
12
- D.
31
- E.
All of these
Answer: Option E
Explanation :
Let the middle number be ‘a’.
According to the question: (a – 1)(a + 1) = a2 - 1
⇒ a2 – 1 = a2 - 1
Now, this is true for all values of ‘a’.
Hence, option (e).
Workspace:
A = 11 × 22 × 33 × 44 × … 100100. How many zeroes will be there at the end of A?
Answer: 1300
Explanation :
A = 11 × 22 × 33 × 44 × … 100100
Product of one 2 and one 5 gives one trailing zero.
Here number of 2’s will definitely be more than number of 5’s, hence number of trailing zeros will be same as number of 5’s.
Powers of 5 will occur at 55, 1010, 1515 … 100100
Multiples of 5
power at least 1 = 5 + 10 + 15 + … + 100 = 1050
additional power of 5 (i.e., power at least 2) = 25 + 50 + 75 + 100 = 250
Total number of 5s = 1050 + 250 = 1300
i.e., there will be 1300 trailing zeros in A.
Hence, 1300.
Workspace:
If x is a prime such that (x4 + 1) is also a prime, then how many values are possible for x.
- A.
0
- B.
1
- C.
2
- D.
more than 2 values
Answer: Option B
Explanation :
One of x and x4 + 1, will definitely be even and the other odd.
The only even prime number is 2. Hence, either x = 2 or x4 + 1 = 2
If x4 + 1 = 2 ⇒ x = 1 (not a prime number, hence rejected)
∴ x = 2 and x4 + 1 = 17.
∴ Only one value of x is possible.
Hence, option (b).
Workspace:
In how many ways can 14 be written as a product of two or more distinct integers, without considering two integers with the same absolute value but opposite signs?
Answer: 8
Explanation :
‘14’ as a product of 2 integers can be written as 1 × 14, 2 × 7, - 1 × -14, -2 × -7 i.e. 4 ways.
‘14’ as a product of 3 integers can be written as 1 × 2 × 7, -1 × -2 × 7, -1 × 2 × -7, 1 × -2 × -7 i.e. 4 ways.
∴ Total number of ways = 4 + 4 = 8.
Hence, 8.
Workspace:
‘N’ is a two-digit number less than 50 such that when the unit’s digit of ‘N’ is erased, the resulting number is a factor of ‘N’. How many possible values of ‘N’ are there?
Answer: 22
Explanation :
Let N = ‘ab’, where a and b are digits
N = 10a + b
Now, since a is a factor of ‘ab’, a should divide (10a + b) completely.
i.e., = 0
⇒ + = 0
⇒ = 0
∴ ‘b’ should be divisible by a.
If a = 1, b can be 0, 1, 2, 3, …, 9 ⇒ 10 possible values
If a = 2, b can be 0, 2, 4, 6, 8 ⇒ 5 possible values
If a = 3, b can be 0, 3, 6, 9 ⇒ 4 possible values
If a = 4, b can be 0, 4, 8 ⇒ 3 possible values
Number of possible values of ‘N’ = 10 + 5 + 4 + 3 = 22.
Hence, 22.
Workspace:
How many integers exist such that not only are they multiples of A = 20202020 but also are factors of B = 20202022?
- A.
36
- B.
48
- C.
45
- D.
None of these
Answer: Option C
Explanation :
The required number should be of the form A × x (where x is any positive integer)
Also, B should be divisible by A × x
Now, 20202022 = 20202020 × 20202
Number of possible values of x will be same as the number of factors of 20202.
20202 = 24 × 52 × 1012
Number of factors of 20202 = (4 + 1) × (2 + 1) × (2 + 1) = 45.
Hence, option (c).
Workspace:
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