a) 25296
b) 24456
c) 23296
d) 26564
We are given a 5 digit number
Let first digit be 'X'
then 5th digit is '3X'
let 2nd digit be 'Y'
then 3rd digit is 'Y - 3'
and 4th digit is 'Y + 4'
then the no is '(X)(Y)(Y - 3)(Y + 4)(3X)'
from the above we can say 3X <= 9
so X<=3 and any of the digit in the number is <= 9
Also Y–3 ≥ 0 and Y+4 ≤ 9
⇒ Y ≥ 3 and Y ≤ 5 i.e.
Y = 3,4 or 5
for x = 1, all the conditions won’t be satisfied.
For x = 2,
Let Y = 5
∴ The number is 25296
2. The least number which when divided by 4, 6, 8, 12 and 16 leaves a remainder of 2 in each case is
a) 50
b) 48
c) 46
d) 52
Let’s find a number that is completely divisible by 4, 6, 8, 12, and 16.
To find such number, we need to find the LCM of all these numbers
LCM of (4,6) , 8 , 12 , 16
= 12 , 8 , 12 , 16
= LCM of (12 , 8) , LCM of (12 , 16)
= 24 , 48
= LCM of (24 , 48)
= 48
Hence 48 + 2 i.e.
Source: safalta.com
50 will leave a remainder of 2 if divided by any of the following numbers: 4, 6, 8, 12 and 163. Maximum numbers that can be formed using all the 4 digits 6, 4, 8, 1 without repetition and which is divisible by 9 ?
a) 0
b) 150
c) 450
d) 900
Since we need to use all 4 digits of the given number, we need to see that the sum of digits 6,4,8,1 = 19
And divisibility rule of 9 says that
For a number to be divisible by 9, the sum of its digits should be divisible by 9.
Hence, we can say that no numbers can be formed using all 4 digits and still divisible by 9.
4. Find the smallest number which leaves 22, 35, 48 and 61 as remainders when divided by 26, 39, 52 and 65 respectively.
a) 760
b) 776
c) 780
d) 766
Let us have N be a number which is always divisible by 22, 35, 48 and 61.
Hence
⇒N = LCM(26, 39,52, 65)
⇒N = LCM[LCM(26,39) , LCM(52,65)]
⇒N = LCM[78,260]
⇒N = 780
Since difference between each divisor and its respective remainder is 4.
Then N - 4 will suit all the conditions given.
Hence, N - 4 = 780 - 4 = 776
5. Let p and q be two prime numbers such that p is greater than q.
If 319 is their LCM then the difference of thrice of q and p is:
a) 1
b) 4
c) 6
d) 8
We can write 319 as the product of numbers in just one way
⇒ 319 = 29 × 11
Hence we can assume that p = 29 and q = 11 (as p > q)
Now we need to find 3q - p
= 3 × 11 – 29
= 33 – 29
= 4
6. Let 13 and 273 are the HCF and LCM of two numbers respectively, and if one of them is less than 140 and greater than 60.
Then what will be that number?
a) 61
b) 91
c) 71
d) 81
Let those 2 numbers be p and q .
We are given that HCF (p, q) = 13 and
LCM (p, q) = 273.
We know that product of numbers = HCF × LCM
Hence, p × q = 13 × 273
⇒ p × q = 13 × 13 × 3 × 7
P cannot be 13 × 3 or 13 × 13 or 7 × 3 or 13 or 3 or 7 as these numbers do not come between 60 and 140.
Hence we can say that p = 13 × 7 = 91
7. What is the smallest number by which 2880 must be divided in order to make it into a perfect square?
a) 3
b) 4
c) 5
d) 6
2880 = 2× 2× 2× 2× 2× 2× 3× 3× 5 = 26 × 32 × 5
Clearly all factors except for 5 is in pair
So dividing 2880 by 5 will convert it into a perfect square.
8. Which of the following are prime numbers?a) 241
b) 303
c) 111
d) 141
To check if a number is prime or not, we find a value which is slightly more than or equal to the square root of the number.
Then, we start dividing the number; from 2 for all prime numbers up to the square root of the number.
If it is divisible by any prime number, we see that it is not prime.
Hence checking for 241, the nearest number of square root will be 16
Hence, we check if 241 is divisible by any of 2, 3, 5, 7, 11, 13.
Since it does not, it is a prime number.9. Calculate: (x-a) (x-b) (x-c).... (x-z)?
a) 0
b) 1
c) 26
d) Infinity
In the sequence of multiplication there is term (x-x) = 0
Anything multiplied by zero is always Zero.
So the value of whole term is 0.10. What is the least number to be added to 4321 to make it a perfect square?
a) 15
b) 25
c) 35
d) 45
We need to find the approximate value of square root of 4321 using long division method or trial and error.
We find that 652 = 4225 < 4321 and 662 = 4356 > 4321
So the least number to be added = 4356 – 4321 = 35.
