Numbers and Sequences Ex 2.2-10th Std Maths-Book Back Question And Answer
Question 1.
For what values of natural number n, 4th can end with the digit 6?
Answer:
4n = (22)n = 22n
= 2n × 2n
2 is a factor of 4n
∴ 4n is always even.
Question 2.
If m, n are natural numbers, for what values of m, does 2n × 5n ends in 5?
Solution:
2n × 5m
2n is always even for all values of n.
5m is always odd and ends with 5 for all values of m.
But 2n × 5m is always even and ends in 0.
∴ 2n × 5m cannot end with the digit 5 for any values of m. No value of m will satisfy 2n × 5m ends in 5.
Question 3.
Find the H.C.F. of 252525 and 363636.
Answer:
To find the HCF of 252525 and 363636 by using Euclid’s Division algorithm.
363636 = 252525 × 1 + 111111
The remainder 111111 ≠ 0
By division of Euclid’s algorithm
252525 = 111111 × 2 + 30303
The remainder 30303 ≠ 0
Again by division of Euclid’s algorithm
111111 = 30303 × 3 + 20202
The remainder 20202 ≠ 0
Again by division of Euclid’s algorithm.
30303 = 20202 + 10101
The remainder 10101 ≠ 0
Again by division of Euclid’s algorithm.
20202 = 10101 × 2 + 0
The remainder is 0
∴ The H.C.F. is 10101
Question 4.
If 13824 = 2a × 3b then find a and b?
Answer:
Using factor tree method factorise 13824
13824 = 29 × 33
Given 13824 = 2a × 3b
Compare we get a = 9 and b = 3
Aliter:
13824 = 29 × 33
Compare with
13824 = 2a × 3b
The value of a = 9 b = 3
Question 5.
If p1x1 × p2x2 × p3x3 × p4x4 = 113400 where p1 p2, p3, p4 are primes in ascending order and x1, x2, x3, x4, are integers, find the value of p1,p2,p3,p4 and x1,x2,x3,x4.
Answer:
Given 113400 = p1x1 × p2x2 × p3x3 × p4x4
Using tree method factorize 113400
113400 = 23 × 34 × 52 × 7
compare with
113400 = p1x1 × p2x2 × p3x3 × p4x4
P1 = 2, x1 = 3
P2 = 3, x2 = 4
P3 = 5, x3 = 2
P4 = 7, x4 = 1
Question 6.
Find the L.C.M. and H.C.F. of 408 and 170 by applying the fundamental theorem of Arithmetic.
Answer:
Factorise 408 and 170 by factor tree method
408 = 23 × 3 × 17
170 = 2 × 5 × 17
To find L.C.M. list all prime factors of 408 and 170 of their greatest exponents.
L.C.M. = 23 × 3 × 5 × 17
= 2040
To find the H.C.F. list all common factors of 408 and 170.
H.C.F. = 2 × 17 = 34
L.C.M. = 2040 ; HCF = 34
Question 7.
Find the greatest number consisting of 6 digits which is exactly divisible by 24,15,36?
Answer:
The greatest number of 6 digits is 999999.
The greatest number must be divisible by L.C.M. of 24, 15 and 36
24 = 23 × 3
15 = 3 × 5
36 = 22 × 32
L.C.M = 23 × 32 × 5
= 360
To find the greatest number 999999 must be subtracted by the remainder when 999999 is divided by 360
The greatest number in 6 digits = 999999 – 279
= 99972
Question 8.
What is the smallest number that when divided by three numbers such as 35, 56 and 91 leaves remainder 7 in each case?
Solution:
35 = 5 × 7
56 = 2 × 2 × 2 × 7
91 = 7 × 13
LCM of 35, 56, 91 = 5 × 7 × 2 × 2 × 2 × 13 = 3640
∴ Required number = 3647 which leaves remainder 7 in each case.
Question 9.
Find the least number that is divisible by the first ten natural numbers?
Answer:
Find the L.C.M of first 10 natural numbers
The least number is 2520
Modular Arithmetic
Two integers “a” and “b” are congruence modulo n if they differ by an integer multiple of n. That b – a = kn for some integer k. This can be written as a = b (mod n).
Euclid’s Division Lemma and Modular Arithmetic
Let m and n be integers, where m is positive. By Euclid’s division Lemma we can write n = mq + r where 0 < r < m and q is an integer.
This n = r (mod m)
