10memu2e2

2 Numbers and Sequences Ex 2.2

1. For what values of natural number n, $4^n$ can end with the digit 6?
Solution:
$4^1=4$, $4^2=16$, $4^3=64$, $4^4=256$, ...
$4^n$ ends with the digit 6, when $n=2,4,6,...$.

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2. If m, n are natural numbers, for what values of m, does $2^n \times 5^m$ ends in 5?
Solution:
$2^n \times 5^m$
$2^n$ is always even for all values of n.
$5^m$ is always odd for all values of m.
But $2^n \times 5^m$ is always even and ends in 0.
∴ $2^n \times 5^m$ cannot end with the digit 5 for any values of m.

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3. Find the H.C.F. of 252525 and 363636.
Solution:
By Euclid's Division algorithm,
$363636 = 252525 × 1 + 111111$
$252525 = 111111 × 2 + 30303$
$111111 = 30303 × 3 + 20202$
$30303 = 20202 + 10101$
$20202 = 10101 × 2 + 0$
∴ The H.C.F. is 10101.

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4. If 13824 = 2^a × 3^b then find a and b?
Solution:

13824 = 2⁹ × 3³
Given $13824 = 2^a × 3^b$
$\implies a = 9, \: b = 3$

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5. If ${p_1}^{x_1}\times {p_2}^{x_2}\times {p_3}^{x_3}\times {p_4}^{x_4}=113400$ where ${p_1},{p_2},{p_3},{p_4}$ are primes in ascending order and ${x_1},{x_2},{x_3},{x_4}$ are integers, find the value of ${p_1},{p_2},{p_3},{p_4}$ and ${x_1},{x_2},{x_3},{x_4}$.
Solution:
${p_1}^{x_1}\times {p_2}^{x_2}\times {p_3}^{x_3}\times {p_4}^{x_4}=113400$

$113400=2^3 \times 3^4 \times 5^2 \times 7^1$
Therefore, ${p_1=2},{p_2=3},{p_3=5},{p_4=7}$ and ${x_1=3},{x_2=4},{x_3=2},{x_4=1}$.

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6. Find the L.C.M. and H.C.F. of 408 and 170 by applying the fundamental theorem of Arithmetic.
Solution:

$408 = 2^3 × 3 × 17$
$170 = 2 × 5 × 17$
$L.C.M. = 2^3 × 3 × 5 × 17 = 2040$
$H.C.F. = 2 × 17 = 34$

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7. Find the greatest number consisting of 6 digits which is exactly divisible by 24,15,36?
Solution:
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

The greatest number in 6 digits which is exactly divisible by 24,15,36 $= 999999-279=999720$

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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:

$35 = 5 × 7$
$56 = 2^3 × 7$
$91 = 7 × 13$
LCM of 35, 56, 91 $= 2^3 \times 3 \times 5 \times 7 \times 13 = 3640$
∴ Required number $= 3640+7=3647$.

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9. Find the least number that is divisible by the first ten natural numbers?
Solution:

$LCM=2^3 \times 3^2 \times 5 \times 7=8 \times 9 \times 5 \times 7=2520$

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