Questions are dedicated to “Number System and Number Theory”. These practice problems on Number system are relatively easy to solve and helpful for those who are preparing for exams like Bank PO, Bank clerk, Bank of Baroda PO, SBI Bank PO, PNB Bank PO, LIC etc.
1. Two numbers differ by 5. If their product is 336, then the sum of the two numbers would be:
Let the numbers are x and y. then, ⇒ x – y = 5 ———— (1) ⇒ x × y = 336 ———— (2) Now, (x + y) = (x – y) + 4xy, ⇒ (x + y) = 25 + 4 × 336 = 1369, ⇒ x + y = √1369 = 37
2. The multiplication of two numbers is 9375 and the quotient, when the greater is divided by the smaller is 15. The sum of the numbers is:
Then, a × b = LCM of (a, b) × HCF of (a, b) According to the question a × b = 493 And we know that HCF for two prime numbers is = 1 So, 493 = LCM of (a, b) × 1 ⇒ LCM of (a, b) = 493
Hint: L.C.M. of some prime numbers is always the product of those numbers
4. A number gives a remainder 5 when it is divided by 8. What will be the remainder when the square of the same number is divided by 4?
Let the number be = 8k + 5 where k is positive integer. Square of the number, (8k + 5) = 64k + 25 + 80k Now, we have to find the remainder when the above number is divided by 4. We will see the reminder of each term when it is divided by 4. Since 64k and 80k will give the remainder 0 and 25 will give the remainder 1 when divided by 4. So the required remainder = 0 + 0 + 1 = 1
5. Find the remainder when:
1661 + 1551 + 1441 + 1331 + 1221 is divided by 20.
When x , x , x , …… are divided by ‘d’ individually the respective remainders obtained are R , R , R , …… etc. and when (x + x + x + ……) is divided by ‘d’ the remainders can be obtained by dividing (R + R + R ……) by d
The remainder when 1661, 1551, 1441, 1331, 1221 are divided by 20 are 1, 11, 1, 11, 1. So the required remainder can be obtained just by dividing (1 + 11 + 1 + 11 + 1 = 25) by 20. Hence, the required remainder is 5.
6. The sum of two numbers is 45 and their product is 500. The HCF of the numbers is:
Let the 1st no. = a Let the 2nd no. = b ⇒ a + b = 45 ….. (i) ⇒ a × b = 500 ….. (ii) Substituting Value of “b” from (i) in (ii) ⇒ a × (45 – a) = 500 ⇒ a – 45a + 500 = 0 ⇒ a – 20a -25a + 500 = 0 ⇒ a(a – 20) – 25(a – 20) = 0 ⇒ (a – 25)(a – 20) = 0 ⇒ a = 25 or a =20 From eq (i) we get b = 20 or b = 25 Either Case two numbers are 20 & 25 HCF is greatest common Factor 20 = 2 × 2 × 5 25 = 5 × 5 Clearly seen from prime factorization of both numbers that the HCF = 5
Two numbers are said to be co – primes when there is no other factor common between them except 1. In other words if the H.C.F of the two numbers is 1 then those two numbers are known as co-primes. So, H.C.F. of 18 and 25 is 1. Hence, they are co-primes.
8. The sum of two numbers is 216 and their H.C.F is 27. The numbers are:
Since the H.C.F of two numbers are 27 then we can write both numbers as 27a and 27b where a and b must be co-primes. Then, 27a + 27b = 216 ⇒ a + b = 8 Now, co-primes with sum 8 are (1, 7) and (3, 5). ∴ Required number are (27 × 1, 27 × 7) and (27 × 3, 27 × 5) i.e. , (27, 189) and (81, 135); Out of these, the given one in the answer is the pair (27, 189).
9. Find the unit digit in the solution of following multiplication.
134 × 428 × 317
Given expression: ⇒ 134 × 428 × 317 Unit digit of the given expression will be given by the multiplication of the unit digits of each numbers s: ⇒ Unit digit in 134 × 428 × 317 = Unit digit in 4 × 8 × 7 = 4
10. Find the number which when divides 1265 leaves quotient 84 and remainder 5.